Name 5200 using thousands and hundreds

Find the slope of the line that contains the points named.

8090 [49]

Answer:

M= 0

Step-by-step explanation:

the ys are the same

6 0

3 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

If one of the factors of r ^ 2 + 21r + 98 is r + 7 , what is the other factor?

irina [24]

Answer:

i think factor the expression (r+7)(r+14) is answer

Step-by-step explanation:

8 0

2 years ago

Read 2 more answers

Assume that 300 births are randomly selected and 5 of the births are girls. Use subjective judgment to describe the number of gi

murzikaleks [220]

Answer:

Following are the response to the given choice.

Step-by-step explanation:

Please find the complete question in the attached file.

Subjective opinion = Question of opinion.

Therefore this requires only just opinion and we don't have to do any actual calculations.

Does this seem like a great number of girls, a little number of girls, or a decent number of girls to you but if 1,300 babies were born, 5 of whom were females?

This is a small number of beautiful gals, in my honest opinion. We anticipate boys and girls to be produced about the very same frequency, thus I expect some half of them to be females if there are 1 300 newborns. You should have roughly 650 girls if 50% of the infants are girls, but now we only have five. That appears to me to be considerably low. which is your own opinion.