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3 years ago

Find the factors of 9q²-45pq+8q-40p

Mathematics

1 answer:

Elina [12.6K]3 years ago

5 0

Answer:

(x−5y)(9x+8)

Step-by-step explanation:

9x^2−45yx+8x−40y

Do the grouping 9x^2−45yx+8x−40y=(9x^2−45yx)+(8x−40y), and factor out 9x in the first and 8 in the second group.

9x(x−5y)+8(x−5y)

Factor out common term x−5y by using distributive property.

(x−5y)(9x+8)

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3 years ago

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3 years ago

Let’s assume that, over the years, a paper and pencil test of anxiety yields a mean score of 35 for all incoming college freshme

vovangra [49]

Answer:

|t| = 2.235 > 2.093

we rejected null hypothesis.

It is not coming from population

Step-by-step explanation:

Step:-1

Let’s assume that, over the years, a paper and pencil test of anxiety yields a mean score of 35 for all incoming college freshmen.

μ = 35

small sample size n =20

mean of the sample x⁻ = 30

standard deviation of (S) = 10

Null hypothesis :- The difference between x⁻ and μ is not significant

Alternative hypothesis:- The difference between x⁻ and μ is significant

that is μ ≠ 35

Level of significance :-∝ =0.05

The test statistic:-

$t = \frac{x^{-} - population mean}{\frac{S}{\sqrt{n} } }$

substitute all values in above equation

μ = 35 ,n =20

mean of the sample x⁻ = 30

standard deviation of (S) = 10

$t = \frac{30 - 35}{\frac{10}{\sqrt{20} } }$

t= -2.235

|t| = 2.235

The degrees of freedom γ =n-1 =20-1 =19

By tabulated value of 't' for 19 degrees of freedom at 5% level of significance.= 2.093 for two tailed test.(see attached diagram below)

conclusion:-

|t| = 2.235 > 2.093

we rejected null hypothesis.

It is not coming from population

8 0

3 years ago