4x+5+4x+5+4x+5+4x+5+4x+5 solve for x

Help me please!! I don't know how to do this

Nesterboy [21]

Answer:

probably the last answer

so D

5 0

2 years ago

If you answer I will give your brainliest just plz help me ok so which answer choice is correct?

Ber [7]

Answer:

it would be A and C

Step-by-step explanation:

I used photo math and it used C as its answer, but answer A simplifies to answer C therefore meaning they are both your answers! I hope this helped!

8 0

2 years ago

A battalion of 520 soldiers goes through 16,380 pounds of rations in 3 weeks. If 780 more soldiers join the battalion, how many

Scorpion4ik [409]

They would need 34.125 tons of rations.

In order to find this, we first need to see how many pounds of rations a single soldier eats in a week. To do this, we take the total eaten in 3 weeks and divide by 3. Then we divide by the number of soldiers.

16,380/3 weeks= 5460 lbs per week

5460/520 soldiers = 10.5 lbs per soldier per week

Now we look to see how many soldiers we will have in total after adding. There are 520 to start and we add 780 to get 1300 total. Next we multiply that by the total per soldier per week.

10.5lbs per soldier per week * 1300 soldiers = 13,650lbs per week

Then we have to multiply by the 5 weeks that battalion will be there.

13,650lbs * 5 weeks = 68,250lbs of rations or 24.125 tons.

5 0

3 years ago

According to the Knot, 22% of couples meet online. Assume the sampling distribution of p follows a normal distribution and answe

Ann [662]

Using the <em>normal distribution and the central limit theorem</em>, we have that:

a) The sampling distribution is approximately normal, with mean 0.22 and standard error 0.0338.

b) There is a 0.1867 = 18.67% probability that in a random sample of 150 couples more than 25% met online.

c) There is a 0.2584 = 25.84% probability that in a random sample of 150 couples between 15% and 20% met online.

<h3>Normal Probability Distribution</h3>

In a normal distribution with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

It measures how many standard deviations the measure is from the mean.

After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X.

By the Central Limit Theorem, for a proportion p in a sample of size n, the sampling distribution of sample proportion is approximately normal with mean $\mu = p$ and standard deviation $s = \sqrt{\frac{p(1 - p)}{n}}$ , as long as $np \geq 10$ and $n(1 - p) \geq 10$ .