BRAINLIEST ANSWER TO FIRST ANSWERER

Shane went bowling with his friends. The bowling alley charges $20 for an hour of group bowling plus a fee to rent each pair of

ValentinkaMS [17]

Answer:

4x+20=31

Step-by-step explanation:

So one hour of bowling is 20 dollars and they bowled for one hour hence the 1x20 which is equal to 20. That is the base coast we add to the variable of x that is the cost of renting one pair of shoes. And since they rented 4 pairs of shoes you multiply x by 4 to get 4x. Add the 20 dollars they paid to bowl and it is 4x+20. They paid 31 dollars in total so equal the equation to 31. 4x+20=31. Solution is $2.75 per pair of shoes. (If you want an explanation as to how to get the answer, just ask

4 0

2 years ago

Dylan Rieder is a statistics student investigating whether athletes have better balance than non-athletes for a thesis project.

Kobotan [32]

Answer:

$t=\frac{(3.7-4.1)-0}{\sqrt{\frac{1.1^2}{32}+\frac{1.3^2}{45}}}}=-1.457$

$p_v =P(t_{75}$

Comparing the p value with a significance level for example $\alpha=0.01$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can't say that the population mean for the athletes is significantly lower than the population mean for non athletes.

Step-by-step explanation:

Data given and notation

$\bar X_{A}=3.7$ represent the mean for athletes

$\bar X_{NA}=4.1$ represent the mean for non athletes

$s_{A}=1.1$ represent the sample standard deviation for athletes

$s_{NA}=1.3$ represent the sample standard deviation for non athletes

$n_{A}=32$ sample size for the group 2

$n_{NA}=45$ sample size for the group 2

$\alpha=0.01$ Significance level provided

t would represent the statistic (variable of interest)

Concepts and formulas to use

We need to conduct a hypothesis in order to check if the population mean for athletes is lower than the population mean for non athletes, the system of hypothesis would be:

Null hypothesis: $\mu_{A}-\mu_{NA}\geq 0$

Alternative hypothesis: $\mu_{A} - \mu_{NA}< 0$

We don't have the population standard deviation's, we can apply a t test to compare means, and the statistic is given by:

$t=\frac{(\bar X_{A}-\bar X_{NA})-\Delta}{\sqrt{\frac{s^2_{A}}{n_{A}}+\frac{s^2_{NA}}{n_{NA}}}}$ (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.

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

$t=\frac{(3.7-4.1)-0}{\sqrt{\frac{1.1^2}{32}+\frac{1.3^2}{45}}}}=-1.457$

P value

We need to find first the degrees of freedom given by:

$df=n_A +n_{NA}-2=32+45-2=75$

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

$p_v =P(t_{75}$

Comparing the p value with a significance level for example $\alpha=0.01$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can't say that the population mean for the athletes is significantly lower than the population mean for non athletes.

5 0

3 years ago

How do I find the correct answer

lys-0071 [83]

I think it is a because it shows us the points that the students made. I also think it is d. But I think it is mainly a.

-Dhruva;)

3 0

3 years ago

I need help on this one please and just click one the photo for the question please

Mnenie [13.5K]

The answer would be c

8 0