1. The center of a normal curve is cannot be negätive 6 is the mean of the distribution always equal to zero d. is the standard deviation a. c. 2. In a standard normal distribution, the mean is I and the standard deviation is 0 a. mean and the standard deviation are both 1 mean and the standard deviation can have any value d.) mean is 0 and the standard deviation is 1 b. C. 3. Z is a standard normal random variable. What is the value of Z if the area to the right of Z is 0.0122? 1.25 a. b. -1.25 -2.25 0.0128944 d. 2.25 to

Q: S ds J V3s? + 1

A: To evaluate the given integral

Q: 64y2+7=32

A: 64y2+7=32

Q: 4y-12=0

A: 4y = 12 y = 3

Q: 1+2x=77

A:

Q: 4w=25

A: topic - linear equation

Q: 3b-12 = 30

A:

Q: 3p - p

A: Here Given that, 3p-p

Q: -3w+5=23

A:

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A: Sketch the graph of polynomial y=x3 and line x+y=1.                         From the graph it can…

Q: {A €R?x2 | tr(A) = 0} | tr(A) — 0}

A: Here we take a general matrix of the given set then find the basis.

Q: Age (years) Frequency fM (M- M-RP 10 uo to 20 20 up to 30 30 up to 40 20 40 up to 50 10 a. The mean…

A: We want to find mean and standard deviations.

Q: -4d = -64

A: Given  -4d=-64

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A: -1+n=15

Q: 8P4 11C7

A:

Q: Find fre dz |2e*dx

A:

Q: -5 -2 1

A:

Q: S V25 – 4x²

A: For the solution follow the next step.

Q: -8 = 6+n

A:

Q: 2: Find: L-1[: 1 3 - 3]

A:

Q: 6+12=

A: Given,          6+12

Q: p dx. -3x (h)

A:

Q: A+12=13

A: Given,   A+12=13     We will solve it for the value of A

Q: 4a5b• V18a3b6

A: We have to find

Q: 6+197

A: Given,          6+197

Q: Find L s²-8

A:

Q: S A+ AMQ for M.

A: NOTE: Refresh your page if you can't see any equations. .

Q: dx 4е2я — 8ет + 5

A: This question can be solved using the concept of special integrals.

Q: ZV zV1- z?dz

A: Let the given integral be y - y = ∫z1-z2 dz           ..............(1)Let,z = sinx…

Q: z? - 34z + 818 = 0

A: