I need help with this

6

qwelly [4]

28% of 1500 is 420.

so add it to 1500 and you’ll get 1920

7 0

3 years ago

Please please factor fully.

Alekssandra [29.7K]

Answer: My bad if I butchered any answers. It was hard to read tbh

Step-by-step explanation:

D.) You can only factor out an x here. So you'd get: $x(x^2-2x+4)$

E.) You can factor out a 3x^2y to get: $3x^2y(2-x)$

F.) You can factor out a 3xy to get: $3xy(y+3x-2)$

G.) You can factor out a -4ab to get: $-4ab(a-4b+2)$

4 0

3 years ago

Read 2 more answers

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

Your paycheck this week is $138.25. You put the $42.54 in a savings account that pays an interest rate of 6%, compounded yearly.

ASHA 777 [7]

Amount of money you will have = 42.54(1+6/100)^5=56.9

3 0

3 years ago

Jenifer had 30$ to spend on her self she spent 1/5 Of the money in a sandwich 1/6 for a ticket to a museum and 1/2 Of It On a Bo

denis-greek [22]

Your answer is $29.13 I hope this helped, I apologize for the late response!

3 0

3 years ago