Beginning with the function y = sin x, which would have range from -1 to 1 and period of 2pi:
Vertical compression of 1/2 compresses the range from -1/2 to 1/2
Phase shift of pi/2 to the left
Horizontal stretch to a period of 4pi, as the crests are at -4pi, 0, 4pi
Vertical shift of 1 unit up moves the range to 1/2 to 3/2
So the first choice looks like a good answer.
Answer:
Q6: B Q7: D Q8: C Q9: D Q10: A
Answer:
6.4
Step-by-step explanation:
distance formula 
≈6.403.......
ik it doesnt make any sinse but hope this helps
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!
