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3 years ago

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Inez earns $160 per month by babysitting she save 25% of our monthly salary average salary increases by 10% how much will she no

w save each month

1 answer:

Liula [17]3 years ago

8 0

She will now save $56 every month. Please Mark Brainliest!!!

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Which equation could be used to find the value of x in the triangle below?

Licemer1 [7]

2x+4x+10+6x-10=180 because all the angles of a triangle equal to 180 so u add up all the angles and set to 180. and x equals 15.

5 0

3 years ago

I need to factor 5a^2 + 7a + 6b^2 - 4b but I can't figure it out.......could it possibly be prime?

DiKsa [7]

I kind of forgot how to do it but here is what I know so far : first you separate the two parts into (5a^2 + 7a) + (6b^2-4b). This process is called grouping. Now you find the greatest common factor in both of them which later your equation will be a(5a + 7) +2b(3b - 2). That is all I can remember so hope that helped

7 0

3 years ago

A simple random sample of 20 days in which Parsnip ate seeds was selected, and the mean amount of time it took him to eat the se

bearhunter [10]

Answer:

$(14.5-13.9) -1.69 \sqrt{\frac{3.98^2}{20}+ \frac{4.03^2}{17}} = -1.63$

$(14.5-13.9) +1.69 \sqrt{\frac{3.98^2}{20}+ \frac{4.03^2}{17}} = 2.83$

And the confidence interval for this case $-1.63 \leq \mu_1 -\mu_2 \leq 2.83$

Step-by-step explanation:

We know the following info from the problem

$\bar X_1 = 14.5$ sample mean for the group 1

$s_1 = 3.98$ the standard deviation for the group 1

$n_1= 20$ the sample size for group 1

$\bar X_2 = 13.9$ sample mean for the group 2

$s_1 = 4.03$ the standard deviation for the group 2

$n_2= 17$ the sample size for group 2

We have all the conditions satisifed since we have random samples.

We want to construct a confidence interval for the true difference of means and the correct formula for this case is:

$(\bar X_1 -\bar X_2) \pm t_{\alpha/2}\sqrt{\frac{s^2_1}{n_1} +\frac{s^2_2}{n_2}}$

The degrees of freedom are given :

$df = n_1 +n_2- 2 = 20+17-2=35$

The confidence level is 0.9 or 90% and the significance level is $\alpha=1-0.9=0.1$ and $\alpha/2 = 0.05$ and the critical value for this case is:

$t_{\alpha/2} = 1.69$

And replacing the info given we got:

$(14.5-13.9) -1.69 \sqrt{\frac{3.98^2}{20}+ \frac{4.03^2}{17}} = -1.63$

$(14.5-13.9) +1.69 \sqrt{\frac{3.98^2}{20}+ \frac{4.03^2}{17}} = 2.83$

And the confidence interval for this case $-1.63 \leq \mu_1 -\mu_2 \leq 2.83$

5 0

3 years ago

The dimensions of a pizza pan in the shape of a cylinder are shown in the diagram.

Alex_Xolod [135]

Answer:

b

Step-by-step explanation:

7 0

3 years ago

A salesperson works 40 hours per week at a job where she has two options for being paid. Option A is an hourly wage of 23$. Opt

Gwar [14]

Step-by-step explanation:

A salesperson works 40 hours per week at a job where she has two options for being paid. Option A is an hourly wage of 23$. Option B is a commission rate of 4% on weekly sales. How much does she need to sell this week to earn the same amount with the two options?
A salesperson works 40 hours per week at a job where she has two options for being paid. Option A is an hourly wage of 23$. Option B is a commission rate of 4% on weekly sales. How much does she need to sell this week to earn the same amount with the two options?
A salesperson works 40 hours per week at a job where she has two options for being paid. Option A is an hourly wage of 23$. Option B is a commission rate of 4% on weekly sales. How much does she need to sell this week to earn the same amount with the two options?

7 0

3 years ago

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