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

The beginning balance of Keyshawn's savings account for the month of September was $3900, and it remained this way for the first

musickatia [10]

Simple or compound interest

8 0

3 years ago

In a study of 799 randomly selected medical malpractice lawsuits, it was found that 499 of them were dropped or dismissed. Use

Mademuasel [1]

Answer:

Claim is true that most medical malpractice lawsuits are dropped or dismissed.

Step-by-step explanation:

Given :In a study of 799 randomly selected medical malpractice lawsuits, it was found that 499 of them were dropped or dismissed.

To Find : Use a 0.05 significance level to test the claim that most medical malpractice lawsuits are dropped or dismissed.

Solution:

n = 799

x = 499

We will use one sample proportion test

$\widehat{p}=\frac{x}{n}\\\widehat{p}=\frac{499 }{ 799}\\\widehat{p}=0.6245$

We are given that the claim is most medical malpractice lawsuits are dropped or dismissed.

Probability of dropped or dismissed is 1/2

$H_0:p=0.5\\H_a:p>0.5$

Formula of test statistic = $\frac{\widehat{p}-p}{\sqrt{\frac{p(1-p)}{n}}}$

= $\frac{0.6245-0.5}{\sqrt{\frac{0.5(1-0.5)}{799}}}$

= 7.038

p value (z>7.038) is 0

α =0.05

So, p value < α

So, we reject the null hypothesis

So,claim is true

Hence most medical malpractice lawsuits are dropped or dismissed.

6 0

3 years ago