i’m pretty positive you would just subtract!
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: i dont understand
Step-by-step explanation:
<h2>
The required poynomial equation is 
.</h2>
Step-by-step explanation:
Given,
The zeros of the polynomial are - 4, - 2, 1 and 4
To find, the polynomial equation = ?
We know that,
The polynomial equation = (x - A)(x - B)(x - C)(x - D)
= (x + 4)(x + 2)(x - 1)(x - 4)
= [(x + 4)(x - 4)] [(x + 2)(x - 1)]
= (
)(
)
= ()(
)
= 
=
Thus, the required poynomial equation is .
Can you please provide the image of the triangle. thank you:)
