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3 years ago

I have a LCM and GCF homework can someone help i dont understand??

Mathematics

1 answer:

Sindrei [870]3 years ago

6 0

Alright,

So LCM stands for "least common multiple"
While GCF is "greatest common factor"

Let's look at your number 1 (3&6)

3,6,9,12
6,12,18,24

Both times table has 6 in them making 6 the LCM

To find the GCF we need to know what can go into 3 equally. Since 3 is prime the greatest factor for 3 is....3. 6 goes into 3 two times so 3 is the GCF.

LCM:6
GCF:3

Lets do number 10 (40&4)

For LCM you see whats the smallest number that is in both times tables

4,8,12,16,20,24,28.......40
40,80, 160....

40 is the LCM because 40 is what's the smallest number between the two

GCF?

4 times 1 equals 4. Nothing bigger than 4 can make 4 (if your multiplying). That makes 4 automatically the GCF. :)

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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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c) Calculate the value of the test statistic = 0.991

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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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