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Hitman42 [59]
3 years ago
11

Just please help me I'm done with this work

Mathematics
2 answers:
Alekssandra [29.7K]3 years ago
8 0
V = (1/2)(13)(16)(12) cm^3

V = (1/2)(2,496) cm^3

V = 1248 cm^3

melamori03 [73]3 years ago
5 0
We need to find the area of the triangle first.
We have 13 and 12. We multiply them together and divide by 2 to find the area.
13•12=156
156/2=78
We then multiply that number by 16.
78•16=1248

The volume is 1248 cubic centimeters.









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DochEvi [55]

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Opposite side to <E=BC

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4 0
3 years ago
7xy + 8z - 2z + 12xy + 10z
cestrela7 [59]

Answer:

19xy + 16z

Step-by-step explanation:

combine like terms

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8z - 2z + 10z = 16z

7 0
3 years ago
A tire company paid $4,992 for 64 tires. Which is the most reasonable estimate for the cost of each tire?
4vir4ik [10]
$78 for each tire
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7 0
3 years ago
Which table represents a linear function?<br> Please hurry I have to finish this soon!
denis23 [38]

Answer:

C

Step-by-step explanation:

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6 0
2 years ago
Read 2 more answers
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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