I will give Brainiest

A study was recently conducted to estimate the mean cholesterol for adult males over the age of 55 years. From a random sample o

Komok [63]

Answer: Correct critical value = 2.2622

Step-by-step explanation:

Confidence interval for population mean when population standard deviation is unknown:

$\overline{x}\pm t_{\alpha/2}\dfrac{s}{\sqrt{n}}$ , where $\overline{x}$ = sample mean, n= sample size, s= sample standard deviation, $t_{\alpha/2}$ = two-tailed t value.

As per given: n= 10

degree of freedom : df = n-1=9

$\overline{x}=242.6\\\\s=73.33\\\\\alpha=5\%=0.05$

Critical t-value : $t_{\alpha/2,df}=t_{0.05/2,9}=t_{0.025,9}=2.2622$

So, the 95 percent confidence interval estimate for the mean :

$242.6\pm (2.2622)\dfrac{73.33}{\sqrt{10}}\\\\=242.6\pm (2.2622)(23.19)\\\\\approx242.6\pm52.46\\\\=(242.6-52.46,\ 242.6+52.46)\\\\=(190.14,\ 295.06)$

The 95 percent confidence interval estimate for the mean:(190.14, 295.06)

8 0

3 years ago

tony said that the school playground was 3,600 inches long. Explain how you could write 3,600 inches in a way that might be easi

kirill115 [55]

Well I would change it to feet to make it easier. There are 12 I inches in a foot so I would divide by 12. 3,600÷12=300 so the playground would be 300feet long. Which is easier than 3,600 inches. ANSWER:300 feet

7 0

3 years ago

Solve each equation.<br> Show all steps<br> 4) -8(-6-5k)=-232

Lubov Fominskaja [6]

the answer is k=4.6. hope this helps

6 0

3 years ago

Let ρ = x3 + xe−x for x ∈ (0, 1), compute the center of mass.

hram777 [196]

The center of mass is mathematically given as

$\bar{x}=\left(\frac{44 e-100}{25 e-40}\right)\end{aligned}$

<h3>What is the center of mass.?</h3>

Determine the center of mass in one dimension:

Represent the masses at the respective distances.

$\begin{|c|c|} Masses \ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ Located at \\\rho=x^{3}+x \cdot e^{-x} & \ \ \ \ x \in(0,1)$ \\\end$

We calculate the total mass of the system.

$\begin{aligned}m &=\int_{0}^{1} \rho \cdot d x \\& m =\int_{0}^{1}\left(x^{3}+x \cdot e^{-x}\right) \cdot d x \\&m =\left|\frac{x^{4}}{4}-(x+1) e^{-x}\right|_{0}^{1} \\&m =\left(\frac{5}{4}-\frac{2}{e}\right)\end{aligned}$

Step 03: Calculate the moment of the system.

$\begin{aligned}M &=\int_{0}^{1}(\rho \cdot x) \cdot d x \\& M=\int_{0}^{1}\left(x^{4}+x^{2} \cdot e^{-x}\right) \cdot d x \\&M =\left|\frac{x^{5}}{5}-\left(x^{2}-2 x+2\right) \cdot e^{-x}\right|_{0}^{1} \\&M=\left(\frac{11}{5}-\frac{5}{e}\right)\end{aligned}$

we calculate the center of mass.

$\begin{aligned}\bar{x} &=\left(\frac{M}{m}\right) \\& \bar{x}=\left\{\left(\frac{\left.11-\frac{5}{5}\right)}{\left(\frac{5}{4}-\frac{2}{e}\right)}\right\}\right.\\& \bar{x}=\left(\frac{11 e-25}{5 e}\right) \cdot\left(\frac{4 e}{5 e-8}\right) \\&\bar{x}=\left(\frac{44 e-100}{25 e-40}\right)\end{aligned}$