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Sergio039 [100]
3 years ago
7

Using the least common denominator rename 3/5 and 11/20?

Mathematics
2 answers:
svlad2 [7]3 years ago
8 0
LCD is 20
3/5-> 12/20
11/20-> 11/20
irina1246 [14]3 years ago
3 0

Answer:

12/20 and 11/20.

Step-by-step explanation:

The least common denominator between 3/5 and 11/20 is 20.

For the first equation, we multiple the numerator and denominator by 4 and we get 12/20.

The second equation already has a denominator of 11/20 so we can leave it as is.

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3 years ago
In a study investigating a link between walking and improved​ health, researchers reported that adults walked an average of 873
Rainbow [258]

Answer:

H_0:\mu =873\\\\H_a: \mu

Step-by-step explanation:

Given : In a study investigating a link between walking and improved​ health, researchers reported that adults walked an average of 873 minutes in the past month for the purpose of health or recreation.

Claim : The true average number of minutes in the past month that adults walked for the purpose of health or recreation is lower than 873 minutes.

i.e. \mu

We know that the null hypothesis contains equal sign , then the set of hypothesis for the given situation will be :-

H_0:\mu =873\\\\H_a: \mu

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2 years ago
Teresa runs each lap in 5 minutes. She will run less than
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Answer:

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Step-by-step explanation:

mark brainliest and have a great day!

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2 years ago
ASAP Please! True or False? Triangle FGH is a right triangle.
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8 0
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Read 2 more answers
a. Among a shipment of 5,000 tires, 1,000 are slightly blemished. If one purchases 10 of these tires, what is the probability th
hodyreva [135]

Answer:

<h2>See the explanation.</h2>

Step-by-step explanation:

3 or less than 3 tires among the 10 tires should be blemished.

There are some possibilities.

1000 tires are blemished, (5000 - 1000) = 4000 tires are not blemished.

From the 5000 tires, 10 tires can be chosen in ^{5000}C_{10} ways.

Possibility 1: <u>3 tires are blemished.</u>

Total 10 tires can be chosen with 3 blemished tires in ^{1000}C_3 \times {4000}C_7 ways.

Possibility 2:  <u>2 tires are blemished.</u>

Total 10 tires can be chosen with 2 blemished tires in ^{1000}C_2 \times ^{4000}C_8  ways.

Possibility 3: <u>1 tires are blemished.</u>

Total 10 tires can be chosen with 1 blemished tires in ^{1000}C_1 \times^{4000}C_9 ways.

Possibility 4:  <u>No tires are blemished.</u>

Total 10 tires can be chosen with no blemished tires in ^{1000}C_0 \times^{4000}C_{10} ways.

Hence, the required probability is \frac{^{1000}C_1 \times^{4000}C_9 + ^{1000}C_0 \times^{4000}C_{10} + ^{1000}C_2 \times^{4000}C_8 + ^{1000}C_3 \times^{4000}C_7}{^{5000}C_{10} }.

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3 years ago
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