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3 years ago

Please help I will reward brainly! Need help now

Mathematics

1 answer:

STatiana [176]3 years ago

3 0

1. 3!/5! = 3*2 / 5*4*3*2 = 6/120 = 1/20

2. 5! / 8! = 5*4*3*2 / 8*7*6*5*4*3*2 = 120 / 40320 = 1/336

3. 9!/9! = 1

4. 7! / 8! = 7*6*5*4*3*2 / 8*7*6*5*4*3*2 = 5040 / 40320 = 1/8

5. 5! / 7! = 5*4*3*2 / 7*6*5*4*3*2 = 120 / 5040 = 1/42

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Solve k^2+4k+4=25 by taking the square root of each side. Please show work, thanks

MAVERICK [17]

Answer:

Step-by-step explanation:

k=3 or k=−7 both choices are correct

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3 years ago

There are two factors of -36 such that one factor is 11 less than half of the other factor choose all the pairs of these factors

Amanda [17]

Answer with explanation:

Factors of -36 are

$=\pm 1, \pm 2,\pm 3,\pm 4, \pm 6, \pm 9, \pm 12, \pm 18, \pm 36$

It is given that , among the factors of , -36 , one factor is 11 less than half of the other factor.

If you will take, $=\pm 1,\pm 3,\pm 4, \pm 9,$ , then you can find such pairs.

Pairs are= (3,-4), (9, -1),(-1, -6),

7 0

3 years ago

Help

BigorU [14]

Answer:

$69$

thats it all of them are 69

4 0

2 years ago

Simplify the radical expression.

Fofino [41]

B. 10 square root 5
hope this helps:)
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3 years ago

Assume a simple random sample of 10 BMIs with a standard deviation of 1.186 is selected from a normally distributed population o

kirza4 [7]

Answer:

a) H0: $\sigma = 1.34$

H1: $\sigma \neq 1.34$

b) $df = n-1= 10-1=9$

And the critical values with $\alpha/2=0.005$ on each tail are:

$\chi_{\alpha/2}= 1.735, \chi_{1-\alpha/2}= 23.589$

c) $t=(10-1) [\frac{1.186}{1.34}]^2 =7.05$

d) For this case since the critical value is not higher or lower than the critical values we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true deviation is not significantly different from 1.34

Step-by-step explanation:

Information provided

n = 10 sample size

s= 1.186 the sample deviation

$\sigma_o =1.34$ the value that we want to test

$p_v$ represent the p value for the test

t represent the statistic (chi square test)

$\alpha=0.01$ significance level

Part a

On this case we want to test if the true deviation is 1,34 or no, so the system of hypothesis are:

H0: $\sigma = 1.34$

H1: $\sigma \neq 1.34$

The statistic is given by:

$t=(n-1) [\frac{s}{\sigma_o}]^2$

Part b

The degrees of freedom are given by:

$df = n-1= 10-1=9$

And the critical values with $\alpha/2=0.005$ on each tail are:

$\chi_{\alpha/2}= 1.735, \chi_{1-\alpha/2}= 23.589$

Part c

Replacing the info we got:

$t=(10-1) [\frac{1.186}{1.34}]^2 =7.05$

Part d

For this case since the critical value is not higher or lower than the critical values we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true deviation is not significantly different from 1.34

5 0

3 years ago