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2 years ago

What is the volume of the figure

Mathematics

2 answers:

Marina CMI [18]2 years ago

7 0

Answer:

v = 5,376

Step-by-step explanation:

So the first thing we have to do is figure out the volume for the first part of the cube figure.

To do this we just multiply H x W x L =

In this case v = 4,896

Now we do the same thing with the other rectangular prism:

10 x 4 x 12 = 480

Now that we have the 2 volumes we add them together:

4,896 + 480 = 5,376

Hence the total volume = 5,376

Law Incorporation [45]2 years ago

3 0

Answer:

5,376in^3

Step-by-step explanation:

find the volume of the smaller box first

10×4×12 = 480in^3

find volume of bigger box

17×12×24 = 4,896in^3

add both volumes 4,896 +480 = 5,376in^3

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Sergio [31]

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3 years ago

Drone INC. owns four 3D printers (D1, D2, D3, D4) that print all their Drone parts. Sometimes errors in printing occur. We know

USPshnik [31]

Answer:

Step-by-step explanation:

Hello!

There are 4 3D printers available to print drone parts, then be "Di" the event that the printer i printed the drone part (∀ i= 1,2,3,4), and the probability of a randomly selected par being print by one of them is:

D1 ⇒ P(D1)= 0.15

D2 ⇒ P(D2)= 0.25

D3 ⇒ P(D3)= 0.40

D4 ⇒ P(D4)= 0.20

Additionally, there is a chance that these printers will print defective parts. Be "Ei" represent the error rate of each print (∀ i= 1,2,3,4) then:

P(E1)= 0.01

P(E2)= 0.02

P(E3)= 0.01

P(E4)= 0.02

Ei is then the event that "the piece was printed by Di" and "the piece is defective".

You need to determine the probability of randomly selecting a defective part printed by each one of these printers, i.e. you need to find the probability of the part being printed by the i printer given that is defective, symbolically: P(DiIE)

Where "E" represents the event "the piece is defective" and its probability represents the total error rate of the production line:

P(E)= P(E1)+P(E2)+P(E3)+P(E4)= 0.01+0.02+0.01+0.02= 0.06

This is a conditional probability and you can calculate it as:

$P(A/B)= \frac{P(AnB)}{P(B)}$

To reach the asked probability, first, you need to calculate the probability of the intersection between the two events, that is, the probability of the piece being printed by the Di printer and being defective Ei.

P(D1∩E)= P(E1)= 0.01

P(D2∩E)= P(E2)= 0.02

P(D3∩E)= P(E3)= 0.01

P(D4∩E)= P(E4)= 0.02

Now you can calculate the probability of the piece bein printed by each printer given that it is defective:

$P(D1/E)= \frac{P(E1)}{P(E)} = \frac{0.01}{0.06}= 0.17$

$P(D2/E)= \frac{P(E2)}{P(E)} = \frac{0.02}{0.06}= 0.33$

$P(D3/E)= \frac{P(E3)}{P(E)} = \frac{0.01}{0.06}= 0.17$

$P(D4/E)= \frac{P(E4)}{P(E)} = \frac{0.02}{0.06}= 0.33$

P(D2)= 0.25 and P(D2/E)= 0.33 ⇒ The prior probability of D2 is smaller than the posterior probability.

The fact that P(D2) ≠ P(D2/E) means that both events are nor independent and the occurrence of the piece bein defective modifies the probability of it being printed by the second printer (D2)

I hope this helps!

8 0

3 years ago

Joan had 198 dollars to spend on 9 books.After buying them she had 18 dollars. How much did each book cost?

Fed [463]

She spent 20 dollars on each book because if you subtract 18 from 198 you'll get 180 then you divide that by 9 n it leaves you with 20.

3 0

3 years ago

A study was conducted looking at the association between the number of polyps in the colon and mean grams of fiber consumed per

rewona [7]

Answer:

C) The Spearman correlation results should be reported because at least one of the variables does not meet the distribution assumption required to use Pearson correlation.

Explanation:

The following multiple choice responses are provided to complete the question:

A) The Pearson correlation results should be reported because it shows a stronger correlation with a smaller p-value (more significant).

B) The Pearson correlation results should be reported because the two variables are normally distributed.

C) The Spearman correlation results should be reported because at least one of the variables does not meet the distribution assumption required to use Pearson correlation.

D) The Spearman correlation results should be reported because the p-value is closer to 0.0556.

Further Explanation:

A count variable is discrete because it consists of non-negative integers. The number of polyps variable is therefore a count variable and will most likely not be normally distributed. Normality of variables is one of the assumptions required to use Pearson correlation, however, Spearman's correlation does not rest upon an assumption of normality. Therefore, the Spearman correlation would be more appropriate to report because at least one of the variables does not meet the distribution assumption required to use Pearson correlation.

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3 years ago

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hammer [34]

Answer:

lines a and b are parallel. The slopes are -1/3

None of the lines are perpendicular to each other.

Step-by-step explanation:

To figure out if any of the lines are parallel or perpendicular to each other, you have to find the slopes of each line. To find the slope look at the graph find the rise over run for all of the lines:

line a: This line goes down one every time it goes over 3, which can be represented by -1/3

line b: This lines goes down one every time it goes over 3, which can also be written as -1/3

line c: This line goes up 5 every time it goes over 2, which makes the slope 5/2

When two lines are parallel, they have the same slope. Line a and line b have the same slope, so they are parallel.

When two lines are perpendicular, their slopes are negative reciprocals of each other. Since none of the slopes are a negative reciprocal of another slope, we have no perpendicular lines.

Hope this helps :)

7 0

3 years ago

What is the volume of the figure​

What is the volume of the figure