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mrs_skeptik [129]
3 years ago
10

It is known that a cable with a​ cross-sectional area of 0.300.30 sq in. has a capacity to hold 2500 lb. If the capacity of the

cable is proportional to its​ cross-sectional area, what size cable is needed to hold 70007000 ​lb?
Mathematics
2 answers:
Leni [432]3 years ago
8 0

Answer:

0.84 square in.

Step-by-step explanation:

Since the capacity of the cable is proportional to its​ cross-sectional area.

Mathematically,

C = k * A

A1 = 0.3 sq

C1 = 2500 lb

C2 = 7000 lb

C1/A1 = C2/A2

2500/0.3 = 7000/C2

= 7000 / 8333.33

= 0.84 square in.

vesna_86 [32]3 years ago
5 0

Answer:

0.84 square in

Step-by-step explanation:

Since the capacity of the cable is proportional to its​ cross-sectional area. If a cable that is 0.3 sq in can hold 2500 lb then per square inch it can hold

2500 / 0.3 = 8333.33 lb/in

To old 7000 lb it the cross-sectional area would need to be

7000 / 8333.33 = 0.84 square in

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Answer:

~q→~p

Step-by-step explanation:

The contrapositive of a statement is formed by negating the hypothesis and conclusion of a statement and switching them (the hypothesis becomes the conclusion and vice versa).

With this definition we can conclude that the contrapositive of p→q would be ~q→~p

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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
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.

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What is the values of the soul out ion to the inequality<br> 9 &lt; 3x + 4?
HACTEHA [7]

Answer:

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Step-by-step explanation:

9<3x+4

3x+4>9

3x>9-4

3x>5

x>5/3

5 0
3 years ago
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