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2 years ago

(-12)+(-18)+(-30) simplify this question

Mathematics

2 answers:

Marizza181 [45]2 years ago

6 0

Answer:

-60

Step-by-step explanation:

Well just follow the equation from left to right

Add it all up.

Anon25 [30]2 years ago

3 0

(+) + (-) = -

-12-18-30

-30-30

-60 is your answer

Hope this helped

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According to a survey, high school girls average 100 text messages daily (The Boston Globe, April 21, 2010). Assume the populati

Ghella [55]

Answer:

Step-by-step explanation:

Hello!

The variable of interest is:

X: number of daily text messages a high school girl sends.

This variable has a population standard deviation of 20 text messages.

A sample of 50 high school girls is taken.

The is no information about the variable distribution, but since the sample is large enough, n ≥ 30, you can apply the Central Limit Theorem and approximate the distribution of the sample mean to normal:

X[bar]≈N(μ;δ²/n)

This way you can use an approximation of the standard normal to calculate the asked probabilities of the sample mean of daily text messages of high school girls:

Z=(X[bar]-μ)/(δ/√n)≈ N(0;1)

P(X[bar]<95) = P(Z<(95-100)/(20/√50))= P(Z<-1.77)= 0.03836

P(95≤X[bar]≤105)= P(X[bar]≤105)-P(X[bar]≤95)

P(Z≤(105-100)/(20/√50))-P(Z≤(95-100)/(20/√50))= P(Z≤1.77)-P(Z≤-1.77)= 0.96164-0.03836= 0.92328

I hope you have a SUPER day!

3 0

3 years ago

Solve for k<br><br> 3.1+k=5.67

cluponka [151]

Answer:

k =2.57

Step-by-step explanation:

3.1+k=5.67

Subtract 3.1 from each side

3.1-3.1+k=5.67-3.1

k =2.57

3 0

3 years ago

Read 2 more answers

Leila puts an integer into the "In" box of each machine.

kykrilka [37]

The machines represent equivalent expressions.

The output produced by both machines is 372

The equation represented on machine A is:

$\mathbf{y = 2(5x + 3) - 24}$

The equation represented on machine B is:

$\mathbf{y = 4(2(x + 11) - 7)}$

Where: x and y represent the inputs and the outputs, respectively

When the outputs are the same, it means:

$\mathbf{y = y}$

So, we have:

$\mathbf{2(5x + 3) - 24 = 4(2(x + 11) - 7)}$

Open brackets

$\mathbf{10x + 6 - 24 = 8x + 88 - 28}$

$\mathbf{10x - 18 = 8x + 60}$

Collect like terms

$\mathbf{10x - 8x = 18 + 60}$

$\mathbf{2x = 78 }$

Divide both sides by 2

$\mathbf{x = 39 }$

Substitute 39 for x in $\mathbf{y = 2(5x + 3) - 24}$

$\mathbf{y = 2 \times (5 \times 39 + 3) - 24}$

$\mathbf{y = 372}$

Hence, the output produced by both machines is 372

Read more about equivalent expressions at:

brainly.com/question/15715866

4 0

3 years ago

I think ik the answer but I’m not so can someone help me

Pavlova-9 [17]

Answer:

25%

Step-by-step explanation:

I would think its 25

hope this helps, good luck :)

8 0

3 years ago

Read 2 more answers

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

3 years ago

(-12)+(-18)+(-30) simplify this question​

(-12)+(-18)+(-30) simplify this question