Given a sample mean difference of 23, a standard error of 4.2, and a critical t value of 2.262, what are the values for the confidence interval?

(13.4996, 32.5004)

Step-by-step explanation:

The confidence interval for mean difference is given by :-

,

where = sample mean difference .

= critical t value (two-tailed)

SE= Standard error.

Given :

SE= 4.2

Now, the confidence interval for mean difference will be :-

Hence, the required confidence interval : (13.4996, 32.5004)

theres no picture: /

step-by-step explanation:

x = ±sqrt(10)

step-by-step explanation:

5 - 2x² = -15

subtract 5 from each side

5-5 - 2x² = -15-5

-2x^2   = -20

divide by -2 on each side

-2x^2   /-2= -20/-2

x^2 =10

take the square root on each side

sqrt(x^2 )=±sqrt(10)

x = ±sqrt(10)

check:

5 - 2x² = -15

+sqrt(10)

5 - 2(sqrt(10))² = -15

5 -2(10) = -15

5 -20 = -15

-15 = -15

-sqrt(10)

5 - 2(-sqrt(10))² = -15

5 -2(10) = -15

5 -20 = -15

-15 = -15