I'm assuming you want us to solve for the unknown variable z
Combine like terms on the right side
Use the distributive property on the left side
Add both sides by 20 to cancel out the "-20" on the left side
Divide both sides by 10
That is the value of the known variable, z, in this equation. Let me know if you need any clarifications, thanks!
~ Padoru
Answer:
(5,-6) is a solution
Step-by-step explanation:
Given
Required
Tell whether (5,-6) is a solution
In (5,-6): x = 5 and y = -6.
Substitute these values in the given equations
becomes
becomes
Because both solutions are true, then (5,-6) is a solution
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!
The correct answer for the question that is being presented above is this one: "28 in^2."
We have to get the individual area of each surface.
A1 = 1*3*2 = 6
A2 = 1*3*2 = 6
A3 = 1*2*2 = 4
A4 = 2*3*2 = 12
Area = A1 + A2 + A3 + A4
Area = 6 + 6 + 4 + 12
Area = 28 in^2
The distribution is uniform for 50 ≤ x ≤ 52.
where x = minutes per class.
The probability P(x < 50.6) is the shaded portion of the distribution.
Its value is
(50.6 - 50)/(52 - 50) = 0.3
Answer:
The probability is 0.3 or 30%
