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FrozenT [24]
3 years ago
6

Qs bisect PQR and MRSQ =71

Mathematics
1 answer:
Soloha48 [4]3 years ago
3 0

Answer: If ∡PQR = 82°, and the ray QS bisects it, it cuts ∡PQR into two equal halves, ∡PQS and ∡RQS, each of which is then 82/2, or 41°.

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Select the number line model that matches the expression 1/4 - 5/4
Yuri [45]

Answer:

The answer would be -5/4

3 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
Unit 4 Lesson 13 exit ticket (please help me im getting stressed with this)​
Illusion [34]

Answer:

1. is A

and

2. is c

Step-by-step explanation:

hope this helps

5 0
3 years ago
Read 2 more answers
Pettit printing company has a total market value of $100 million, consisting of 1 million shares selling for $50 per share and $
amm1812

The weighted average cost of capital for the firm will be 11.25%.

<h3>How to calculate the WACC?</h3>

The weighted average cost of capital is the calculation of the cost of capital for a firm where each category of capital is weighted.

Here, the weighted average cost of capital will be:

= 0.5(10%)(1 - 15%) + 0.5(14%)

= 0.5(0.1)(0.85) + 0.5(0.14)

= 11.25%

The corporate value at 70% debt when WACC is 11.94% will be:

= (EBIT)(1 - T)/WACC

= (13.24)(1 - 0.15)/0.1194

= $94.26 million

The corporate value at 30% debt when WACC is 11.14% will be:

= (EBIT)(1 - T)/WACC

= (13.24)(1 - 0.15)/0.1114

= $101.02 million

Learn more about WACC on:

brainly.com/question/25566972

#SPJ1

4 0
2 years ago
Is it possible for any angle A, sinA=7/5Explain.​
Sedaia [141]

Answer:

No

Step-by-step explanation:

note \frac{7}{5} = 1.4

and  sinA is defined for - 1 ≤ sinA ≤ 1

Thus sinA ≠ \frac{7}{5}

4 0
3 years ago
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