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Natali5045456 [20]
2 years ago
11

Compare the processes for finding common denominators for two, three and four fractions please help

Mathematics
1 answer:
Tema [17]2 years ago
5 0
12 because you count by 4s and 3s
and see what number is common 
common also known as same
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Perform the following food service calculation. Size of juice can = 54 oz. Portions of juice served = 6 oz. Number of cases used
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2 cases x 12 cans = 24 cans

24 cans x 54 oz per can = 1296 oz

1296 oz / 6 oz per serving = 216 guests served
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You drop a ping pong ball into a measuring cup. The water started at 200 mL but after you drop the ping pong ball the water rise
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Step-by-step explanation:

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I need help ASAP!!Please explain how to solve the problem
rjkz [21]

Answer:

( x - 2 )^2 + ( y - 1 )^2 = 1

Step-by-step explanation:

As i previously explained,

The general form of equation for a circle is ( x - h )^2 + ( y - k )^2 = r^2.

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( x - h )^2 + ( y - k )^2 = r^2

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( x - 2 )^2 + ( y - 1 )^2 = 1^2

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2 years ago
Draw a model to help solve 5/6 + 1/4. Write your anserw as a mixed number
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Step-by-step explanation:

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2 years ago
The time needed to complete a final examination in a particular college course is normally distributed with a mean of 77 minutes
Komok [63]

Answer:

a. The probability of completing the exam in one hour or less is 0.0783

b. The probability that student will complete the exam in more than 60 minutes but less than 75 minutes is 0.3555

c. The number of students will be unable to complete the exam in the allotted time is 8

Step-by-step explanation:

a. According to the given we have the following:

The time for completing the final exam in a particular college is distributed normally with mean (μ) is 77 minutes and standard deviation (σ) is 12 minutes

Hence, For X = 60, the Z- scores is obtained as follows:

Z=  X−μ /σ

Z=60−77 /12

Z=−1.4167

Using the standard normal table, the probability P(Z≤−1.4167) is approximately 0.0783.

P(Z≤−1.4167)=0.0783

Therefore, The probability of completing the exam in one hour or less is 0.0783.

b. In this case For X = 75, the Z- scores is obtained as follows:

Z=  X−μ /σ

Z=75−77 /12

Z=−0.1667

Using the standard normal table, the probability P(Z≤−0.1667) is approximately 0.4338.

Therefore, The probability that student will complete the exam in more than 60 minutes but less than 75 minutes is obtained as follows:

P(60<X<75)=P(Z≤−0.1667)−P(Z≤−1.4167)

=0.4338−0.0783

=0.3555

​

Therefore, The probability that student will complete the exam in more than 60 minutes but less than 75 minutes is 0.3555

c. In order to compute  how many students you expect will be unable to complete the exam in the allotted time we have to first compute the Z−score of the critical value (X=90) as follows:

Z=  X−μ /σ

Z=90−77 /12

Z​=1.0833

UsING the standard normal table, the probability P(Z≤1.0833) is approximately 0.8599.

Therefore P(Z>1.0833)=1−P(Z≤1.0833)

=1−0.8599

=0.1401

​

Therefore, The number of students will be unable to complete the exam in the allotted time is= 60×0.1401=8.406

The number of students will be unable to complete the exam in the allotted time is 8

6 0
3 years ago
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