What is 100 x 1.5 plz help me

Of the 84 parents who attended a meeting at a school, 35 volunteered to supervise children during the school picnic and 11 volun

Tatiana [17]

Answer:

The number of parents who volunteered to bring refreshments is 16.

Step-by-step explanation:

The information provided is:

N = 84

n (S) = 35

n (S ∩ R) = 11

n (R) = 1.5 × n (S' ∪ R')

Compute the value of n (S ∪ R) as follows:

$n(S\cup R)=n(S)+n(R)-n (S\cap R)\\\\n(S\cup R)=n(S)+[1.5(N-n(S\cup R)]-n (S\cap R)\\\\n(S\cup R)=n(S)+1.5N-1.5n(S\cup R)-n (S\cap R)\\\\2.5n(S\cup R)=n(S)+1.5N-n (S\cap R)\\\\2.5n(S\cup R)=35+(1.5\times84)-11\\\\2.5n(S\cup R)=150\\\\n(S\cup R)=\frac{150}{2.5}\\\\n(S\cup R)=60$

Now compute the value of n (R) as follows:

$n (R) = 1.5 \times n (S' \cup R') \\\\n (R) = 1.5 \times [N-n (S \cup R)]\\\\n (R) = 1.5 \times [84-60]\\\\n(R)=16$

Thus, the number of parents who volunteered to bring refreshments is 16.

5 0

3 years ago

The table shows the weights and costs of various wood purchases.

White raven [17]

You have to add then subtract

3 0

3 years ago

Read 2 more answers

According to Nielsen Media Research. of all the U.S. households that owned at least one television set, 83% had two or more sets

Rudik [331]

Answer:

The proportion of U.S. households that owned two or more televisions is 83%.

Step-by-step explanation:

To determine whether the proportions of U.S. households that owned two or more televisions is less than 83% or not let us perform a hypothesis test for single proportion.

Assumptions:

The sample size (n) selected by the local cable company is 300 which is quite large. Then according to the Central limit theorem the sampling distribution of sample proportion follows a normal distribution with mean p and standard deviation $\sqrt{\frac{p(1-p)}{n} }$ .

Since the sampling distribution of sample proportions follows a normal distribution use the z-test for one proportion to perform the test.

The hypothesis is:

$H_{0}$ : The proportion of U.S. households that owned two or more televisions is 83%, i.e. $p=0.83$

$H_{1}$ :The proportion of U.S. households that owned two or more televisions is less than 83%, i.e. $p< 0.83$

Decision Rule:

At the level of significance α = 0.05 the critical region for a one-tailed z-test is:

$\\ Z\leq -1.645\\$

**Use the z table for the critical values.

So, if $\\ Z\leq -1.645\\$ the null hypothesis will be rejected.

Test statistic value:

$z=\frac{\hat{p}-p}{\sqrt{\frac{p(1-p)}{n}}}$

Here $\hat{p}$ is the sample proportion.

Compute the value of $\hat{p}$ as follows:

$\hat{p}=\frac{X}{n} \\=\frac{240}{300}\\ =0.80$

Now compute the value of the test statistic as follows:

$z=\frac{\hat{p}-p}{\sqrt{\frac{p(1-p)}{n}}}\\=\frac{0.80-0.83}{\sqrt{\frac{0.83*(1-0.83)}{300} } } \\=-1.383$

The test statistic is -1.383 which is more than -1.645.

Thus, the test statistic lies in the acceptance region.

Hence we fail to reject the null hypothesis.

Conclusion:

At 0.05 level of significance we fail to reject the null hypothesis stating that the proportion of U.S. households that owned two or more televisions is 83%.

6 0

3 years ago

Which expressions represent Ernie's problem? Select each correct answer. 1500525 5251500 1500 ÷ 525 525 ÷ 1500 5251500 1500525

Kay [80]

Answer:

525 ÷ 1500 5251500

Step-by-step explanation:

3 0

3 years ago

Please help! i really need this

Katen [24]

Answer:

1/6

Step-by-step explanation:

on the cube, each number is listed one time. if he rolls the cube one time, he has 1/6 chance of it landing on 4.

6 0

4 years ago

Read 2 more answers