Help help help help help

What is 4×5/8 please write the correct answer

wel

The answer would be 2.5.

hope this helps (;

4 0

3 years ago

A nationwide survey claimed that at least 65% of parents with young children condone spanking their child as a regular form of p

liq [111]

Answer:

here p=0.65 and n=100 hence standard error =(p*(1-.

Step-by-step explanation:

4 0

2 years ago

Simplify: ( 1^0+2^1+3^2) ÷ 22/ 5^0

Aleksandr-060686 [28]

Step-by-step explanation:

( 1 + 2 + 9 ) ÷ 22/ 1

12 ÷ 22

6 ÷ 11

cc: Anything raised to the power of zero is 1

7 0

2 years ago

Read 2 more answers

Tier: Homework 1<br> Solve 2(4y - 5) = 14 983

Lostsunrise [7]

Answer:

y = 1874.25

Step-by-step explanation:

2(4y - 5) = 14 983

2*4y + 2*-5 = 14983

8y - 10 = 14983

8y = 14983 + 10

8y = 14993

y = 14993/8

y = 1874.25

8 0

3 years ago

Read 2 more answers

According to an article in Newsweek, the natural ratio of girls to boys is 100:105. In China, the birth ratio is 100:114 (46.7%

mojhsa [17]

Answer:

$z=\frac{0.42 -0.467}{\sqrt{\frac{0.467(1-0.467)}{150}}}=-1.154$

$p_v =2*P(z$

So the p value obtained was a very high value and using the significance level given $\alpha=0.05$ we have $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the proportion of girls born is not significantly different from 0.467

Step-by-step explanation:

Data given and notation

n=150 represent the random sample taken

X=63 represent the number of girls born

$\hat p=\frac{63}{150}=0.42$ estimated proportion of girls born

$p_o=0.467$ is the value that we want to test

$\alpha=0.05$ represent the significance level

Confidence=95% or 0.95

z would represent the statistic (variable of interest)

$p_v$ represent the p value (variable of interest)

Concepts and formulas to use

We need to conduct a hypothesis in order to test the claim that the true proportion if girls is 0.467.:

Null hypothesis: $p=0.467$

Alternative hypothesis: $p \neq 0.467$

When we conduct a proportion test we need to use the z statistic, and the is given by:

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

The One-Sample Proportion Test is used to assess whether a population proportion $\hat p$ is significantly different from a hypothesized value $p_o$ .

Calculate the statistic

Since we have all the info requires we can replace in formula (1) like this:

$z=\frac{0.42 -0.467}{\sqrt{\frac{0.467(1-0.467)}{150}}}=-1.154$

Statistical decision

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.

The significance level provided $\alpha=0.05$ . The next step would be calculate the p value for this test.

Since is a bilateral test the p value would be:

$p_v =2*P(z$

So the p value obtained was a very high value and using the significance level given $\alpha=0.05$ we have $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the proportion of girls born is not significantly different from 0.467

3 0

3 years ago