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3 years ago

PLS HELP me quick i need help

Mathematics

1 answer:

kykrilka [37]3 years ago

7 0

Answer:

B. Austin did not combine -3 and -9 correctly.

Step-by-step explanation:

He solved it like it was 9 - 3 but he needed to solve as -9 - 3.

PLZ THANK AND GIVE FIVE STARS.

AND PLZ GIVE BRAINLIEST.

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Round 22364.0741279 to the nearest hundred

Leona [35]

22400 because the rule for rounding is to round based off of the number next to the place you want to round to. You're rounding to the nearest hundreds so you look at the tens place. If the number is 4 or less you round down, and if the number is 5 or more, you round up. The number in the tens place is 6 so you round up to get 22400. Hope that this answer and explanation help!

3 0

3 years ago

Read 2 more answers

Which is bigger .72 or .725?

aleksley [76]

.72
.725 would be less because it has more behind the decimal.
.72 would be more because there are only 2 numbers behind the decimal

3 0

3 years ago

A certain medicine is given in an amount proportional to a patient's body weight. Suppose a patient weighing 150 pound requires

mart [117]

9514 1404 393

Answer:

182 lbs

Step-by-step explanation:

You can write the proportion for the required patient weight (w) as ...

weight/medicine = w/(174.72 mg) = (150 lb)/(144 mg)

w = (150 lb)(174.72/144) . . . . . multiply by 174.72 mg

w ≈ 182 lb

The patient's weight is 182 pounds.

5 0

3 years ago

Luda [366]

Answer:

a) H0: There is no association between level of education and TV station preference (Independence)

H1: There is association between level of education and TV station preference (No independence)

b) $\chi^2 = \frac{(15-10)^2}{10}+\frac{(15-20)^2}{20}+\frac{(10-10)^2}{10}+\frac{(5-10)^2}{10}+\frac{(25-10)^2}{10}+\frac{(10-20)^2}{20} =33.75$

c) $\chi^2_{crit}=5.991$

d) Since the p value is lower than the significance level we enough evidence to reject the null hypothesis at 5% of significance, and we can conclude that we have dependence between the two variables analyzed.

Step-by-step explanation:

A chi-square goodness of fit test "determines if a sample data matches a population".

A chi-square test for independence "compares two variables in a contingency table to see if they are related. In a more general sense, it tests to see whether distributions of categorical variables differ from each another".

Assume the following dataset:

High school Some College Bachelor or higher Total

Public Broadcasting 15 15 10 40

Commercial stations 5 25 10 40

Total 20 40 20 80

We need to conduct a chi square test in order to check the following hypothesis:

Part a

H0: There is no association between level of education and TV station preference (Independence)

H1: There is association between level of education and TV station preference (No independence)

The level os significance assumed for this case is $\alpha=0.05$

The statistic to check the hypothesis is given by:

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

Part b

The table given represent the observed values, we just need to calculate the expected values with the following formula $E_i = \frac{total col * total row}{grand total}$

And the calculations are given by:

$E_{1} =\frac{20*40}{80}=10$

$E_{2} =\frac{40*40}{80}=20$

$E_{3} =\frac{20*40}{80}=10$

$E_{4} =\frac{20*40}{80}=10$

$E_{5} =\frac{40*40}{80}=20$

$E_{6} =\frac{20*40}{80}=10$

And the expected values are given by:

High school Some College Bachelor or higher Total

Public Broadcasting 10 20 10 40

Commercial stations 10 10 20 40

Total 20 30 30 80

Part b

And now we can calculate the statistic:

$\chi^2 = \frac{(15-10)^2}{10}+\frac{(15-20)^2}{20}+\frac{(10-10)^2}{10}+\frac{(5-10)^2}{10}+\frac{(25-10)^2}{10}+\frac{(10-20)^2}{20} =33.75$

Now we can calculate the degrees of freedom for the statistic given by:

$df=(rows-1)(cols-1)=(2-1)(3-1)=2$

Part c

In order to find the critical value we need to look on the right tail of the chi square distribution with 2 degrees of freedom a value that accumulates 0.05 of the area. And this value is $\chi^2_{crit}=5.991$

Part d

And we can calculate the p value given by:

$p_v = P(\chi^2_{3} >33.75)=2.23x10^{-7}$

And we can find the p value using the following excel code:

"=1-CHISQ.DIST(33.75,2,TRUE)"

Since the p value is lower than the significance level we enough evidence to reject the null hypothesis at 5% of significance, and we can conclude that we have dependence between the two variables analyzed.

7 0

4 years ago

Does anyone know the answer to this?

Alex

Answer:

10.5.

Step-by-step explanation:

By similar triangles:

IJ / 21 = 8 /(8+8) = 1/2

IJ / 21 = 1/2

IJ = 1/2 * 21

= 10.5.

6 0

3 years ago