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9966 [12]
3 years ago
11

Help can u answer this

Mathematics
1 answer:
umka21 [38]3 years ago
3 0

Answer:

c) 82

d) - 11

g)27

h)1000

k)11

l) 8

Step-by-step explanation:

{9}^{2}  + 1  \\  = (9 \times 9) + 1 \\  = 81 + 1 \\  = 82

{6}^{2}  -  {5}^{2}  \\  = (6 \times 6) - (5 \times 5) \\  = 36 - 25 \\  =  - 11

(1 + 2 {)}^{3}  \\  =  {3}^{3}  \\  = 3 \times 3  \times 3\\  = 27

(2 + 8 {)}^{3}  \\  =  {10}^{3}  \\  = (10 \times 10 \times 10) \\  = 1000

\sqrt{4}  +  {3}^{2}  \\  = 2 +  {3}^{2}  \\  = 2 + (3 \times 3) \\  = 2 + 9 \\  = 11

2 \times 5 -  \sqrt{4}  \\  = 10 -  \sqrt{4}  \\  = 10 - 2 \\  = 8

<h3>Hope it is helpful....</h3>
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If 10 miles≈16 km​, about how many miles would be equal to 92 km​? Show how you decided.
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A sample of 11001100 computer chips revealed that 62b% of the chips fail in the first 10001000 hours of their use. The company's
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Answer:

Rule

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P-value < significance level --- reject Null hypothesis

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Since z at 0.10 significance level is between -1.645 and +1.645 and the z score for the test (z = 1.35) falls with the region bounded by Z at 0.1 significance level. And also the two-tailed hypothesis P-value is 0.177 which is greater than 0.1. Then we can conclude that we don't have enough evidence to FAIL or reject the null hypothesis, and we can say that at 0.10 significance level the null hypothesis is valid.

Question; A sample of 1100 computer chips revealed that 62% of the chips fail in the first 1000 hours of their use. The company's promotional literature states that 60% of the chips fail in the first 1000 hours of their use. The quality control manager wants to test the claim that the actual percentage that fail is different from the stated percentage. Determine the decision rule for rejecting the null hypothesis, H0, at the 0.10 level.

Step-by-step explanation:

Given;

n=1100 represent the random sample taken

Null hypothesis: H0 = 0.60

Alternative hypothesis: Ha <> 0.62

Test statistic z score can be calculated with the formula below;

z = (p^−po)/√{po(1−po)/n}

Where,

z= Test statistics

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po = Null hypothesized value = 0.60

p^ = Observed proportion = 0.62

Substituting the values we have

z = (0.62-0.60)/√{0.60(1-0.60)/1100}

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