A home with an area of 2,200 square feet has 660 square feet of wood flooring. What percent of the home's flooring is wood

Setler79 [48]

The answer is 20 because 292+20=312

8 0

3 years ago

The average height of middle school students at Seton Grove School is found to be 5.25 feet. Which sample makes this generalizat

Ede4ka [16]

Answer:

every other student who enters the school

Step-by-step explanation:

It is given that the middle school students have an average height of 5.25 feet. The students belongs to Seton Grove School.

It means that every student who comes to this school or enters the school has a height that is around 5 feet or more.

Thus we can say that every other student who enters the school has a average height of 5.25 feet.

This sample makes the generalization about the heights of the school students valid.

6 0

3 years ago

A plane flies from Houston to New York City in 4 hours the distance I traveled was 1627 miles what is the average speed of the p

Nikolay [14]

1627 / 4 = 406.75 miles per hr

5 0

4 years ago

Greatest common factor for 8x and 40

CaHeK987 [17]

The GCF of 8x and 40 is 8

40 | 2
20 | 2
10 | 2
5 | 5
1

8 | 2
4 | 2
2 | 2
1

40 = 2³ * 5
8 = 2³

We take only common factors with lowest exponent.

So 2³ = 8

4 0

3 years ago

Integrated circuits consist of electric channels that are etched onto silicon wafers. A certain proportion of circuits are defec

boyakko [2]

Answer:

$z=\frac{0.033 -0.05}{\sqrt{\frac{0.05(1-0.05)}{1000}}}=-2.467$

$p_v =P(Z>-2.467)=0.0068$

So the p value obtained was a very low value and using the significance level assumed for example $\alpha=0.05$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of circuits that show evidence of undercutting is significantly less than 0.05.

Step-by-step explanation:

1) Data given and notation

n=1000 represent the random sample taken

X=33 represent the number of circuits that show evidence of undercutting

$\hat p=\frac{33}{1000}=0.033$ estimated proportion of circuits that show evidence of undercutting

$p_o=0.05$ is the value that we want to test

$\alpha$ represent the significance level

z would represent the statistic (variable of interest)

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

2) Concepts and formulas to use

We need to conduct a hypothesis in order to test the claim that they reduce the rate of undercutting to less than 5%.:

Null hypothesis: $p\geq 0.05$

Alternative hypothesis: $p < 0.05$

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 $p_o$ .

3) Calculate the statistic

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

$z=\frac{0.033 -0.05}{\sqrt{\frac{0.05(1-0.05)}{1000}}}=-2.467$

4) 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 next step would be calculate the p value for this test.

Since is a left tailed test the p value would be:

$p_v =P(Z>-2.467)=0.0068$

So the p value obtained was a very low value and using the significance level assumed for example $\alpha=0.05$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of circuits that show evidence of undercutting is significantly less than 0.05.

4 0

3 years ago