What is the area of the circle below, in terms of ?90 metersO 457O 907O 1807O 81007134

A line has slope −34 and y–intercept 5. Which answer is the equation of the line? y=−5x+34 y=−34x+5 y=34x−5 y=5x−34?

Dafna1 [17]

The answer would be the first equation!!!!

3 0

3 years ago

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You bought jeans last week for $55. Today you see that the jeans are on

PtichkaEL [24]

Yes I think it’s that

7 0

3 years ago

A photoconductor film is manufactured at a nominal thickness of 25 mils. The product engineer wishes to increase the mean speed

Natali [406]

Answer:

$t=\frac{1.17-1.04}{\sqrt{\frac{0.11^2}{8}+\frac{0.09^2}{8}}}}=2.587$

$df=n_{1}+n_{2}-2=8+8-2=14$

Since is a one sided test the p value would be:

$p_v =P(t_{(14)}>2.587)=0.0108$

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, we have enough evidence to reject the null hypothesis on this case and the 25 mil film have a mean greater than the 20 mil film so then the claim is not appropiate

Step-by-step explanation:

Data given and notation

$\bar X_{1}=1.17$ represent the mean for the sample 1 (25 mil film)

$\bar X_{2}=1.04$ represent the mean for the sample 2 (20 mil film)

$s_{1}=0.11$ represent the sample standard deviation for the sample 1

$s_{2}=0.09$ represent the sample standard deviation for the sample 2

$n_{1}=8$ sample size selected for 1

$n_{2}=8$ sample size selected for 2

$\alpha=0.05$ represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

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

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if reducing the film thickness increases the mean speed of the film, the system of hypothesis would be:

Null hypothesis: $\mu_{1} \leq \mu_{2}$

Alternative hypothesis: $\mu_{1} > \mu_{2}$

If we analyze the size for the samples both are less than 30 so for this case is better apply a t test to compare means, and the statistic is given by:

$t=\frac{\bar X_{1}-\bar X_{2}}{\sqrt{\frac{s^2_{1}}{n_{1}}+\frac{s^2_{2}}{n_{2}}}}$ (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other".

Calculate the statistic

We can replace in formula (1) the info given like this:

$t=\frac{1.17-1.04}{\sqrt{\frac{0.11^2}{8}+\frac{0.09^2}{8}}}}=2.587$

P-value

The first step is calculate the degrees of freedom, on this case:

$df=n_{1}+n_{2}-2=8+8-2=14$

Since is a one sided test the p value would be:

$p_v =P(t_{(14)}>2.587)=0.0108$

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, we have enough evidence to reject the null hypothesis on this case and the 25 mil film have a mean greater than the 20 mil film so then the claim is not appropiate

4 0

2 years ago

As part of the Pew Internet and American Life Project, researchers conducted two surveys in late 2009. The first survey asked a

REY [17]

Answer:

The 95% confidence interval for the difference between the proportion of all U.S. teens and adults who use social networking sites is (0.223, 0.297). This means that we are 95% sure that the true difference of the proportion is in this interval, between 0.223 and 0.297.

Step-by-step explanation:

Before building the confidence interval we need to understand the central limit theorem and the subtraction of normal variables.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean $\mu = p$ and standard deviation $s = \sqrt{\frac{p(1-p)}{n}}$

Subtraction between normal variables:

When two normal variables are subtracted, the mean is the difference of the means, while the standard deviation is the square root of the sum of the variances.

Sample of 800 teens. 73% said that they use social networking sites.

This means that:

$p_T = 0.73, s_T = \sqrt{\frac{0.73*0.27}{800}} = 0.0157$

Sample of 2253 adults. 47% said that they use social networking sites.

This means that:

$p_A = 0.47,s_A = \sqrt{\frac{0.47*0.53}{2253}} = 0.0105$

Distribution of the difference:

$p = p_T - p_A = 0.73 - 0.47 = 0.26$

$s = \sqrt{s_T^2+s_A^2} = \sqrt{0.0157^2+0.0105^2} = 0.019$

Confidence interval:

Is given by:

$p \pm zs$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$ .

95% confidence level

So $\alpha = 0.05$ , z is the value of Z that has a pvalue of $1 - \frac{0.05}{2} = 0.975$ , so $Z = 1.96$ .

Lower bound:

$p - 1.96s = 0.26 - 1.96*0.019 = 0.223$

Upper bound:

$p + 1.96s = 0.26 + 1.96*0.019 = 0.297$

The 95% confidence interval for the difference between the proportion of all U.S. teens and adults who use social networking sites is (0.223, 0.297). This means that we are 95% sure that the true difference of the proportion is in this interval, between 0.223 and 0.297.

3 0

2 years ago

What is the completely factored form of 8x2 - 50?

bogdanovich [222]

Answer:

2(2x-5)(2x+5)

Step-by-step explanation:

I think you meant to say 8x^2-50. If so then factoring this down should be easy. Since there is no x value in the middle the equation will have a positive and a negative number. This is also a perfect square therefore, this factors down to: 2(2x-5)(2x+5).

3 0

2 years ago

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