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aalyn [17]
3 years ago
8

3(u+5)=6u+30 please help​

Mathematics
2 answers:
ch4aika [34]3 years ago
7 0

Answer:

3 ≠ 6

The equation does not work

<u>Ignore my answer, I know it is wrong</u>

Step-by-step explanation:

3(u + 5) = 6u + 30

3(u + 5) = 6(u + 5)

3(u + 5)/(u + 5) = 6(u + 5)/(u + 5)

3 ≠ 6

liq [111]3 years ago
6 0

Answer:

-5

Step-by-step explanation:

3(u+5)=6u+30

3u+15=6u+30

15=3u+30

-15=3u

u=-5

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Tickets sold at a school football game cost $1.00 each if purchased before the day of the game. They cost 1.50 each if bought at
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200

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Let number of tickets sold before game be x while those at the gate be y

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Subtracting equation 1 and 2

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Hugh bought some magazines that cost $3.95 each and some books that cost $8.95 each. He spent a total of $47.65. Let m represent
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8 0
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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
4 years ago
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