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3 years ago

Evaluate the permutation P(36,19)=____

1 answer:

6 0

Easy

nPr= $\frac{n!}{(n-r)!}$
here
36P19= $\frac{36!}{(36-19)!}=$
$\frac{36!}{(17)!}=$
$\frac{36*35*34*33*32*31*30*29*28*27*26*25*24*23*22*21*20*19*18}{1}=$
1045843337171591729971200000

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Kayden is the youngest of four siblings whose ages are consecutive integers. if the sum of their ages is 90, find kayden's age.

NISA [10]

Based on the fact that Kayden is the youngest and their ages are consecutive and add up to 90, the Kayden is 21.

<h3>How old is Kayden?</h3>

To find Kayden's age, assume that it is x and use the formula:

90 = x + (x + 1) + (x +2) + (x +3)

Solving for x gives:

90 = 4x + 6

4x = 90 - 6

x = 84 / 4

x = 21

Find out more on solving for age at brainly.com/question/11757623

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6 0

1 year ago

Please just state if they are agasent, supplementary, or neither!!! I need nothing else!!!

Rina8888 [55]

Step-by-step explanation:

1) supplementary

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3) neither

4) neither

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7) supplementary

8) neither

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3 0

3 years ago

Susan threw a softball 42 yards on her first try and 51 1/2 yards on her second try. How much farther did she throw the softball

kozerog [31]

9ft and 12 inches

That's the first answer of it

7 0

3 years ago

Construct a 95% confidence interval for the population mean, mu. Assume the population has a normal distribution. A random sam

aniked [119]

Given Information:

Number of lithium batteries = n = 16

Mean life of lithium batteries = μ = 645 hours

Standard deviation of lithium batteries = σ = 31 hours

Confidence level = 95%

Required Information:

Confidence Interval = ?

Answer:

$CI = 628.5 \: to \: 661.5 \: hours$

Step-by-step explanation:

The confidence interval is given by

$CI = \mu \pm t_{\alpha/2} (\frac{\sigma}{\sqrt{n}})$

Where μ is the mean life of lithium batteries, σ is the standard deviation, n is number of lithium batteries selected, and t is the critical value from the t-table with significance level of

tα/2 = (1 - 0.95) = 0.05/2 = 0.025

and the degree of freedom is

DoF = n - 1 = 16 - 1 = 15

The critical value (tα/2) at 15 DoF is equal to 2.131 (from the t-table)

$CI = 645 \pm 2.131(\frac{31}{\sqrt{16}})$