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2 years ago

8.

Mathematics

2 answers:

Sholpan [36]2 years ago

7 0

D because of the distribution property

zalisa [80]2 years ago

4 0

Answer:

4y + 12 is equivalent to D

Step-by-step explanation:

4(y+3)

4(y) = 4y

4(3) = 12

so it's D

Hope this worked !

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The following 5 questions are based on this information: An economist claims that average weekly food expenditure of households

liq [111]

Answer:

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

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

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

b) 2.70

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

b. 1.85

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

b. 0.03

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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3 years ago

The size of a typical virus is about 0.000008 centimeters, which can be

Reptile [31]

I wanna say it’s d because of the way it’s set up, i hope this helps if not i’m really really sorry

5 0

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Valentin [98]

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. The concert tickets were $57 plus two souvenirs for your friend and yourself. You spent $94

Lena [83]

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4 0

3 years ago

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A bank in the Bay area is considering a training program for its staff. The probability that a new training program will increas

WITCHER [35]

Answer:

$P(B' \cup A') = P((A \cap B)') = 1-P(A \cap B)= 1-0.32=0.68$

See explanation below.

Step-by-step explanation:

For this case we define first some notation:

A= A new training program will increase customer satisfaction ratings

B= The training program can be kept within the original budget allocation

And for these two events we have defined the following probabilities

$P(A) = 0.8, P(B) = 0.2$

We are assuming that the two events are independent so then we have the following propert:

$P(A \cap B ) = P(A) * P(B)$

And we want to find the probability that the cost of the training program is not kept within budget or the training program will not increase the customer ratings so then if we use symbols we want to find:

$P(B' \cup A')$