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oksano4ka [1.4K]
3 years ago
14

-) 2x(4+4-49-1-2)x2 =

Mathematics
1 answer:
WITCHER [35]3 years ago
3 0

Answer:

-88x^3

Step-by-step explanation:

1) distribute the "2x"

(8x+8x-98x-2x-4x)x^2

a) add all like terms in parantheses

(-88x)x^2

b) multiply -88x and x^2

-88x*x^2=-88x^3

hope this helps!

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Jillian walked 0.5 miles before she started jogging at an average pace of 5 miles per hour. The equation d = 0.5 + 5t can be use
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Yes, since the test statistic value is greater than the critical value.

Step-by-step explanation:

Data given and notation

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s_{1}=50000 represent the sample standard deviation for downtown

s_{2}=40000 represent the sample standard deviation for the freeway

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\alpha=0.05 significance level provided

Develop the null and alternative hypotheses for this study?

We need to conduct a hypothesis in order to check if the true mean of revenue for downtown is higher than for freeway restaurant, the system of hypothesis would be:

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

Since we don't know the population deviations for each group, 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.

Determine the critical value.

Based on the significance level\alpha=0.05 and \alpha/2=0.025 we can find the critical values from the t distribution dith degrees of freedom df=36+36-2=55+88-2=70, we are looking for values that accumulates 0.025 of the area on the right tail on the t distribution.

For this case the value is t_{1-\alpha/2}=1.66

Calculate the value of the test statistic for this hypothesis testing.

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

t=\frac{360000-340000}{\sqrt{\frac{50000^2}{36}+\frac{40000^2}{36}}}}=1.874

What is the p-value for this hypothesis test?

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

p_v =P(t_{70}>1.874)=0.033

Based on the p-value, what is your conclusion?

Comparing the p value with the significance level given \alpha=0.05 we see that p_v so we can conclude that we reject the null hypothesis, and the true mean for the downtown revenue restaurant seems higher than the true mean revenue for the freeway restaurant.

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Yes, since the test statistic value is greater than the critical value.

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Answer:

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Step-by-step explanation:

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