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:
A ≈ 65.8 m²
Step-by-step explanation:
Given 2 sides and the angle between them then the area (A) is
A = 
bc sinA
= 0.5 × 19.9 × 10.7 × sin38.17°
≈ 65.8 m² ( to 1 dec. place )
y = -1/2x + 5 should be your equation because it is 5 ft above sea level and its eroding, or going down, 1 ft per 20 years. So every year, the sand dune erodes 1/20 ft more.
Hope this helps!
