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3 years ago

Distributive property of 36+ 9y

2 answers:

6 0

Find the greatest common factor (GCF)
in this case it would be 9

divide each term by the GCF
36 / 9 = 4
9y / 9 = y

put the two "new" terms in parenthesis and the GCF in front of it
9 (4 + y)
it could also be 9 (y + 4) because of Commutative Property of Addition

lions [1.4K]3 years ago

5 0

This problem in the distributive form would be 9(4+y) because the greatest common factor, or GCF, is 9. When you divide 36 by 9 you get 4 and 9 divided by 9 is one, therefore leaving the y.
To conclude the answer would be 9(4+y).
I hope this helped! :-)

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2 years ago

The proportion of defective computers built by Byte Computer Corporation is 0.15. In an attempt to lower the defective rate, the

FinnZ [79.3K]

Answer:

a) Null hypothesis : H₀: the proportion of defective item of computer has been lowered. That is P < 0.15

Alternative hypothesis: H₁: The proportion of defective item of computer

has been higher. That is P> 0.15 (Right tailed test)

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d) The critical value at 0.01 level of significance = Z₀.₀₁ = 2.57

e) Null hypothesis accepted at 0.01 level of significance

f) we accepted null hypothesis.

Hence the proportion of defective item of computer has been lowered.

Step-by-step explanation:

Step(i):-

Given the sample size 'n' = 42

Given random sample of 42 computers were tested revealing a total of 4 defective computers.

The defective computers 'x' = 4

The sample proportion of defective computers

$p = \frac{x}{n} = \frac{4}{42} = 0.095$

Given The Population proportion 'P' = 0.15

The level of significance ∝=0.01

Step(ii):-

a) Null hypothesis : H₀: the proportion of defective item of computer has been lowered. That is P < 0.15

Alternative hypothesis: H₁: The proportion of defective item of computer

has been higher. That is P> 0.15 (Right tailed test)

Test statistic $Z = \frac{p-P}{\sqrt{\frac{PQ}{n} } }$

$Z = \frac{0.095-0.15}{\sqrt{\frac{0.15(0.85)}{42} } }$

$z = \frac{-0.055}{\sqrt{0.00303} } = - 0.9991$

Calculate the value of the test statistic Z = - 0.9991

|Z| = |- 0.9991| = 0.991

Step(iii):-

The critical value at 0.01 level of significance = Z₀.₀₁ = 2.57

e) Calculate the value of the test statistic Z = 0.991 < 2.57 at 0.01 level of significance.

Conclusion:-

Hence the null hypothesis is accepted at 0.01 level of significance.

The proportion of defective item of computer has been lowered.

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Answer:

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