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Rudiy27
3 years ago
13

In Chad's reading class, all the students are reading the same book. The school bought each student a book at $7 per book. If th

ere are 27 students in Chad's class, which of the following expressions could not be used to calculate the total cost?
a. 7(30) - 7(7)
b. 7(20) + 7(7)
c. 7(30) - 7(3)
Mathematics
2 answers:
Molodets [167]3 years ago
8 0

It would be A. 7(30) - 7(7)


natka813 [3]3 years ago
5 0

Answer:

a. 7(30) - 7(7)

Step-by-step explanation:

Given,

The total number of students = 27,

∵ Every student gets a book of $ 7,

So, the total cost ( in dollars ) = price of each book × total student  

= 7 × 27

= 7(27)

∵ 7(30) - 7(7) = 7(30-7) = 7(23) ≠ 7(27)

7(20) + 7(7) = 7(20+7) = 7(27)

7(30) - 7(3) = 7(30-3) = 7(27)

Hence, option 'a' is correct.

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A door delivery florist wishes to estimate the proportion of people in his city that will purchase his flowers. Suppose the true
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Answer:

99.74% probability that the sample proportion will be less than 0.1

Step-by-step explanation:

I am going to use the binomial approximation to the normal to solve this question.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

E(X) = np

The standard deviation of the binomial distribution is:

\sqrt{V(X)} = \sqrt{np(1-p)}

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that \mu = E(X), \sigma = \sqrt{V(X)}.

In this problem, we have that:

n = 276, p = 0.06

So

\mu = E(X) = np = 276*0.06 = 16.56

\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{276*0.06*0.94} = 3.9454

What is the probability that the sample proportion will be less than 0.1

This is the pvalue of Z when X = 0.1*276 = 27.6. So

Z = \frac{X - \mu}{\sigma}

Z = \frac{27.6 - 16.56}{3.9454}

Z = 2.8

Z = 2.8 has a pvalue of 0.9974

99.74% probability that the sample proportion will be less than 0.1

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4 years ago
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Answer:

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The null and alternative hypothesis are:

H_0: \pi_1-\pi_2=0\\\\H_a:\pi_1-\pi_2< 0

being the subindex 1 for the experimental group and subindex 2 for the control group.

2) Test statistic z=-1.26

Step-by-step explanation:

This is a hypothesis test for the difference between proportions.

The claim is that the vaccine was effective.

Then, the null and alternative hypothesis are:

H_0: \pi_1-\pi_2=0\\\\H_a:\pi_1-\pi_2< 0

The significance level is 0.01.

The sample 1 (experimental group), of size n1=100000 has a proportion of p1=0.0002.

p_1=X_1/n_1=21/100000=0.0002

The sample 2 (control group), of size n2=100000 has a proportion of p2=0.0003.

p_2=X_2/n_2=30/100000=0.0003

The difference between proportions is (p1-p2)=-0.0001.

p_d=p_1-p_2=0.0002-0.0003=-0.0001

The pooled proportion, needed to calculate the standard error, is:

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The estimated standard error of the difference between means is computed using the formula:

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Then, we can calculate the z-statistic as:

z=\dfrac{p_d-(\pi_1-\pi_2)}{s_{p1-p2}}=\dfrac{-0.00009-0}{0.000071}=\dfrac{-0.00009}{0.000071}=-1.26

This test is a left-tailed test, so the P-value for this test is calculated as (using a z-table):

P-value=P(z

As the P-value (0.104) is bigger than the significance level (0.01), the effect is not significant.

The null hypothesis failed to be rejected.

There is  not  enough evidence to support the claim that the vaccine was effective.

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