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miskamm [114]
3 years ago
15

There are 165 million full-time and voluntary part-time workers, 7 million underemployed workers, and 5 million discouraged work

ers. The labor force is the sum of those that are employed plus the unemployed. Based on the given info, what is the unemployment rate?
Mathematics
1 answer:
algol133 years ago
4 0

Answer:

Step-by-step explanation:

It is given that the labor force is the sum of those that are employed plus the unemployed.

Also it is given that full-time and voluntary part-time workers are 165 million

Unemployed people are 7 million

Discouraged people are 5 million

So the labor force = 165million + 7 million = 172 million

Unemploymenr rate = unemployed workers/ labour force * 100 = 7/172 *100=

0.04069767441  * 100 = 4%

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There is a line that includes the point (4, 1) and has a slope of 1/4. what is its equation in
Lelechka [254]

Answer:

y=1/4x

Step-by-step explanation:

y=mx+b

m=1/4

so, y=1/4x+b

Now, look at our line's equation so far: . b is what we want, the 1/4 is already set and x and y are just two "free variables" sitting there. We can plug anything we want in for x and y here, but we want the equation for the line that specfically passes through the the point (4,1).

So, why not plug in for x the number 4 and for y the number 1? This will allow us to solve for b for the particular line that passes through the point you gave!.

(4,1). y=mx+b or 1=1/4 × 4+b, or solving for b: b=1-(1/4)(4). b=0.

y=1/4x+0

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3 years ago
Can someone please help me?
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3 years ago
Brian is walking 2 3/4 of a mile everyday. He does this for one week.
laiz [17]

Answer:

1. 11/4

2. 19 1/4

Step-by-step explanation:

To change into an improper fraction:

((2 times 4)+3)/4

=11/4

To calculate total distance:

(11/4) times 7

=77/4

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6 0
3 years ago
What is the solution of the equation 11 – 4x = 5? round your answer to the nearest ten-thousandth?
OLga [1]
11-4x=5
4x=6
x=6/4=1.5000
5 0
3 years ago
Read 2 more answers
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
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