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Luda [366]
3 years ago
10

(7-4n) x 6= I need the answer as quick as possible

Mathematics
2 answers:
Paraphin [41]3 years ago
4 0

Answer:

42 - 24n

Step-by-step explanation:

(7-4n)  6

Distribute

7*6 - 4n*6

42 - 24n

NeTakaya3 years ago
3 0

Answer:

42 x - 24 x n - i = 0

Step-by-step explanation:

Expand the following:

6 x (7 - 4 n) = i

6 x (7 - 4 n) = 6 x×7 + 6 x (-4 n):

6×7 x - 4×6 x n = i

6×7 = 42:

42 x - 4×6 x n = i

6 (-4) = -24:

42 x + -24 x n = i

Subtract i from both sides of 42 x - 24 x n = i:

42 x - 24 x n - i = i - i

-i + i = 0:

Answer:  42 x - 24 x n - i = 0

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Answer:

L - 10 . W - 5

Step-by-step explanation:

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0 = 3w^2 - 5w -50

0= (3w+10)(w-5)

w = 5 , w can't equal -10/3 because width can't be negative

width is 5

Length = 3 * 5 - 5

Length = 10 , width = 5

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Ok so I want anyone who knows anything about riverdale to answer these
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Step-by-step explanation:

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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:

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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
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Step-by-step explanation:

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Step-by-step explanation:

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