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3 years ago

Marie factored 11x^3y^5 as (8x^3)(3y^5). Stanley factored 11x^3y^5 as (11xy)(x^2y^4). Which of them factored 11x^3y^5 correctly?

Mathematics

1 answer:

tangare [24]3 years ago

6 0

<u>Answer:</u>

Stanley factored it correctly.

<u>Step-by-step explanation:</u>

$11x^3y^5$

We are given that Marie and Staley factored the above expression as follows and we are to determine who factored it correctly:

Marie: $11x^3y^5$ ---> $(8x^3)(3y^5)$

$(8x^3)(3y^5)$ = $24x^3y^3$ so this is wrong.

Stanley: $11x^3y^5$ ---> $(11xy)(x^2y^4)$

$( 1 1 x y ) ( x ^ 2 y ^ 4 )$ = $1 1 x ^ 3 y ^ 5$

Therefore, Stanley factored it correctly.

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Answer:

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Alternative hypothesis: $\mu_{1}-\mu_{2}>0$

$SE=\sqrt{\frac{12.5^2}{35}+\frac{9.25^2}{30}}=2.705$

b) 2.70

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a. We can reject H(0) in favor of H(a)

Step-by-step explanation:

Data given and notation

$\bar X_{1}=164$ represent the mean for the sample 1

$\bar X_{2}=159$ represent the mean for the sample 2

$\sigma_{1}=12.5$ represent the population standard deviation for the sample 1

$s_{2}=9.25$ represent the population standard deviation for the sample B2

$n_{1}=35$ sample size selected 1

$n_{2}=30$ sample size selected 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 the expenditure of households in City 1 is more than that of households in City 2, the system of hypothesis would be:

Null hypothesis: $\mu_{1}-\mu_{0}\leq 0$

Alternative hypothesis: $\mu_{1}-\mu_{2}>0$

We know the population deviations, so for this case is better apply a z test to compare means, and the statistic is given by:

$z=\frac{(\bar X_{1}-\bar X_{2})-0}{\sqrt{\frac{\sigma^2_{1}}{n_{1}}+\frac{\sigma^2_{2}}{n_{2}}}}$ (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".

Standard error

The standard error on this case is given by:

$SE=\sqrt{\frac{s^2_{1}}{n_{1}}+\frac{s^2_{2}}{n_{2}}}$

Replacing the values that we have we got:

$SE=\sqrt{\frac{12.5^2}{35}+\frac{9.25^2}{30}}=2.705$

b. 2.70

Calculate the statistic

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

$t=\frac{(164-159)-0}{\sqrt{\frac{12.5^2}{35}+\frac{9.25^2}{30}}}=1.850$

P-value

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

$p_v =P(Z>1.85)=0.032$

b. 0.03

Conclusion

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.

a. We can reject H(0) in favor of H(a)

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Answer:

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