Answer:
ag=74 and ed=83
Step-by-step explanation:
super simple actually
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!
989.75060139
100.023424234 divided by 10.32342 = 9.68898138737
9.68898138737 + 1000.293920 = 1009.98290139
1009.98290139 - 20.2323 = 989.75060139
Yo sup??
the answer is option A ie
4.19*10-3
0.38+0.039=0.419
just do the basic multiplication
Hope this helps
Answer:
x = - 5
Step-by-step explanation:
5x + 23 = - 2x - 12 ( add 2x to both sides )
7x + 23 = - 12 ( subtract 23 from both sides )
7x = - 35 ( divide both sides by 7 )
x = - 5
