Someone plz help me

jim is planning a party. he spends $150 on a room for the party. he also spends $16.50 per person for food. jim finds the total

meriva

$577.50is the amount he paid.

8 0

3 years ago

The confidence interval based on the two-sample design is wider because there is considerable variation in mileage between the c

vovikov84 [41]

Answer:

The confidence interval based on the paired design is wider because there is little variation in mileage between the cars.

Step-by-step explanation:

The sample size randomly collected is matched sample and since the confidence interval is based on the paired design is large and wide. There is small variation between the mileage of the two cars indicating the cars have mileage based on the fuel.

5 0

3 years ago

Sarah claims that the thickness of the spearmint gum she produces is 7.5 one-hundredths of an inch. A quality control specialist

satela [25.4K]

Answer:

$t=\frac{7.55-7.5}{\frac{0.103}{\sqrt{10}}}=1.539$

$p_v =2*P(t_{(9)}>1.539)=0.158$

If we compare the p value and the significance level assumed $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to fail reject the null hypothesis.

Step-by-step explanation:

Data given and notation

Data: 7.65, 7.60, 7.65, 7.7, 7.55, 7.55, 7.4, 7.4, 7.5, 7.5

We can begin calculating the sample mean given by:

$\bar X = \frac{\sum_{i=1}^n X_i}{n}$

And the sample deviation given by:

$s=\sart{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}$

And we got:

$\bar X=7.55$ represent the sample mean

$s=0.103$ represent the sample standard deviation

$n=10$ sample size

$\mu_o =7.5$ represent the value that we want to test

$\alpha=0.05$ 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 the mean is equal to 7.5 or no, the system of hypothesis would be:

Null hypothesis: $\mu = 7.5$

Alternative hypothesis: $\mu \neq 7.5$

If we analyze the size for the sample is < 30 and we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

Calculate the statistic

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

$t=\frac{7.55-7.5}{\frac{0.103}{\sqrt{10}}}=1.539$

P-value

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

$df=n-1=10-1=9$

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

$p_v =2*P(t_{(9)}>1.539)=0.158$

Conclusion

If we compare the p value and the significance level assumed $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to fail reject the null hypothesis.

3 0

4 years ago

Pre Calculus Help! Will give Brainliest!

Sveta_85 [38]

It's a quotient identity

6 0

3 years ago

Choose the correct simplification of (9xy^5)^2.

Grace [21]

Answer:

D) 81x²y¹⁰

Step-by-step explanation:

(9xy⁵)²

Because all of 9xy⁵ are in the parentheses, each component is raised to the 2nd degree.

(9)² = 81

(x)² = x²

(y⁵)² = y¹⁰ -- When you raise an exponent to another exponent, you multiply the exponents.

Now combine the components.

D) 81x²y¹⁰

5 0

3 years ago