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3 years ago

14

Make the following conversions in the metric system by multiplying by the appropriate conversion factors. Write your answers as

a whole numbers or decimals.
25 cm to millimeters
mm

25 mm to centimeters
cm

1 answer:

creativ13 [48]3 years ago

7 0

25cm = 250mm
because 25 x 10 = 250

25mm = 2.5cm
because 25/10 = 2.5.

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Which function in Excel is used to perform a test of independence? a. T.TEST b. CHISQ.TEST c. NORM.S.DIST d. Z.TEST

denis23 [38]

Answer:

b. CHISQ.TEST

Step-by-step explanation:

Notation and previous concepts

A chi-square test is "used to test if the variance of a population is equal to a specified value. This test can be either a two-sided test or a one-sided test. The two-sided version tests against the alternative that the true variance is either less than or greater than the specified value"

The statistic is given by:

$\chi^2 = \sum_{i=1}^n \frac{(O_i-E_i)^2}{E_i}$

In order to calculate the p value we need to have in count the degrees of freedom , on this case $df=(rows-1)(#cols-1)$ . And we can calculate the p value would be given by:

$p_v =P(\chi^2 >\chi^2_{calc})$

But the correct comand is given by:

b. CHISQ.TEST(Actual range, Expected range)

We just need to put in one column the Observed values and in other the expectec values.

3 0

3 years ago

The producer of a certain bottling equipment claims that the variance of all its filled bottles is .027 or less. A sample of 30

o-na [289]

Answer:

$p_v = P(\chi^2_{29}>42.963)=1-0.954=0.0459$

b. between .025 and .05

Step-by-step explanation:

Previous concepts and notation

The chi-square test is used to check if the standard deviation of a population is equal to a specified value. We can conduct the test "two-sided test or a one-sided test".

$\bar X$ represent the sample mean

n = 30 sample size

s= 0.2 represent the sample deviation

$\sigma_o =\sqrt{0.027}=0.164$ the value that we want to test

$p_v$ represent the p value for the test

t represent the statistic

$\alpha=$ significance level

State the null and alternative hypothesis

On this case we want to check if the population standard deviation is less than 0.027, so the system of hypothesis are:

H0: $\sigma \leq 0.027$

H1: $\sigma >0.027$

In order to check the hypothesis we need to calculate the statistic given by the following formula:

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

This statistic have a Chi Square distribution distribution with n-1 degrees of freedom.

What is the value of your test statistic?

Now we have everything to replace into the formula for the statistic and we got:

$t=(30-1) [\frac{0.2}{0.164}]^2 =42.963$

What is the approximate p-value of the test?

The degrees of freedom are given by:

$df=n-1= 30-1=29$

For this case since we have a right tailed test the p value is given by:

$p_v = P(\chi^2_{29}>42.963)=1-0.954=0.0459$

And the best option would be:

b. between .025 and .05

6 0

3 years ago

One hundred employees of a company are asked how they get to work and whether they work full time or part time. The table below

inna [77]

The probability is 56/100, or 14/25 = 0.56.

These events are not mutually exclusive, meaning they can happen at the same time. This means we use

P(A or B) = P(A) + P(B) - P(A and B)

P(carpool or full time) = P(carpool) + P(full time) - P(carpool & full time)

There are 6+9=15 people out of 100 that carpool.
There are 7+4+30+6=47 people out of 100 that work full time.
There are 6 people out of 100 that carpool and work full time.

This gives us
15/100 + 47/100 - 6/100 = 56/100

5 0

3 years ago

13 1/10-7 9/10<br><br> i need help

a_sh-v [17]

Answer:

52/ 10 = 5 2/10 or 5 1/5

Step-by-step explanation:

first turn the mixed fractions into improper fraction

13 1/10= 131/10

7 9/10= 79/10

since the two fractions already have a common denominator just subtract

131 - 79 = 52

52/ 10 = 5 2/10 or 5 1/5

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5 gallons 11 quarts divided by 5

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1.55 us gallons.
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Thunderclan89

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3 years ago

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