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!
12,870 dollars. Let me know if you want me to show my work.
Y = 1/3x + b
Plug in the point
3 = 1/3(2) + b
3 = 2/3 + b, b = 7/3
Solution: y = 1/3x + 7/3
Answer:
Infinitely many solutions. Each real number is a solution of this equation.
Step-by-step explanation:
-5 + (-5r) + 10 = -5r + 5
5 - 5r = -5r + 5 <em>add 5r to both sides</em>
5 = 5 TRUE
