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Olegator [25]
3 years ago
11

Factor the expression by finding the GCF.

Mathematics
1 answer:
Deffense [45]3 years ago
8 0

Answer:

4m(4m-3)

Step-by-step explanation:

Factor 4m out of the statement because 4 is a factor of both 16 and -12, and m is a factor in m^2 and m.

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The product of 9 and t squared, increased by the sum of the square of t and 2
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9t^2+(t+2)^2

Step-by-Step:

t is 9 X t, or the product of 9 and t. ^2 is squared while + is increased t+2 is the sum of t and t. the ^2 after the parenthesis is indicating the "square of t and 2"

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Answer this question very very fast first to have correct answer will get brainiest :)
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Option d.

Amelie read 30 pages of a book. she read 1/5 of the pages on monday. how many pages did she read on monday?

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Solve the following for y. <br> x - 2y = -23
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all work is shown and pictured

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3 years ago
Read 2 more answers
The 2014 Community College Survey of Student Engagement (CCSSE) included a question that asked faculty how much of their coursew
Kisachek [45]

Answer:

Step-by-step explanation:

Hello!

The objective is to test if the courses emphasize memorizing facts, ideas or methods following the same distribution as before.

The variable of interest is

X: Opinion of students in how much of their coursework emphasized memorizing facts, ideas, or methods. Categorized: 1_"Very little", 2_" Some", 3_" Quite a bit" and 4_"Very much"

It is known for a survey made in 2014 that the percentages for each category are: 1_Very little: 21.5%, 2_Some: 33.7%, 3_Quite a bit: 27.7% and 4_Very much: 17.1%

To test if the current situation follows the same distribution as the historical data (from 2014) you have to conduct a Goodness to Fit Chi-Square test.

The hypotheses are:

H₀: P₁= 0.215; P₂= 0.337, P₃= 0.277 and P₄= 0.171

H₁:

α: 0.01

X^2= sum[\frac{(O:i-E_i)^2}{E_i} ]~~X^2_{k-1}

k= number of categories of the variable.

This test is always one-tailed (right), which means that you will reject the null hypothesis to high values of X² (when the observed and expected frequencies for each category are too different)

The critical value is:

X^2_{k-1;1-\alpha }= X^2_{3;0.99}= 11.345

You will reject the null hypothesis if X^2_{H_0} \geq  11.345

You will not reject the null hypothesis if X^2_{H_0} < 11.345

Before calculating the statistic under the null hypothesis, you have to calculate the expected value for each category following the formula:

E_i= n*P_i

n= 400

E₁= n*P₁= 400*0.215= 86

E₂= n*P₂= 400*0.337= 134.8

E₃= n*P₃= 400*0.277= 110.8

E₄=n*P₄= 400*0.171= 68.4

The observed frequencies are:

O₁= 39

O₂= 139

O₃= 148

O₄= 74

X^2_{H_0}= (\frac{(39-86)^2}{86} )+(\frac{(139-134.8)^2}{134.8} )+(\frac{(148-110.8)^2}{110.8} )+(\frac{(74-68.4)^2}{68.4} )= 38.76

As said before, this test is one-tailed to the right (always) and so is its p-value:

P(X₃²≥38.76)= 1 - P(X²₃<38.76)= 1 - 1 ≅ 0

p-value < 0.00001

Using both approaches (p-value and critical value) the decision is to reject the null hypothesis.

With a level of significance of 1% the decision is to reject the null hypothesis, then the actual courses do not emphasize the memorization of facts, ideas, and methods following the historical percentages.

I hope this helps!

8 0
3 years ago
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