$\overline{x}\pm t_{\alpha/2}\dfrac{s}{\sqrt{n}}$ , 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

2 years ago

I need help finding ac, cb, and ab

Taya2010 [7]

AC is 2x+1
CB is 3x-4
AB= 2x+1+3x-4

6 0

3 years ago

28. What is the solution to 12-(2x+6) = -4x +10 +2(x-8) *

S_A_V [24]

Answer:

A-no solution

Step-by-step explanation:

There are no values of x that make the equation true.

7 0

2 years ago

Read 2 more answers

On a coordinate plane, triangle T V W has points (negative 4, 8), (0, 4), and (4, 4).

Vanyuwa [196]

The coordinates of the endpoints of line segments T'V' are; T'(-1, 2) and V'(0, 1).

<h3>What are the coordinates of the endpoints of the segment T'V'?</h3>

It follows from the task content that the transformation involved in the formation of the image from the pre-image is dilation by a scale factor of 1/4.

On this note, given that the coordinates of T and V from the task content are; (-4, 8) and (0,4), it follows that the coordinates of the endpoints as required are; T'(-1, 2) and V'(0, 1).