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!
(a) * (a+1) = 420
a^2 + a = 420
a^2 + a - 420 = 0
(a+21)(a-20) = 0
a = -21 OR 20
The two integers can be -21 and -20, OR 20 and 21.
The answer is:
A. 10
This is because segment AB is parallel to segment DC, meaning they are the same length.
Hope this helps!
Answer:
y = 0.15x+77 is the equation linear connecting total cost y and miles driven x
Step-by-step explanation:
Given that the leasing company charged a flat rental fee for the week, plus a charge for each mile driven.
Let flat rental fee be c and cost per mile driven = m and miles driven = x
Total cost =y
Then y = mx+C is the linear equation.
to find m and c, we use the fact that y =110.30 when x = 222
i.e. 110.30 = c+222x
and similarly 99.05 = c+147x
Subtract to eliminate c
11.25 = 75 x
0.15 =x
Substitute in I equation
110.30 = c+222(0.15)
c = 77
Hence y = 0.15x+77 is the equation linear connecting total cost y and miles driven x
Answer:
Since an equilateral triangle has all three side equal then one of its side is 38mm
