Answer:
See proof below, the converse is true
Step-by-step explanation:
Let f be a function whose domain is the set A, and X,Y be arbitrary subsets of A. One way to prove the equality of sets f(X)∩f(Y)=f(X∩Y) is, by definition, to prove the inclusions f(X)∩f(Y) ⊆ f(X∩Y) and f(X)∩f(Y )⊇f(X∩Y).
First, let a∈f(X)∩f(Y). Then a∈f(X) and a∈f(Y) by definition of intersection. Thus, a=f(t) for some t∈X (because a is an element of the direct image of X) and a=f(s) for some s∈Y. Then a=f(t)=f(s). We know that f is 1-1, so the previous equality implies that t=s. Now, t∈X and t=s∈Y, thus t∈X∩Y. Therefore, a=f(t) for some t∈X∩Y, which means that a is an element of the direct image of X∩Y, that is, a∈f(X∩Y). This holds for all a, thus f(X)∩f(Y) ⊆ f(X∩Y).
For the second inclusion, let b∈f(X∩Y), then there exists some z∈X∩Y such that b=f(z). Since z∈X∩Y, we have that z∈X and z∈Y. Thus, b=f(z) for some z∈X, that is, b∈f(X). Similarly, b∈f(Y) because b=f(z) and z∈Y. Therefore b∈f(X)∩f(Y) for all b∈f(X∩Y), then f(X)∩f(Y )⊇f(X∩Y). Note that this holds for any function f, as we did not use that f is 1-1.
To prove the converse, suppose that f(X)∩f(Y)=f(X∩Y) for all X, Y⊆ A. We will prove that f is 1-1, that is, for all a,b∈A if f(a)=f(b) then a=b.
Define the sets X:={a} and Y:={b}. Then f(X)={f(a)} and f(Y)={f(b)}. Assuming that f(a)=f(b), we obtain f(X)=f(Y). We know that f(X)∩f(Y)=f(X∩Y), which means {f(a)}=f(X∩Y). Then, there exists some s∈X∩Y such that f(s)=f(a), thus X∩Y is a non-empty set, which implies that a∈Y. But Y has only one element, b, therefore a=b which shows that f is 1-1.
To illustrate the effects of driving under the influence (DUI) of alcohol, a police officer brought a DUI simulator to a local high school. Student reaction time in an emergency was measured with unimpaired vision and also while wearing a pair of special goggles to simulate the effects of alcohol on vision. For a random sample of nine teenagers, the time (in seconds) required to bring the vehicle to a stop from a speed of 60 miles per hour was recorded.
Subject 1 2 3 4 5 6 7 8 9
Normal, Xi 4.47 4.24 4.58 4.65 4.31 4.80 4.55 5.00 4.79
Impaired, Yi 5.77 5.67 5.51 5.32 5.83 5.49 5.23 5.61 5.6
(a) Whether the student had unimpaired vision or wore goggles first was randomly selected. Why is this a good idea in designing the experiment?
(b) Use a 95% confidence interval to test if there is a difference in braking time with impaired vision and normal vision where the differences are computed as "impaired minus normal. "Note: A normal probability plot and boxplot of the data indicate that the differences are approximately normally distributed with no outliers.
Answer:
a) If we design the experiment on this way we can check if we have an improvement with the method used.
We assume that we have the same individual and we take a value before with the normal impaired condition and the final condition is the normal case.
b) [tex]-0.96-2.306\frac{0.359}{\sqrt{9}}=-1.24[/tex]
[tex]-0.96+2.306\frac{0.359}{\sqrt{9}}=-0.69[/tex]
The 95% confidence interval would be given by (-1.24;-0.69)
Step-by-step explanation:
Part a
If we design the experiment on this way we can check if we have an improvement with the method used.
We assume that we have the same individual and we take a value before with the normal impaired condition and the final condition is the normal case.
Part b
For this case first we need to find the differences like this :
Normal, Xi 4.47 4.24 4.58 4.65 4.31 4.80 4.55 5.00 4.79
Impaired, Yi 5.77 5.67 5.51 5.32 5.83 5.49 5.23 5.61 5.6
Let [tex]d_i = Normal -Impaired[/tex]
[tex] d_i : -1.3, -1.43, -0.93, -0.67,-1.52, -0.69, -0.68, -0.61, -0.81[/tex]
The second step is calculate the mean difference
[tex]\bar d= \frac{\sum_{i=1}^n d_i}{n}=-0.96[/tex]
The third step would be calculate the standard deviation for the differences, and we got:
[tex]s_d =\frac{\sum_{i=1}^n (d_i -\bar d)^2}{n-1} =0.359[/tex]
The confidence interval for the mean is given by the following formula:
[tex]\bar d \pm t_{\alpha/2}\frac{s_d}{\sqrt{n}}[/tex] (1)
In order to calculate the critical value [tex]t_{\alpha/2}[/tex] we need to find first the degrees of freedom, given by:
[tex]df=n-1=9-1=8[/tex]
Since the Confidence is 0.95 or 95%, the value of [tex]\alpha=0.05[/tex] and [tex]\alpha/2 =0.025[/tex], and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,8)".And we see that [tex]t_{\alpha/2}=2.306[/tex]
Now we have everything in order to replace into formula (1):
[tex]-0.96-2.306\frac{0.359}{\sqrt{9}}=-1.24[/tex]
[tex]-0.96+2.306\frac{0.359}{\sqrt{9}}=-0.69[/tex]
So on this case the 95% confidence interval would be given by (-1.24;-0.69)
Random selection for first testing condition (impaired or unimpaired) was used to avoid order effects. The confidence interval on whether braking times under impaired and unimpaired conditions were significantly different can be determined using a paired t-test and if the interval includes zero, we can say that there is no significant difference.
Explanation:(a) Random selection of whether the student had unimpaired or impaired vision was a good idea because it helps to prevent any order effects. An order effect occurs if the order in which the tests are performed can influence the results. For example, if the unimpaired test was always done first, the driver might be more cautious in the second test as they have learned from the first test.
(b) The confidence interval for a difference between two means (in this case the braking times) can be calculated with a paired t-test. We will compare the average of differences (impaired vision braking time - normal vision braking time) to zero, assuming that they follow a normal distribution.
The formula to calculate the confidence interval for paired data is:
(Avg(D) - (t * StdDev(D) / sqrt(n)), Avg(D) + (t * StdDev(D) / sqrt(n)))
Where Avg(D) is the average of the differences, StdDev(D) is the standard deviation of the differences, n is the sample size (9 in this case), and t is the t-value from the t-distribution table (which will be 2.306 considering 95% confidence for 8 degrees of freedom).
After calculating you'll get the confidence interval for the differences. If this interval includes zero, we can say there is no significant difference for the braking time under impaired and unimpressed conditions using the 95% confidence interval.
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A piece of paper in the shape of an equilateral triangle with side length 3 and a circular piece of paper with radius 1 are glued together so that their centers coincide. How long is the outer perimeter of the resulting 2-dimensional shape?
Answer:
Outer perimeter of the resulting 2-dimensional shape will be 18.84 units
Step-by-step explanation:
From the figure, we could see
AE = FB = 1
And minor arc EF
=> [tex]\frac{2 \pi (1) }{6}[/tex]
=>[tex]= \frac{2 \pi}{ 6}[/tex]
=>[tex]= \frac{ \pi}{ 3}[/tex]
So by symmetry the perimeter is
=>[tex]3(2 + \frac { \pi}{3} )[/tex]
=>[tex]3(\frac {6 \pi}{3} )[/tex]
=>[tex]6 \pi[/tex]
=> [tex]6 \times 3.14[/tex]
=> 18.84 units
Answer:
6+pi or 9.14
Step-by-step explanation:
Your watch beeps every 15 seconds, and your moms watch beeps every 25 seconds. If they beep together at 3:00 pm, at what time will they beep together for the 20th time?
Answer:
The two watches will beep together for 20th time at 3:25 pm.
Step-by-step explanation:
My watch beeps every 15 seconds and mom's watch beeps every 25 seconds.
Thus both the watches beep at the same time at an interval of 75 seconds.
75 is the smallest multiple of 15 and 25.
They beep together at 3 pm and when they beep together for the 20th time , we have to add 20 times the time taken for both the watches to beep together.
This time interval = 20 [tex]\times[/tex] 75 = 1500 seconds = 25 minutes.
The two watches will beep together for 20th time at 3:25 pm.
The length, L, of a rectangle is decreasing at the rate of 2cm/sec, while the width, W, is increasing at the rate of 2 cm/sec.When L = 12 cm and W = 5 cm, find the rates of change of :______________
A) The area
B) The Perimeter
C) The lengths of the diagonals of the rectangle
Answer:
A) 14 cm² per sec
B) 0 cm per sec
C) -28 cm per sec
Step-by-step explanation:
We know that,
If l = length of a rectangle and w = width of the rectangle
A) Area of a rectangle,
[tex]A=l\times w[/tex]
Differentiating with respect to t ( time )
[tex]\frac{dA}{dt}=l\frac{dw}{dt}+w\frac{dl}{dt}[/tex]
We have,
[tex]l=12\text{ cm}, w=5\text{ cm}, \frac{dw}{dt}=2\text{ cm per sec}, \frac{dl}{dt}=-2\text{ cm per sec}[/tex]
[tex]\frac{dA}{dt}=12\times 2+5\times -2[/tex]
[tex]\frac{dA}{dt}=24-10[/tex]
[tex]\frac{dA}{dt}=14\text{ square cm per sec}[/tex]
B) Perimeter of the rectangle,
[tex]P=2(l+w)[/tex]
Differentiating with respect to t ( time ),
[tex]\frac{dP}{dt}=2(\frac{dl}{dt}+\frac{dw}{dt})=2(-2+2)=0[/tex]
C) Length of the diagonal,
[tex]D=\sqrt{l^2+w^2}[/tex]
[tex]D^2 = l^2 + w^2[/tex]
Differentiating with respect to t ( time ),
[tex]2D\frac{dD}{dt}=2l\frac{dl}{dt}+2w\frac{dw}{dt}[/tex]
Since, if l = 12 cm, w = 5 cm,
[tex]D=\sqrt{12^2+5^2}=\sqrt{144+25}=\sqrt{169}=13\text{ cm}[/tex]
[tex]\implies 2\times 13 \frac{dD}{dt}=2(12)(-2)+2(5)(2)[/tex]
[tex]26\frac{dD}{dt}=-48+20=-28\text{ cm per sec}[/tex]
Using implicit differentiation, it is found that:
a) The rate of change of the area of the rectangle is: 14 cm²/sec.
b) The rate of change of the perimeter of the rectangle is: 0 cm/sec.
c) The rate of change of the lengths of the diagonal of the rectangle is: [tex]\mathbf{-\frac{14}{13}}[/tex] cm/sec.
Item a:
The area of a rectangle of length l and width w is given by:
[tex]A = lw[/tex]
Applying implicit differentiation, the rate of change is of:
[tex]\frac{dA}{dt} = w\frac{dl}{dt} + l\frac{dw}{dt}[/tex]
For this problem, the values are:
[tex]w = 5, \frac{dl}{dt} = -2, l = 12, \frac{dw}{dt} = 2[/tex]
Then:
[tex]\frac{dA}{dt} = 5(-2) + 12(2)[/tex]
[tex]\frac{dA}{dt} = 14[/tex]
The rate of change of the area is of 14 cm²/sec.
Item b:
The perimeter is given by:
[tex]P = 2l + 2w[/tex]
Applying implicit differentiation, the rate of change is of:
[tex]\frac{dP}{dt} = 2\frac{dl}{dt} + 2\frac{dw}{dt}[/tex]
Then
[tex]\frac{dP}{dt} = 2(-2) + 2(2) = 0[/tex]
The rate of change of the perimeter is of 0 cm/sec.
Item c:
The diagonal is the hypotenuse of a right triangle in which the sides are the length and the width, then, applying the Pythagorean Theorem:
[tex]d^2 = l^2 + w^2[/tex]
The value of the diagonal is:
[tex]d^2 = 5^2 + 12^2[/tex]
[tex]d^2 = 169[/tex]
[tex]d = \sqrt{169}[/tex]
[tex]d = 13[/tex]
The rate of change is:
[tex]2d\frac{dd}{dt} = 2l\frac{dl}{dt} + 2w\frac{dw}{dt}[/tex]
Then
[tex]26\frac{dd}{dt} = -48 + 20[/tex]
[tex]26\frac{dd}{dt} = -28[/tex]
[tex]\frac{dd}{dt} = -\frac{14}{13}[/tex]
The rate of change of the lengths of the diagonals of the rectangle is of [tex]\mathbf{-\frac{14}{13}}[/tex] cm/sec.
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Consider the following code segment: primes = {2, 3, 5, 7} odds = {1, 3, 5, 7} Which line of code will result in x containing {1}? a. x = primes.difference(odds) b. x = odds.difference(primes) c. x = primes.intersection(odds) d. x = odds.intersection(primes)
Answer:
a. x = primes.difference(odds)
Step-by-step explanation:
Given the list of numbers primes and odds, the list x is made subtracting odds to primes. To do that in a object oriented language you use: x = primes.difference(odds) which is equivalent to x = primes - odds
The correct line of code that will result in x containing {1} is option b: x = odds.difference(primes). This provides the set of elements in the 'odds' set that are not in the 'primes' set, which is {1}.
Explanation:The question revolves around the concept of set operations in mathematics. Specifically, it focuses on the difference and intersection operations between two sets named primes and odds. To find the set containing only the number {1}, we should look for the difference between the odds and primes sets because 1 is in the odds but not in the primes.
Answer option a, x = primes.difference(odds), would result in a set that contains elements present in primes but not in odds, which would be {2}. However, that's not what we are looking for.
Answer option b, x = odds.difference(primes), is correct. It would result in a set containing elements that are in odds but not in primes, which is exactly {1}.
Answer options c and d, which refer to the intersection of the two sets, would result in the set {3, 5}, which are elements common to both primes and odds, and therefore not the correct answer.
For visual representation, you could draw a Venn diagram with two circles, one for primes containing {2, 3, 5, 7} and another for odds containing {1, 3, 5, 7}. The number 1 would be in the part of the odds circle that does not overlap with the primes circle.
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Two thirds of a number reduced by 11 is equal to 4 more than the number. Find the number. n= Answer
Answer:
-45
Step-by-step explanation:
2/3x - 11 = x + 4
I created an inequality representing the above statement first. This makes things look less complicated than what the question is asking.2/3x = x + 15
solve for x. I started by adding 11 to both sides.-1/3x = 15
multiply both sides by -3x = -45
Answer:
-45.
Step-by-step explanation:
2/3 x - 11 = x + 4
2/3 x - x = 4 + 11
-1/3x = 15
x = 15 * -3
x = -45.
7a-3=3-2a(If there is no solution, type in "no solution") a= Answer
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How many ways can I put down two indistinguishable pieces on an ordinary 8 times 8 chessboard, if the pieces must either be in the same row or be in the same column?
Answer:
we have 448 ways of putting them.
Step-by-step explanation:
For the first piece we have no restrictions, so we have 8*8 = 64 ways of puting it. For the second piece we have 7 ways to put it in the same row and 7 ways of put it in the same column, so we have 14 ways.
This gives us a total of 14*64 = 896 ways.
However, since the pieces are indistinguishable, we need to divide the result by two, because we were counting each possibility twice, (if we swap the pieces, then it counts as the same way), thus we have 896/2 = 448 of putting two pieces on the board.
Which of the following represents the slope of the line?
Help ASAP
Answer:option 3 is the correct answer.
Step-by-step explanation:
Slope, m =change in value of y on the vertical axis / change in value of x on the horizontal axis
change in the value of y = y2 - y1
Change in value of x = x2 -x1
y2 = final value of y
y 1 = initial value of y
x2 = final value of x
x1 = initial value of x
From the graph given,
y2 = 3
y1 = 1
x2 = 0
x1 = - 1
Slope = (3 - 1)/(0 - - 1)
Slope = 2/1 = 2
A 150-pound person burns 5.1 calories per minute when walking at a speed of 4 miles per hour. While walking, this person eats a snack that has 60 calories. This snack subtracts from the calories burned while walking. How long must the person walk at this speed to burn at least 150 calories? Use pencil and paper. Describe what values the person could change so that the amount of time spent walking would be less.
Answer:
He needs to walk 41 mins or 2,76 miles to burn 150 calories.
If he increases the speed of walking or eats snack with less calories, he will spend less time for walking.
Step-by-step explanation:
The person is burning
[tex]5,1*60=306[/tex]
calories per hour.
He needs to burn 150 calories plus 60 calories that comes from the snack. In total 210 calories to burn.
[tex]210/306=0,69[/tex]
He needs to walk 0,69 hours (aprrox. 41 mins) to burn 210 calories
[tex]0,69*4=2,76[/tex]
In total he need to walk 2,76 miles to burn 210 calories.
Final answer:
A 150-pound person must walk approximately 17.65 minutes at 4 mph to burn at least 150 calories, after accounting for a 60-calorie snack. To decrease walking time, they can increase activity intensity or choose lower-calorie snacks.
Explanation:
To calculate the time required for a 150-pound person to burn at least 150 calories at a rate of 5.1 calories per minute while walking at 4 miles per hour, we must account for the 60-calorie snack they consumed. We first subtract the 60 calories from the 150-calorie goal, which leaves 90 calories to be burned through exercise. Dividing 90 calories by the 5.1 calories burned per minute gives us the time needed in minutes.
Calculation:
Total calories to burn = 150 - 60 (from snack) = 90 calories
Calories burned per minute = 5.1
Time (in minutes) = Total calories to burn ÷ Calories burned per minute = 90 ÷ 5.1 = 17.65 minutes
So, the person must walk approximately 17.65 minutes to burn at least 150 calories, minus the calories from the snack. To reduce the time spent walking, the person could increase their walking speed, engage in a more vigorous exercise, or consume a lower-calorie snack.
Tran has a credit card with a spending limit of $2000 and an APR (annual percentage rate) of 12%. During the first month, Tran charged $450 and paid $150 of that in his billing cycle. Which expression will find the amount of interest Tran will be charged after the first month?
Step-by-step explanation:
Given data:
Tran has a credit card with a spending limit of $2000 and an APR (annual percentage rate) of 12%.
During the first month, Tran charged $450 and paid $150 of that in his billing cycle.
The expression which will find the amount of interest Tran will be charged after the first month is (0.012)(300)
Here 0.01 because it is 1 month tax.
300 is remaining amount as Tran used $450 but paid $150.
Answer:
its 0.01 [300]
Step-by-step explanation:
There are 225 students at March Middle School. On Friday, 135 students wore spirit shirts. What percent of the students did NOT wear spirit shirts on Friday?
Answer:
40%
Step-by-step explanation:
135 of 225 is equal to 60%. Therefore, the remaining percent, which should equal to 100, did not wear spirit shirts. 100 - 60 = 40. Therefore, 40% of students did not wear spirit shirts on Friday.
The percentage of students who did NOT wear spirit shirts on Friday is 40%
Explanation:To find the percentage of students who did NOT wear spirit shirts on Friday, we subtract the number of students who wore spirit shirts from the total number of students and divide by the total number of students.
Then, multiply by 100 to get the percentage.
Number of students who did NOT wear spirit shirts = Total number of students - Number of students who wore spirit shirts = 225 - 135 = 90.
Percentage of students who did NOT wear spirit shirts = (90 / 225) x 100 = 40%
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A croissant shop has plain croissants, cherry croissants, chocolate croissants, almond croissants, apple croissants, and broccoli croissants. How many ways are there to choose
a)a dozen croissants?
b)three dozen croissants?
c)two dozen croissants with at least two of each kind?
d)two dozen croissants with no more than two broccoli croissants?
e)two dozen croissants with at least five chocolate croissants and at least three almond croissants?
f)two dozen croissants with at least one plain croissant, at least two cherry croissants, at least three chocolate croissants, at least one almond croissant, at least two apple croissants, and no more than three broccoli croissants?
Answer:
A. 6 188
B. 749 398
C. 6 188
D. 52 975
E. 20 349
F. 11 316
Explanation:
(a) The shop has 6 types of croissants of which a dozen(12) has to be selected
Therefore n=6, r=12
Repetition of croissants is permitted
And C(n+r-1, r)
C(6+12-1, 12) = C(17, 12) = 17!÷ 12!(17-12)! = 17!÷12! 5! =6 188
(b) The shop has 6 types of croissants of which three dozen(36) has to be selected
Therefore n=6, r=36
Repetition of croissants is permitted
And C(n+r-1, r)
C(6+36-1, 12) = C(41, 36) = 41!÷ 36!(41-36)! = 41!÷36! 5! = 749 398
(c) The shop has 6 types of croissants of which two dozen(24) has to be selected
Let us first select 2 of each kind which 12 croissants in total. Then we still need to select the remaining 12 croissants
Therefore n=6, r=12
Repetition of croissants is permitted
And C(n+r-1, r)
C(6+12-1, 12) = C(17, 12) = 17!÷ 12!(17-12)! = 17!÷12! 5! =6 188
(d) The shop has 5 types of croissants of which two dozen(24) has to be selected
Therefore n=5, r=24
Repetition of croissants is permitted
And C(n+r-1, r)
C(5+24-1, 24) = C(28, 24) = 28!÷ 24!(28-24)! = 28!÷24! 4! = 20 475
This problem involves calculating the number of combinations with repetition where the order of choosing croissants does not matter and requires applying the combinatorics formula for such scenarios.
Explanation:The question asked is concerning the number of ways to choose different quantities of croissants from a croissant shop that has six different types. This is a classic example of a combinatorics problem which falls under the category of counting the number of combinations with repetition.
The counting principle states that if there are 'n' ways to do one thing, and 'm' ways to do another, then there are n * m ways to do both. However, when choosing with repetition where the order does not matter, we use the formula for combinations with repetition: (n + r - 1)! / r!(n - 1)!, where 'n' is the number of types and 'r' is the number of items to be chosen.
How many elements does each of these sets have where a and b are distinct elements?
a) P({a, b, {a, b}})
b) P({∅, a, {a}, {{a}}})
c) P(P(∅))
Answer:
a) P({a, b, {a, b}})=[tex] 2*2*2=8[/tex]
b) P({∅, a, {a}, {{a}}})=[tex] 2*2*2*2=16[/tex]
c) P(P(∅))=P(P{∅})=2
Step-by-step explanation:
Previous concepts
For this case we need to remember the concpts of subset and the power set.
A is a subset of B if every element of A is an eelement if B and we write [tex]A \subseteq B[/tex]
The power set of a set R is th set of all the possible subsets of R and we writ this as P(R)
Solution to the problem
Part a
P({a, b, {a, b}}) as we can see this set contains 3 elements a,b and {a,b}
And since the power set contains all the possible subsets of the elements. For this case for each element we have 3-1 = 2 options in order to combine, then the number of possible subsets would be equal to the product of possible options and we got:
P({a, b, {a, b}})=[tex] 2*2*2=8[/tex]
Part b
P({∅, a, {a}, {{a}}}) as we can see this set contains 4 elements ∅,a,{a} and {{a}}
And since the power set contains all the possible subsets of the elements. For this case for each element we have 3-1 = 2 options in order to combine, then the number of possible subsets would be equal to the product of possible options and we got:
P({∅, a, {a}, {{a}}})=[tex] 2*2*2*2=16[/tex]
Part c
P(P(∅)) for this case we need to remember that P(∅) have all the possible subsets empty, but the power set for the empty set is also empty.
P(∅)={∅}
And we see that this last one set have just one element.
For this special case we have again 2 options since we can have the element in the subset or the element that is not on the subset, so then we have this:
P(P(∅))=P(P{∅})=2
The number of elements of the set are 8, 16, and 2.
How to solve the elements?It should be noted that the objects in a set are known as the elements or members of the set.
The number of elements in the set will be calculated thus:
P({a, b, {a, b} = 2³ = 2 × 2 × 2 = 8
P({∅, a, {a}, {{a}}) = 2⁴ = 16
P(P(∅)) = 2¹ = 2
In conclusion, the correct options are 8, 16, and 2.
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Suppose the expression a(b)n models the approximate number of people who visited an aquarium each day since an aquarium opened, where a is the initial number of people who visited, b is the rate of increase in the number of people who visited each day, and n is the number of days since the aquarium opened.
If the expression below models the number of visitors of a particular aquarium, what is the correct interpretation of the second factor?
54(1.3)^7
A.
There were 9.1 times as many people who visited the aquarium on the 7th day as on the first day.
B.
There were 1.3 times as many people who visited the aquarium on the 7th day as on the first day.
C.
There were 6.27 times as many people who visited the aquarium on the 7th day as on the first day.
D.
There were 10.2 times as many people who visited the aquarium on the 7th day as on the first day.
Answer:
C. There were 6,27 times as many people who visited the aquarium on the seventh day as on the first day.
Explanation:
[tex]\displaystyle 1,3^7 = 6,2748517 ≈ 6,27[/tex]
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A researcher measures the time (in seconds) it takes a sample of five participants to complete a memory task. It took four of the participants 7, 8, 8, and 9 seconds. If M = 9, then what must be the fifth time recorded?
Using the mean formula and the data given, the fifth participant’s time to complete the memory task must be 13 seconds to result in a mean (M) of 9 seconds for all five participants.
To find the fifth time recorded when four participants took 7, 8, 8, and 9 seconds to complete a memory task and the mean (M) time is 9 seconds, we need to use the formula for the mean of a sample which is the sum of all values divided by the number of values.
Here's the equation based on the data provided:
(7 + 8 + 8 + 9 + x) / 5 = 9
First, calculate the sum of the times we know:
7 + 8 + 8 + 9 = 32 seconds
Now, insert this sum into the equation and solve for x, where x represents the fifth time:
(32 + x) / 5 = 9
32 + x = 45
x = 45 - 32
x = 13 seconds
The fifth participant must have taken 13 seconds to complete the memory task to have a mean time of 9 seconds.
When solving for an unknown variable that has a number preceding it, you will divide both sides of the equation by this number, which is known as the ______________.
Answer: Coefficient
Step-by-step explanation:
When solving for an unknown variable that has a number preceding it, you will divide both sides of the equation by this number, which is known as the coefficient.
A coefficient is a number preceding any variable in a function for example, given the function 4x, the variable is 'x' and the number preceding it is 4. This number preceding the variable is what we call 'coefficient' of the variable 'x'
The Norwood Tigers are having team shirts made. One option is to pay Harold's Tees a $41 setup fee and then buy the shirts for $10 each. Another option is to go to City Printing, paying $36 for a setup fee and an additional $15 per shirt. The team parent in charge of the project notices that, with a certain number of shirts, the two options have the same cost. What is the cost? How many shirts is that?
The cost will be the same when 1 shirt is ordered.
Explanation:Cost Comparison
Option 1: Cost = $41 (setup fee) + $10 (shirt price)
Option 2: Cost = $36 (setup fee) + $15 (shirt price)
Let's assume the number of shirts is 'x'.
Equating the costs of both options:
$41 + $10x = $36 + $15x
$5x = $5
x = 1
Therefore, the cost will be the same for 1 shirt.
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Kira goes running Kira goes running Monday, Wednesday, and Friday of each week. Each Monday she runs 2.5 miles, each Wednesday she runs 5.75 miles, and each Friday she runs 7.75 miles. How many total miles does she run in 9 weeks?
Kira runs a total of 144 miles in 9 weeks.
Step-by-step explanation:
Given,
Distance covered on Monday = 2.5 miles
Distance covered on Wednesday = 5.75 miles
Distance covered on Friday = 7.75 miles
Total distance covered per week = 2.5+5.75+7.75 = 16 miles
Total distance in 9 weeks = Distance per week * 9
Total distance in 9 weeks = 16*9 = 144 miles
Kira runs a total of 144 miles in 9 weeks.
Keywords: addition, multiplication
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write a polynomial function with the given zeros x= -2,1,4
Answer:
With the given zeros x=-2,1,4 the polynomial function is [tex]x^3-3x^2-6x-8[/tex]
Step-by-step explanation:
Given zeros are x=-2,1,4
Now to find the polynomial function:
With the given zeros we can write it as below:
x=-2 implies that x+2=0
x=-1 implies that x-1=0
x=4 implies that x-4=0
Then we can the zeros or factors by (x+2)(x-1)(x-4)
Now expanding the zeros or factors:
[tex](x+2)(x-1)(x-4)[/tex]
[tex](x+2)(x-1)(x-4)=(x^2-x+2x-2)(x-4)[/tex] ( multiply each term with each term of another factor)
[tex]=(x^2+x-2)(x-4)[/tex] ( adding the like terms)
[tex]=x^3-4x^2+x^2-4x-2x+8[/tex] ( multiply each term with each term of another factor)
[tex]=x^3-3x^2-6x+8[/tex] ( adding the like terms)
[tex](x+2)(x-1)(x-4)=x^3-3x^2-6x+8[/tex]
Therefore the polynomial function is [tex](x+2)(x-1)(x-4)=x^3-3x^2-6x+8[/tex]
With the given zeros x=-2,1,4 the polynomial function is [tex]x^3-3x^2-6x-8[/tex]
Listen Two construction cranes are used to lift identical 1200-kilogram loads of bricks the same vertical distance. The first crane lifts the bricks in 20. seconds and the second crane lifts the bricks in 40. seconds. Compared to the power developed by the first crane, the power developed by the second crane's the same twice as greathalf as greatfour times as great?
The power developed in the first crane will be twice the power developed in the second crane.
Given information:
Two construction cranes are used to lift identical 1200-kilogram loads of bricks at the same vertical distance.
The first crane lifts the bricks in 20 seconds and the second crane lifts the bricks in 40 seconds.
Now, the load is the same and the vertical lift is also the same. So, the work done W by both the lifts will also be the same.
The power developed by first crane will be,
[tex]P_1=\dfrac{W}{20}[/tex]
The power developed by second crane will be,
[tex]P_2=\dfrac{W}{40}[/tex]
So, the ratio of power developed in two cranes will be,
[tex]P_1:P_2=\dfrac{W}{20}:\dfrac{W}{40}\\P_1:P_2=2:1[/tex]
Therefore, the power developed in the first crane will be twice the power developed in the second crane.
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Final answer:
The power developed by the first crane is twice as great as the power developed by the second crane because it lifts the bricks in half the time.
Explanation:
To compare the power developed by the first crane with the power developed by the second crane, we can use the formula:
Power = Work / time
Since both cranes are lifting identical 1200-kilogram loads of bricks the same vertical distance, the work done by both cranes is the same. Therefore, the power developed by the first crane is twice as great as the power developed by the second crane. This is because the first crane lifts the bricks in 20 seconds, which is half the time it takes the second crane to do the same work in 40 seconds.
How many three-digit positive integers exist, all of whose digits are 2's and/or 5's?
Answer:
8
Step-by-step explanation:
There are two choices for each digit, and 3 digits, so 2^3 = 8 possible numbers. Here's a list:
222, 225, 252, 255
522, 525, 552, 555
Answer:
8
Step-by-step explanation:
The first digit has two possibilities; it can either be a 2 or a 5.
The second digit has two possibilities; it can either be a 2 or a 5.
The third digit has two possibilities; it can either be a 2 or a 5.
2 x 2 x 2 = 8 integers.
(Here are the numbers listed out):
222
225
252
255
522
525
552
555
Consider that the length of rectangle A is 21 cm and its width is 7 cm. Which rectangle is similar to rectangle A?
Question is incomplete, complete question is given below.
Consider that the length of rectangle A is 21 cm and its width is 7 cm. Which rectangle is similar to rectangle A? A) A rectangle with a length of 9 cm and a width of 3 cm. B) A rectangle with a length of 20 cm and a width of 10 cm. C) A rectangle with a length of 18 cm and a width of 9 cm. D) A rectangle with a length of 27 cm and a width of 3 cm.
Answer:
A) A rectangle with a length of 9 cm and a width of 3 cm.
Step-by-step explanation:
Given,
Length = 21 cm Width = 7 cm
[tex]Ratio = \frac{Length}{Width}\\\\Ratio = \frac{21}{7}=3:1[/tex]
For the similarity of rectangle the ratio of length to width of any two rectangles should be equal.
So we have four options given;
A) A rectangle with a length of 9 cm and a width of 3 cm.
[tex]Ratio =\frac{9}{3}=3:1[/tex]
Here the ratio is equal as of rectangle A. So this rectangle is similar to rectangle A.
B) A rectangle with a length of 20 cm and a width of 10 cm.
[tex]Ratio =\frac{20}{10}=2:1[/tex]
Here the ratio is not equal as of rectangle A. So this rectangle is not similar to rectangle A.
C) A rectangle with a length of 18 cm and a width of 9 cm.
[tex]Ratio =\frac{18}{9}=2:1[/tex]
Here the ratio is not equal as of rectangle A. So this rectangle is not similar to rectangle A.
D) A rectangle with a length of 27 cm and a width of 3 cm.
[tex]Ratio =\frac{27}{3}=9:1[/tex]
Here the ratio is not equal as of rectangle A. So this rectangle is not similar to rectangle A.
Thus the correct option is A) A rectangle with a length of 9 cm and a width of 3 cm.
A property's taxes are $9,540 and are paid for in arrears. The property was sold and closing took place on March 7. The seller was responsible for the day of closing. Using the 360 day method, how will taxes be prorated?
1 - Credit the buyer $1,776 and debit the seller $1,776.
2 - Credit the seller $186 and debit the buyer $186.
3 - Credit the buyer $1,590 and debit the seller $1,590.
4 - Credit the buyer $1,675.50 and debit the seller $1,675.50.
Answer:
Credit the buyer $1,776 and debit the seller $1,776.
Step-by-step explanation:
In accounting books the buyer will be credit from $1,776 and seller will be debit from $1,776.
Answer:
1) Credit the buyer $1,776 and debit the seller $1,776.
Step-by-step explanation:
For this, we are to simply understand and follow the calculation of proration. Proration is the allocation or dividing of certain money items at the closing.
The buyer pays:
Total taxes = $ 9540
Total Time Span before closing took place = 2 months and 7 days
Per month charge = [tex]\frac{9540}{12}[/tex] = $795 per month
Charges per day = [tex]\frac{795}{30}[/tex] =$ 26.5 per day
2 months charge = 795 x 2 = $1590
Charges for 7 days = 7 x 26.5 = $185.5
Total Charges = 1590 + 185.5 = $ 1775.5 ≈ $1776 (Rounding off)
As the property's taxes were paid for by the buyer and the seller was responsible for the day of closing, and buyer paid for the arrears too of the previous 2 months and 7 days, the buyer gets a credit and seller gets a debit. The seller owes the buyer $1776.
Hence,
The buyer gets a credit of $1,776 and Seller gets a debit of $1,776
Jose has 5/6 hour to spend on his homework. He wants to spend 1/3 of his time on math homework, 15 minutes of his time on reading homework, and then the remaining time on science homework. How much time in minutes, does Jose have to spend on his science homework?
Jose will spend 18 minutes on his science homework.
Step-by-step explanation:
Given,
Time Jose have to spend on homework = 5/6 hour
We know that 1 hour = 60 minutes
Therefore;
[tex]\frac{5}{6}\ hour = \frac{5}{6}*60\ minutes\\\\\frac{5}{6}\ hour = 50\ minutes[/tex]
Time Jose have to spend on homework = 50 minutes
Time spent on math home work = [tex]\frac{1}{3}\ of\ his\ time[/tex]
Time spent on math home work = [tex]\frac{1}{3}*50=16.66\ minutes[/tex]
Rounding off to nearest whole minute;
Time spent on math homework = 17 minutes
Time spent on reading homework = 15 minutes
Total time spent = Mathematics + Reading + Science
50 = 17+15+Science
50 = 32 + Science
Science = 50-32 = 18 minutes
Jose will spend 18 minutes on his science homework.
Keywords: addition, fraction
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The dot plot represents a sampling of ACT scores:
(picture below)
Which box plot represents the dot plot data?:
(choices in second picture below)
Answer:
Option D.
Step-by-step explanation:
Form the given line plot, first we need to find the data set. So, our data set is
24, 26, 26, 26, 27, 27, 27, 28, 28, 28, 28, 28, 30, 30, 30, 32, 32, 32, 35
Divide the data in two equal parts.
(24, 26, 26, 26, 27, 27, 27, 28, 28), 28, (28, 28, 30, 30, 30, 32, 32, 32, 35)
Divide each of the parenthesis in two equal parts.
(24, 26, 26, 26), 27, (27, 27, 28, 28), 28, (28, 28, 30, 30), 30, (32, 32, 32, 35)
Now, we get
Minimum value = 24
First quartile = 27
Median = 28
Third quartile = 30
Maximum value = 35
It means the box lies between 27 and 30. The line inside the box at 28. Left point of the line isi 24 and right point of the line 35.
This description represented by the box plot in option D.
Hence, the correct option is D.
There are 2 numbers and the sum of those numbers are 40 and the difference of those numbers is 10 what are the numbers that equal the sum and difference
Answer:
15 and 25
Step-by-step explanation:
15+25=40
10+20=30+5+5=10+30=40
Answer:one number is 15 and the other number is 25
Step-by-step explanation:
Let x represent one of the numbers.
Let y represent the other number.
There are 2 numbers and the sum of those numbers are 40. This means that
x + y = 40 - - - - - - - -1
The difference of those numbers is 10. This means that
x - y = 10 - - - - - - - - -2
We would eliminate x by subtracting equation 2 from equation 1, it becomes
2y = 50
y = 50/2 = 25
Substituting y = 25 into equation 1, it becomes
x + 25 = 40 - 25 = 15
Plzz help gahhhhh :\
Answer:
The product of 'x' and 'y' is [tex]\boxed 8[/tex].
Step-by-step explanation:
Given:
[tex]\log_{5\sqrt5}125=x\\\\\log_{2\sqrt2}64=y[/tex]
We need to determine the product of 'x' and 'y'.
Using the following logarithmic property:
[tex]\log_ab=\frac{\log b}{\log a}[/tex]
Here, [tex]a=5\sqrt5\ and\ 2\sqrt2[/tex]
[tex]b=125\ and\ 64[/tex]
So, [tex]log_{5\sqrt5}125=\frac{\log 125}{\log 5\sqrt{5}}\\\\log_{5\sqrt5}125=\frac{\log 5^3}{\log 5\times5^{1/2}}.......[\sqrt5=5^{1/2}][/tex]
[tex]log_{2\sqrt2}64=\frac{\log 64}{\log 2\sqrt{2}}\\\\log_{2\sqrt2}64=\frac{\log 2^6}{\log 2\times2^{1/2}}.......[\sqrt2=2^{1/2}][/tex]
Now, we use another property of log and exponents.
[tex]\log a^m=m\log a\\a^m\times a^n=a^{m+n}[/tex]
[tex]log_{5\sqrt5}125=\frac{3\log 5}{\log 5^{1+{1/2}}}=\frac{3\log 5}{\log 5^{\frac{3}{2}}}=\frac{3\log 5}{\frac{3}{2}\log 5}=2\\\\\\\\log_{2\sqrt2}64=\frac{6\log 2}{\log 2^{1+{1/2}}}=\frac{6\log 2}{\log 2^{\frac{3}{2}}}=\frac{6\log 2}{\frac{3}{2}\log 2}=\frac{12}{3}=4[/tex]
So, [tex]x=2\ and\ y=4[/tex]
The product of 'x' and 'y' = [tex]2\times 4=8[/tex]
Therefore, the product of 'x' and 'y' is 8.
(1.005)[tex]x^{2} 400[/tex]
match the equation with its graph 5x-4y=20
A) First Picture
B) Second Picture
C) Third Picture
D) Fourth Picture
I've attached the correct image to this answer, just in case they're scrambled.
To match the equation 5x-4y=20 with its graph, convert it to slope-intercept form to identify a positive slope of 5/4 and a y-intercept of -5. Look for a graph with these features in the provided pictures.
Explanation:To match the equation with its graph for 5x-4y=20, we first need to rewrite it in slope-intercept form, which is y=mx+b. By isolating y, we get 4y = 5x - 20, or y = (5/4)x - 5. This equation signifies a line with a slope (m) of 5/4, meaning for every increase of 4 units in x, y increases by 5 units. The y-intercept (b) is -5, indicating where the line crosses the y-axis. Now, looking at the provided pictures, we should find a graph with a positive slope of 5/4 and a y-intercept of -5.
The correct graph will rise more than it runs since the slope is greater than 1, and will cross the y-axis below the origin at -5. Without seeing the actual images, we are unable to definitively identify which picture corresponds to the given equation. Typically, you would examine each graph to see which one matches these characteristics.