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astraxan [27]
3 years ago
8

Joe ran 12 laps around the field to run a total of 1 km on Sunday. How many laps will he need to run on Monday to run 4/6 of a k

m?
Mathematics
1 answer:
icang [17]3 years ago
8 0

Answer:

4 \div 6 \times 12 = 4 \times 2 \\    \\ 4 \times 2 = 8

so, 8 laps

please mark it brainliest

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In ΔEFG, the measure of ∠G=90°, GE = 2.1 feet, and EF = 4.8 feet. Find the measure of ∠F to the nearest degree.
trapecia [35]

Answer:

26 degrees

Step-by-step explanation:

8 0
3 years ago
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The nation of Finlandia has 120,000 people. Of these, 20,000 are children under 16, 72,000 have jobs, 8,000 don't have jobs but
stepan [7]

Answer: labor force participation rate  =60%

unemployment rate = 6.67%

Step-by-step explanation:

The labor force participation rate is computed as

\text{labor force participation rate}=\dfrac{\text{Total employed persons}}{\text{Total population}}

i.e. \text{labor force participation rate}= \dfrac{72000}{120000}

= 0.6\ or\ 60 \%

The unemployment rate is computed as:

\text{unemployment rate}=\dfrac{\text{Number of personsdon't have jobs }}{\text{Total population}}\\\\\text{unemployment rate}=\dfrac{8000}{120000}\\\\\text{unemployment rate}=0.0667\ or\ 6.67\%

Hence, labor force participation rate  =60%

unemployment rate = 6.67%

6 0
2 years ago
Suppose a particular type of cancer has a 0.9% incidence rate. Let D be the event that a person has this type of cancer, therefo
natita [175]

Answer:

There is a 12.13% probability that the person actually does have cancer.

Step-by-step explanation:

We have these following probabilities.

A 0.9% probability of a person having cancer

A 99.1% probability of a person not having cancer.

If a person has cancer, she has a 91% probability of being diagnosticated.

If a person does not have cancer, she has a 6% probability of being diagnosticated.

The question can be formulated as the following problem:

What is the probability of B happening, knowing that A has happened.

It can be calculated by the following formula

P = \frac{P(B).P(A/B)}{P(A)}

Where P(B) is the probability of B happening, P(A/B) is the probability of A happening knowing that B happened and P(A) is the probability of A happening.

In this problem we have the following question

What is the probability that the person has cancer, given that she was diagnosticated?

So

P(B) is the probability of the person having cancer, so P(B) = 0.009

P(A/B) is the probability that the person being diagnosticated, given that she has cancer. So P(A/B) = 0.91

P(A) is the probability of the person being diagnosticated. If she has cancer, there is a 91% probability that she was diagnosticard. There is also a 6% probability of a person without cancer being diagnosticated. So

P(A) = 0.009*0.91 + 0.06*0.991 = 0.06765

What is the probability that the person actually does have cancer?

P = \frac{P(B).P(A/B)}{P(A)} = \frac{0.91*0.009}{0.0675} = 0.1213

There is a 12.13% probability that the person actually does have cancer.

3 0
3 years ago
Jennifer deposits $80 into an investment account with a rate of 4%, compounded annually. The equation y = 80(1 + 0.04)x can be u
IceJOKER [234]

Answer:the answer is A: (0,80)

Step-by-step explanation:

8 0
3 years ago
Your are 5 feet tall. What is your height in meters?i
Svetach [21]

Answer: 5 ft = 1.5240 m

3 0
3 years ago
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