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4 years ago

How do I find the value of the variable?

Mathematics

1 answer:

Kryger [21]4 years ago

3 0

93+36+m=180 combine like terms
93+36=129 rewrite
129+m=180 subtract 129 from both sides
180-129=51 rewrite
M=51 and to check this just plug 51 in for m
93+36+51=180
93+36=129
129+51=180
180=180
Hope this helps have a nice nite!

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Marianne is running in a race. This fable shows the time and distance she has already run.

Debora [2.8K]

I’m gonna say your answer would be 50.0 because when I wrote the question and answered it I got 52 so I’m gonna say it would be 50.0 because that’s closest to 52

5 0

3 years ago

Read 2 more answers

Simplify. -9c-(5-3c)

Sidana [21]

Answer:

-12c-5

Step-by-step explanation:

Step 1- Distribute into the parenthesis(Since there's nothing next to the parenthesis to multiply, there is a 1).

-9c-1(5-3c)

-9c-5(1)-3(1)c

Step 2- Multiply,

-9c-5-3c

Step 3- Add common variables.

(-9c-3c)-5

-12c-5

7 0

3 years ago

What number equals 100% if 648 equals 54%

OLEGan [10]

If 648 equals 54%, then 1200 equals 100%.

8 0

3 years ago

-7x-6y=-1

Xelga [282]

Cancel something
we cancel x's

multiply 1st equation by 5 and 2nd by 7 and add them
-35x-30y=-5
<u>35x-28y=7 +</u>
0x-58y=2

-58y=2
divide both sides by -58
y=-1/29
sub back
5x-4(-1/29)=1
5x+4/29=1
minus 4/29 from both sides
note, 1=29/29
5x=25/29
divide bot sides by 5 (or times 1/5)
x=5/29

(5/29,-1/29)

6 0

3 years ago

A book claims that more hockey players are born in January through March than in October through December. The following data sh

astra-53 [7]

Answer:

$\chi^2 = \frac{(67-47.5)^2}{47.5}+\frac{(56-47.5)^2}{47.5}+\frac{(30-47.5)^2}{47.5}+\frac{(37-47.5)^2}{47.5}=18.295$

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

$df=(categories-1)=4-1=3$

And we can calculate the p value given by:

$p_v = P(\chi^2_{3} >18.295)=0.00038$

Since the p value is very low we have enough evidence to reject the null hypothesis and we can conclude that the players' birthdates are not uniformly distributed throughout the year

Step-by-step explanation:

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

H0: There is no difference of birthdates distributed throughout the year

H1: There is a difference between birthdates distributed throughout the year

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 $E_i = \frac{total}{4}$

And replacing we got:

$E_{1} =\frac{67+56+30+37}{4}=47.5$

And now we can calculate the statistic:

$\chi^2 = \frac{(67-47.5)^2}{47.5}+\frac{(56-47.5)^2}{47.5}+\frac{(30-47.5)^2}{47.5}+\frac{(37-47.5)^2}{47.5}=18.295$

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

$df=(categories-1)=4-1=3$

And we can calculate the p value given by:

$p_v = P(\chi^2_{3} >18.295)=0.00038$

Since the p value is very low we have enough evidence to reject the null hypothesis and we can conclude that the players' birthdates are not uniformly distributed throughout the year

3 0

4 years ago