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:
Step-by-step explanation:
1/2(7x - 6) = 6x - 10
7/2x - 3 = 6x - 10 .....multiply by 2
7x - 6 = 12x - 20
-6 + 20 = 12x - 7x
14 = 5x
14/5 = x <===
P(at least 5 rolls until 1) = P(4 rolls are not 1) = 5/6 x 5/6 x 5/6 x 5/6 = 0.4823 (4sf)
Fewer than 7 rolls to get second 1 after first takes 3 rolls means second occurs on 4th, 5th or 6th roll
The probability of each of these is 1/6, 5/6 x 1/6 and 5/6 x 5/6 x 1/6 respectively.
P(second 1 on 4th, 5th or 6th roll) = 1/6 + 5/36 + 25/216 = 91/216 = 0.4213 (4sf)
If you add the two numbers, 10 +8 is 18. The fractional parts are 1/4 and 2/4 which equals 3/4. So the total length is 18 and 3/4 feet.
Answer:
Step-by-step explanation:
The first term is 22. It looks like each term after that is the term before, plus a negative 3.
a-sub-one = 22
common difference = d= -3
Formula for 
:
= 25 - 3n
