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3 years ago

How do I solve 5 + 6y = 9y + 2 + 3(1 - y)

2 answers:

5 0

First, simplify this equation as much as you can.

<span>5 + 6y = 9y + 2 + 3(1 - y)
Use the distributive property on the </span>3(1 - y) part.
3 • 1 = 3
3 • -y = -3y

5 + 6y = 9y + 2 + 3 + -3y
Combine like terms.
The like terms are:
9y and -3y
2 and 3
9y - 3y = 6y
2 + 3 = 5

5 + 6y = 6y + 5
This can not be simplified further, since doing so would only result in 0.
So no matter what y is, this is true.
If you were to fully simplify this, it would be 0 = 0.

Hope this helps!

Stella [2.4K]3 years ago

3 0

Actually you can't solve this. There is no solutions.

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A math club has 40 members. The number of girls is 5 less than 2 times the number of boys. How many members are boys? How many m

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A personnel manager is concerned about absenteeism. She decides to sample employee records to determine if absenteeism is distri

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Answer:

$df=categor-1=6-1=5$

The critical value can be founded with the following Excel formula:

=CHISQ.INV(1-0.05,5)

And we got $\chi^2_{critc}= 11.0705$

a. 11.070

And since our calculated value is lower than the critical we FAIL to reject the null hypothesis at 5% of significance

Step-by-step explanation:

Previous concepts

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".

Solution to the problem

For this case we want to test:

H0: Absenteeism is distributed evenly throughout the week

H1: Absenteeism is NOT distributed evenly throughout the week

We have the following data:

Monday Tuesday Wednesday Thursday Friday Saturday Total

12 9 11 10 9 9 60

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

The statistic to check the hypothesis is given by:

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

The table given represent the observed values, we just need to calculate the expected values with the following formula $E_i = \frac{60}{6}= 10$ and the expected value is the same for all the days since that's what we want to test.

now we can calculate the statistic:

$\chi^2 = \frac{(12-10)^2}{10}+\frac{(9-10)^2}{10}+\frac{(11-10)^2}{10}+\frac{(10-10)^2}{10}+\frac{(9-10)^2}{10}+\frac{(9-10)^2}{10}=0.8$

Now we can calculate the degrees of freedom (We know that we have 6 categories since we have information for 6 different days) for the statistic given by:

$df=categor-1=6-1=5$

The critical value can be founded with the following Excel formula:

=CHISQ.INV(1-0.05,5)

And we got $\chi^2_{critc}= 11.0705$

a. 11.070

And since our calculated value is lower than the critical we FAIL to reject the null hypothesis at 5% of significance

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3 years ago

There are 750 toothpicks in a regular sized box. If a jumbo box is made by doubling all the dimensions of the regular sized box,

Serga [27]

To answer this, let the 3 sides of the regular sized rectangular box be x, y, and z.

When we double all the sides, the new sides have lengths 2x, 2y, and 2z, respectively.

Since the regular box (with dimensions x,y, and z) have 750 toothpicks, we can write

$xyz=750$

Similarly we can write for the jumbo box as,

$(2x)(2y)(2z)$

which is equal to $8(xyz)$

So this is 8 times the original of $xyz$ . Hence the jumbo box will have 8 times 750 number of toothpicks.

ANSWER: $8*750=6000$ toothpicks

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