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3 years ago

How fast is 300 kilometers in miles per hour?

Mathematics

1 answer:

forsale [732]3 years ago

8 0

186 miles per hours. 1 kilometer is = to .62 miles so .62*300 is 186

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$19.92 for 2.4 pounds of walnuts what is the price per pound of walnuts

professor190 [17]

Answer:

$8.3

Step-by-step explanation:

Given

2.4 pounds of walnuts=$19.92

Solution

the price per pound of walnuts=19.92/2.4

=$8.3

Price of per pound walnuts is $8.3

5 0

3 years ago

In a manufacturing plant that produces new computers, a 0.15 probability exists that a computer will be defective. if five compu

Sindrei [870]

The probability that all of them will be defective is 0.0000759375

Explanation

The general Binomial Probability formula is....

$P= ^nC_{r}*p^r*(1-p)^n^-^r$ , where p is the probability of success, n is the total number of trials and r is the desired numbers of trials.

Given, the probability that a computer will be defective is 0.15 , so p = 0.15

Five computers are manufactured and we need to find the probability that all of them will be defective. That means, n = 5 and r = 5

Now according to the above formula....

$P= ^5C_{5}*(0.15)^5*(1-0.15)^5^-^5 \\ \\ P= 1*(0.15)^5* (0.85)^0\\ \\ P=(0.15)^5 =0.0000759375$

So, the probability that all of them will be defective is 0.0000759375

7 0

4 years ago

PLEASE HELP ASAP! :(

Law Incorporation [45]

I’ll help but u where is the question

4 0

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

C= 5/9 (F−32)

Grace [21]

Answer:

C = 5/9 (F - 32)

This is a temperature scale change comparison.

By this we understand, that ,

1°C = 5/9°F

Which is equal to, 5/9°F = 1°C

These two statements are satisfied by I. and II.

So, the correct answers are, I. and II.

If my answer helped, kindly mark me as the brainliest!

Thank You!!

5 0

3 years ago