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stepladder [879]
3 years ago
10

Yesterday, there were 56 problems assigned for math homework. Desmond got 14 problems correct and 42 problems incorrect. What pe

rcentage did Desmond get correct?
Mathematics
1 answer:
Vika [28.1K]3 years ago
6 0

Answer: 25%

100/56 = 1.785714285714286

1.785714285714286 x 14 = 25

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Jerry randomly selected a point in the rectangle shown below. What is the probability the point he selected is in section A
vlabodo [156]
Jerry had a 3 in 10 chance or a 30 percent chance to hit area A randomly
5 0
3 years ago
A sample of 11001100 computer chips revealed that 62b% of the chips fail in the first 10001000 hours of their use. The company's
STALIN [3.7K]

Answer:

Rule

If;

P-value > significance level --- accept Null hypothesis

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

Z score > Z(at 90% confidence interval) ---- reject Null hypothesis

Z score < Z(at 90% confidence interval) ------ accept Null hypothesis

Null hypothesis: H0 = 0.60

Alternative hypothesis: Ha <> 0.62

z score = 1.35

P value = P(Z<-1.35) + P(Z>1.35) = 0.0885 + 0.0885= 0.177

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

n = Sample size = 1100

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}

z = 1.354

z = 1.35

To determine the p value (test statistic) at 0.01 significance level, using a two tailed hypothesis.

P value = P(Z<-1.35) + P(Z>1.35) = 0.0885 + 0.0885= 0.177

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.

3 0
4 years ago
7) At the beginning of the weekend a swimming pool contains 1,800 gallons of water. Every hour 7 gallons are splashed
MaRussiya [10]

Answer:

1,800g=7h

(I don't see answer choices, I don't know if there's supposed to be any.)

4 0
3 years ago
Which expression is equivalent to 7a2b + 10a2b2 + 14a2b3?
Ugo [173]
The given expression can be simplified in many ways by grouping like terms. The simplest form is obtained by factoring out a²b which gives us the following expression.

a²b(7 + 10b +14b²)
4 0
4 years ago
Read 2 more answers
What’s the answer??(SOMEONE PLEASE HELP ME)
Evgesh-ka [11]

To find f^{-1}, you can switch the "x" and "f(x) or y" in the equation.

f(x) = \sqrt[3]{x-2} +8

y = \sqrt[3]{x-2}+ 8

x = \sqrt[3]{y-2}+8

Now you need to isolate the "y"

x = \sqrt[3]{y-2}+8   Subtract 8 on both sides

x - 8 = \sqrt[3]{y-2}  Cube ( ³ ) each side to get rid of the ∛

(x-8)^{3} = (\sqrt[3]{y-2}) ^{3}

(x-8)^{3} = y -2   Add 2 on both sides

(x-8)^{3}+2 = y


f^{-1} = (x-8)^{3} + 2


5 0
3 years ago
Read 2 more answers
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