Answer:
10.04 × 8.8 =?, ? = 88.352 :)
Answer:
Step-by-step explanation:
Hello!
The variable of interest is:
X: number of daily text messages a high school girl sends.
This variable has a population standard deviation of 20 text messages.
A sample of 50 high school girls is taken.
The is no information about the variable distribution, but since the sample is large enough, n ≥ 30, you can apply the Central Limit Theorem and approximate the distribution of the sample mean to normal:
X[bar]≈N(μ;δ²/n)
This way you can use an approximation of the standard normal to calculate the asked probabilities of the sample mean of daily text messages of high school girls:
Z=(X[bar]-μ)/(δ/√n)≈ N(0;1)
a.
P(X[bar]<95) = P(Z<(95-100)/(20/√50))= P(Z<-1.77)= 0.03836
b.
P(95≤X[bar]≤105)= P(X[bar]≤105)-P(X[bar]≤95)
P(Z≤(105-100)/(20/√50))-P(Z≤(95-100)/(20/√50))= P(Z≤1.77)-P(Z≤-1.77)= 0.96164-0.03836= 0.92328
I hope you have a SUPER day!
Answer:
i think it's 16
Step-by-step explanation:
Answer:
x = 25
Step-by-step explanation:
.20x + .68(120 - x) = .54(120) (Distribute .68 to 120 and -x)
.20x + 81.6 - .68x = .54(120) (multiply .54 and 120)
.20x + 81.6 - .68x = 69.6 ( add .20x and -.68x)
-.48x + 81.6 = 69.6 ( subtract 81.6 on each side)
-.48x = -12 (divide -12 by -.48)
x = 25
