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3 years ago

Complete the steps to factor 12x3-9x2+8x-6 by grouping

Mathematics

1 answer:

iogann1982 [59]3 years ago

6 0

Answer:

(4x−3)(3x²+2)

Step-by-step explanation:

Factor 12x3−9x2+8x−6

12x3−9x2+8x−6

=(4x−3)(3x2+2)

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A private school $860,000 in one year to pay all of its teachers and administrators. the school has a total of twenty-four teach

ehidna [41]

Hi there!

Let t be the number of teachers
Let a be the number of administrators

t+a=24
35,000t+40,000a=860,000
a=24-t
35,000t+40,000(24-t)=860,000
35,000t+960,000-40,000t=860,000
35,000t-40,000t=860,000-960,000
-5,000t=-100,000
t=-100,000/-5,000
t=20
a=24-t
a=24-20
a=4

Therefore there are 4 administrators and 20 teachers.

Hope this helps!

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3 years ago

Remove the largest possible common factor. Check your answer by multiplication.

Nimfa-mama [501]

Answer:

Step-by-step explanation:

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3 years ago

Solve for x. Round to the nearest tenth.

skelet666 [1.2K]

Answer:

D. 14.1

Step-by-step explanation:

Use sin, opposite over hypotenuse.

sin67 = 13/X

solve for X

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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:

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

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

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4 years ago

Find the measure of the missing angles.

8090 [49]

Answer:

d= 99 e= 81 f= 99

Step-by-step explanation:

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3 years ago

Read 2 more answers