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mihalych1998 [28]
3 years ago
7

a normal distribution has a mean of 10 and a standard deviation of 1 what is the probability of selecting a number between 8 and

11
Mathematics
1 answer:
rusak2 [61]3 years ago
5 0

Answer:

0.815

Step-by-step explanation:

First, find the z-scores.

z = (x − μ) / σ

z₁ = (8 − 10) / 1

z₁ = -2

z₂ = (11 − 10) / 1

z₂ = 1

P(-2 < Z < 1) = P(Z < 1) − P(Z < -2)

Use a chart, calculator, or the empirical rule to find the probability.

Using the empirical rule:

P(-2 < Z < 1) = 0.84 − 0.025

P(-2 < Z < 1) = 0.815

Using a chart:

P(-2 < Z < 1) = 0.8413 − 0.0228

P(-2 < Z < 1) = 0.8185

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3 0
2 years ago
Find XY if X(0, -7) and Y(-3, 2).<br> d =
guapka [62]

9514 1404 393

Answer:

  d = 3√10 ≈ 9.487

Step-by-step explanation:

The distance formula is appropriate.

  d = √((x2 -x1)^2 +(y2 -y1)^2)

  d = √((-3 -0)^2 +(2 -(-7))^2) = √(9 +81) = √90

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2 years ago
Cosx+1/sin^3x=cscx/1-cosx
ANTONII [103]
<span> I am assuming you want to prove:
csc(x)/[1 - cos(x)] = [1 + cos(x)]/sin^3(x).

 </span>
<span>If we multiply the LHS by [1 + cos(x)]/[1 + cos(x)], we get:
LHS = csc(x)/[1 - cos(x)]
= {csc(x)[1 + cos(x)]/{[1 + cos(x)][1 - cos(x)]}
= {csc(x)[1 + cos(x)]}/[1 - cos^2(x)], via difference of squares
= {csc(x)[1 + cos(x)]}/sin^2(x), since sin^2(x) = 1 - cos^2(x).

 </span>
<span>Then, since csc(x) = 1/sin(x):
LHS = {csc(x)[1 + cos(x)]}/sin^2(x)
= {[1 + cos(x)]/sin(x)}/sin^2(x)
= [1 + cos(x)]/sin^3(x)
= RHS.

 </span>
<span>I hope this helps! </span>
8 0
3 years ago
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zubka84 [21]

Answer:(9,-3)

Step-by-step explanation:

You just have to flip the numbers

3 0
2 years ago
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Andrews [41]

Answer:

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Step-by-step explanation:

If you have -9 and you get to -6 that means you added an interval of 3

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3 0
3 years ago
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