Solve 3u + 5u+ 30 = 12 for u. Write your answer as a fraction,

The Barnes family has a coupon for 10% off their dinner. If their bill comes to $82.50 and they wish to tip 15%, what will they

satela [25.4K]

Answer:

Total cost= $86.25

Step-by-step explanation:

Giving the following information:

Discount= 10%

Nill= $82.5

Tip= 15%

First, we need to calculate the discounted price:

Discount price= 82.5 / 1.1= $75

Now, the tip:

Tip= 75*0.15= $11.25

Finally, the total cost:

Total cost= 75 + 11.25

Total cost= $86.25

3 0

4 years ago

Two college roommates have each committed to donating to charity each week for the next year. The roommates’ weekly incomes are

schepotkina [342]

Answer:

C. 5 weeks.

Step-by-step explanation:

In this question we have a random variable that is equal to the sum of two normal-distributed random variables.

If we have two random variables X and Y, both normally distributed, the sum will have this properties:

$S=X+Y\\\\\ \mu_S=\mu_X+\mu_Y=30+60=90\\\\\sigma_S=\sqrt{\sigma_X^2+\sigma_Y^2}=\sqrt{10^2+20^2}=\sqrt{100+400}=\sqrt{500}=22.36$

To calculate the expected weeks that the donation exceeds $120, first we can calculate the probability of S>120:

$z=\frac{S-\mu_S}{\sigma_S} =\frac{120-90}{22.36}=\frac{30}{22.36}=1.34\\\\P(S>120)=P(z>1.34)=0.09012$

The expected weeks can be calculated as the product of the number of weeks in the year (52) and this probability:

$E=\#weeks*P(S>120)=52*0.09012=4.68$

The nearest answer is C. 5 weeks.

6 0

3 years ago

WILL GIBE BRAINLIEST ASAP

Romashka [77]

Answer:

A and D have whole grid squares that are the same size and aren't over lapping

C has overlapping grid squares making it hard to count

B can't be used to find area because some of the grid squares are different sizes

You still could use B because four of the smaller squares seems to be equivalent to one of the larger squares

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

A survey conducted in July 2015 asked a random sample of American adults whether they had ever used online dating (either an onl

Natasha2012 [34]

Answer:

The 99% confidence interval is (0.0787, 0.1613).

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of \pi, and a confidence interval 1-\alpha, we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

Z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$ .

For this problem, we have that:

The survey included 411 adults between the ages of 55 and 64, and 50 of them said that they had used online dating. This means that $n = 411, \pi = \frac{50}{411} = 0.12$