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e-lub [12.9K]
3 years ago
7

Assume the random variable X is normally distributed with mean muequals50 and standard deviation sigmaequals7. Find the 87 th pe

rcentile.

Mathematics
1 answer:
serious [3.7K]3 years ago
7 0

Answer:

  57.885

Step-by-step explanation:

For such calculations a probability calculator is very helpful. The one in the attachment shows the 87th percentile to be 57.885.

___

A table of the standard normal distribution will tell you the 87th percentile corresponds to a z-value of 1.12639. Then the X value is ...

  X = Zσ +μ = 1.12639(7) +50 = 57.885

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How do I solve this?
jok3333 [9.3K]

Answer:

h(x-11)=-5

Step-by-step explanation:

just put the equetion from the top

h(x-11)=-5

8 0
3 years ago
Xavier made 25 pounds of roasted almonds for a fair. He has 3 1/2 pounds left at the end of the fair. How many pounds of roasted
VLD [36.1K]
25-3.5(3 1/2)=21.5. 21.5 pounds of roasted almonds were sold
 
4 0
4 years ago
Which can be used to describe the expression? Check all that apply.
Burka [1]

Answer:

B and D

Step-by-step explanation:

One thing is for certain. The final answer will be 1/r^12 or r^(-12)

When written as the given of (r^-4)^3, the powers are multiplied.

The first one is false. There are 4 factors of r^-4 If you get an answer that says it is true, then the question means that r^-4 is 3 equal factors of r^-12

B is true

C: not true. The exponents are multiplied, not added.

D is true

E: not true. This is the same thing as C.

6 0
3 years ago
What must the radius of C be for PR to be tangent to it at Q?
Setler79 [48]

Answer:

D. 8

Step-by-step explanation:

Draw segment CQ which will be perpendicular to tangent PR.

So, triangle CQR will be right angled triangle right angle at Q.

Let radius of the circle be r.

Therefore,

CQ = CS = r

By Pythagoras Theorem:

{(r + 9)}^{2}  =  {r}^{2}  +  {15}^{2}  \\  \\  {r}^{2}  + 18r + 81 =  {r}^{2}  + 225 \\  \\ 18r = 225 - 81 \\  \\ 18r = 144 \\  \\ r =  \frac{144}{18}  \\  \\ r = 8

5 0
3 years ago
Consider the infinite geometric series (∞/Σ/n=1) * -4(1/3)^{n-1}.
raketka [301]
The first four terms of the series probably refer to the first four partial sums.

\displaystyle\sum_{n=1}^1-4\left(\frac13\right)^{n-1}=-4\left(\frac13\right)^{1-1}=-4
\displaystyle\sum_{n=1}^2-4\left(\frac13\right)^{n-1}=-\frac{16}3
\displaystyle\sum_{n=1}^3-4\left(\frac13\right)^{n-1}=-\frac{52}9
\displaystyle\sum_{n=1}^4-4\left(\frac13\right)^{n-1}=-\frac{160}{27}

which you can compute either by adding one term at a time, or using the well-known formula,

\displaystyle\sum_{n=1}^kar^{n-1}=a\frac{1-r^{k+1}}{1-r}

(I can provide a link to a derivation I gave in a nearly identical question in the comments)

The series converges as k\to\infty if and only if |r|, which is certainly the case here, since -1.

Extrapolating from the formula above, the sum of the convergent series is

\displaystyle\sum_{n=1}^\infty ar^{n-1}=\frac a{1-r}

so the sum for this series is \dfrac{-4}{1-\frac13}=-6.
5 0
4 years ago
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