In a random sample of 90 students in first grade, the mean hours of sleep per night was 9.2 with a standard deviation of 0.37 in a random sample of 70 students in the 12th grade, the mean hours of sleep per night was 7.3 with a standard deviation of 0.58. construct a 99% confidence interval for the difference between the mean hours of sleep. a. (1.82 h, 1.98 h)b. (1.79 h, 2.01 h)c. (1.74 h, 2.06 h)d. (1.69 h, 2.11 h)

Option C is right

Step-by-step explanation:

Given are two samples

I group has sample size 90, mean = 9.2 and std dev =0,.37

II group has sample size 70, mean = 7.3 and std dev = 0.58

we find that mean difference i.e. difference between mean hours of sleep between two groups

= 9.2-7.3 = 1.9

Std error for difference = 0.0620

Since sample size is sufficiently large we can use Z critical for 99%

i.e. 2.58

Margin of error =±2.58(0.062)

=0.16

Hence confidence interval 99% for difference in hours of sleep

= (1.9-0.16,1.9+0.16)

=(1.74, 2.06)

exa:

)

3+11

14

step-by-step explanation:

from this example, we can see that subtracting with integers is the same as adding the opposite. we can solve using a number line or by using the saying "keep, change, change" to us solve. it is important to remember that we can subtract a negative and end up with a larger number than we started with.

< xcy ≅ < xby

step-by-step explanation:

in δcax and δbax,

so, δcax ≅ δbax (side-angle-side or sas)

since δcax ≅ δbax, we can conclude < acx ≅ < abx

in δacy and δaby,

so, δacy ≅ δaby (side-angle-side or sas)

since δacy ≅ δaby, we can conclude < acy ≅ < aby

now,

< acy ≅ < aby

=> < acx + < xcy ≅ < abx + < xby

=> < acx + < xcy ≅ < acx + < xby   (since < acx ≅ < abx already proved above)

subtracting < acx from both sides, we get

< acx + < xcy -< acx ≅ < acx + < xby -< acx

cancelling out < acx and -< acx from both sides, we get

< xcy ≅ < xby (proved)