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3 years ago

Help. Me. Please. I don't understand it!! Asap please!

Mathematics

2 answers:

vladimir1956 [14]3 years ago

6 0

Answer:

He has 286 dollars in his bank account at the end of the month.

Step-by-step explanation:

Deposit Meaning- basically to add

Withdrawal- basically to subtract

154+237+45= 436

-125-25= -150

436-150= 286

Bond [772]3 years ago

6 0

I think the last answer is wrong. No offense. If he starts with $154 and I deposited 237 bring it Up to 391 then he withdraws $25 so you subtract that and you get 366 then he withdraws another 125 bringing it down to 241 then he deposits $45 bringing him up to 286. 286 is your answer and make sure you put dollars as the unit. Hope this helps!

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PLZ dont ignore me! I need help on this :(

uysha [10]

Answer: It’s the second one!

Explanation: The dot is open and pointing to the side of the numbers greater than -3. It has to be an open dot because it is “greater than -3” not “greater than or equal to-3”! :)

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3 years ago

The price of a community pool membership has a one-time sign-up fee and a monthly fee. The price can be modeled by the function

Misha Larkins [42]

Your answer is option 1.

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2 years ago

Solve for y, z, x or t:

Rus_ich [418]

Answer:

Step-by-step explanation:

$if~it ~is ~y+\frac{y}{a} =b\\then~\frac{ay+y}{a} =b\\ay+y=ab\\ y(a+1)=ab\\ y=\frac{ab}{a+1} \\$

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2 years ago

Which of the following would be a correct interpretation of a 99% confidence interval such as 4.1 less than muless than5.6? Ch

Harrizon [31]

Answer: A. We are 99% confident that the interval from 4.1 to 5.6 actually does contain the true value of mu.

Step-by-step explanation:

The correct interpretation of a 99% confidence interval is that we are 99% confident that the true population mean falls in it.

Given 99% confidence interval : $4.1$

Then, the correct interpretation : We are 99% confident that the interval from 4.1 to 5.6 actually does contain the true value of mu.

8 0

2 years ago

A restaurant chain has two locations in a medium-sized town and, believing that it has oversaturated the market for its food, is

scoray [572]

Answer:

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

Step-by-step explanation:

Data given and notation

$\bar X_{1}=360000$ represent the mean for the downtown restaurant

$\bar X_{2}=340000$ represent the mean for the freeway restaurant

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

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

$n_{1}=36$ sample size for the downtown restaurant

$n_{2}=36$ sample size for the freeway restaurant

t would represent the statistic (variable of interest)

$\alpha=0.05$ 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:

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

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.

The best option would be:

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

5 0

2 years ago