$\begin{gathered} \sqrt[25]{1580} \\ 1\text{ and 15 cannot divide 25 therefore, we use the first three numbers to divide 25} \\ \frac{158}{25}\text{ = }6\text{ remainder 8} \\ \text{You have to carry that remainder which is 8 and combine it with the 0 in 1580.} \\ \frac{80}{25}\text{ = }3\text{ remainder 5} \\ The\text{ remainder 5 will also divide 25} \\ \frac{5}{25}=0.2 \\ The\text{ quotient values are 63 + 0.2 = 63.2} \end{gathered}$

4 0

1 year ago

Please help me ASAP

vovikov84 [41]

1018262910101101 explanation:

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