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alexira [117]
3 years ago
14

Assume that advanced students average 93% on an achievement test and regular students averaged 75%. If 100 advanced students and

300 regular students took the test, what would you expect the average to be? How many of each kind of student would be needed to get a group of 90 students who average 87% on the test? Please include work/explaination, I along with a few of my classmates are confused by this problem. We are in a chapter about systems of equations.
Mathematics
1 answer:
Lemur [1.5K]3 years ago
7 0
Average (mean) = (sum of all the data) / (# of data)

sum of all the data = (average)(# of data)

Thus for 100 students with an average of 93,
sum of all data = (93)(100) = 9300

and for 300 students with an average of 75,
sum of all data = (75)(300) = 22500

Therefore you would expect the overall average to be
(9300 + 22500) / (100 + 300) = 79.5 %

Now if there are x # of advanced students and y # of regular students, then

x + y = 90 (total # of students) and 93x + 75y = 87(x + y) (overall average)

The second equation can be simplified to x - 2y = 0

Subtracting the two equations yields

x = 60 and y = 90

Therefore you would need 60 advanced and 30 regular students.
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I think the answer is negative 8 square root 726

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5 0
3 years ago
Which of the following combined with the equation −9x + 3y = 12 creates a system of linear equations with no solution?
Mumz [18]

Answer:

Step-by-step explanation:

You can't get an exact answer to this because there are no choices.

Add 9x to both sides.

3y = 9x + 12              Divide by 3

3y/3 = 9x/3 + 12/3    Combine

y = 3x + 4

Now to get anything at all that is parallel to this and providing no answer because there is no intersection of the two lines, just change the 4 to something else. Here's one example

y = 3x + 8

As long as the number in front of the x's is the same, they are parallel with no point in common.

8 0
3 years ago
A television network is deciding whether or not to give its newest television show a spot during prime viewing time at night. If
musickatia [10]

Answer:

a. The test statistic for this hypothesis would be z=1.71

b. Critical value at α = 0.10: z=1.282

c. Reject H0, the television network should not keep its current lineup.

d. H0: p ≤ 0.50; HA: p > 0.50

Step-by-step explanation:

We have to performa an hypothesis test on a proportion.

The claim is that the proportion of viewers that prefer the new show is bigger than 50% (meaning that most viewers prefer the new show).

The null hypothesis will state that both shows have the same proportion.

Then, the null and alternative hypothesis are:

H_0: \pi\leq0.50\\\\H_a: \pi>0.5

The sample proportion is p=0.53

p=X/n=438/827=0.53

The standard error of the proportion is:

\sigma_p=\sqrt{\dfrac{p(1-p)}{n}}=\sqrt{\dfrac{0.5*0.5}{827}}= 0.017

Then, the z-statistic can be calculated as:

z=\dfrac{p-\pi-0.5/n}{\sigma_p}=\dfrac{0.53-0.50-0.5/827}{0.017}=\dfrac{ 0.029 }{0.017} =1.706

The critical value for a right tailed test at a significance level of 0.10 is zc=1.282. This value can be looked up in a standard normal distribution table.

As the statistic is bigger than the critical value, it lies in the rejection region. The null hypothesis is rejected. There is evidence to support the claim that the proportion of viewers which support the new show is larger than 0.50.

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