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TiliK225 [7]
2 years ago
12

Let R be the triangular region in the first quadrant with vertices at points (0,0), (a,0), and (0,b), where a and b are positive

constants. Write dow the volume of the solid generated when region R is revolved about the x-axis?
Mathematics
1 answer:
Salsk061 [2.6K]2 years ago
6 0

Answer:

volume of the solid generated when region R is revolved about the x-axis is π ₀∫^a ( \frac{-b}{a}x + b )² dx

Step-by-step explanation:

Given the data in the question and as illustrated in the image below;

R is in the region first quadrant with vertices; 0(0,0), A(a,0) and B(0,b)

from the image;

the equation of AB will be;

y-b / b-0 = x-0 / 0-a

(y-b)(0-a) = (b-0)(x-0)

0 - ay -0 + ba = bx - 0 - 0 + 0

-ay + ba = bx  

ay = -bx + ba

divide through by a

y = \frac{-b}{a}x + ba/a

y = \frac{-b}{a}x + b

so R is bounded by  y = \frac{-b}{a}x + b and y =0, 0 ≤ x ≤ a

The volume of the solid revolving R about x axis is;

dv = Area × thickness

= π( Radius)² dx

= π ( \frac{-b}{a}x + b )² dx

V = π ₀∫^a ( \frac{-b}{a}x + b )² dx

Therefore,  volume of the solid generated when region R is revolved about the x-axis is π ₀∫^a ( \frac{-b}{a}x + b )² dx

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Answer:

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2 years ago
An English professor assigns letter grades on a test according to the following scheme. A: Top 14% of scores B: Scores below the
NeX [460]

Answer:

Grades between 62 and 64 result in a D grade.

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

\mu = 71.9, \sigma = 7.8

Find the numerical limits for a D grade.

D: Scores below the top 84% and above the bottom 10%

So below the 100-84 = 16th percentile and above the 10th percentile.

16th percentile:

This is the value of X when Z has a pvalue of 0.16. So X when Z = -0.995.

Z = \frac{X - \mu}{\sigma}

-0.995 = \frac{X - 71.9}{7.8}

X - 71.9 = -0.995*7.8

X = 64

10th percentile:

This is the value of X when Z has a pvalue of 0.1. So X when Z = -1.28.

Z = \frac{X - \mu}{\sigma}

-1.28 = \frac{X - 71.9}{7.8}

X - 71.9 = -1.28*7.8

X = 62

Grades between 62 and 64 result in a D grade.

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-2x-2+5x=-2x+5x-2=3x-2

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See the attached image for the explanation.

Hope this helps! Please make me the brainliest, it’s not necessary but appreciated, I put a lot of effort and research into my answers. Have a good day, stay safe and stay healthy.

8 0
1 year ago
Read 2 more answers
1 2 3 4 5 6 7 8 9 10 11 12 13
Rina8888 [55]

Answer:

The mean will increase more than the median, but both will increase.

[third option listed]

Step-by-step explanation:

the <em>median </em>of a data set is the number in the middle [when listed from lowest to highest in value]

1 2 3 4 5 6 7 8 9 10 11 12 13

is the current median

let's consider what adding 12 would mean--it would mean that we move the median slightly higher [further along in the data set] because there are more numbers (but let's try this out to confirm:)

1 2 3 4 5 6 7 8 9 10 11 12 12 13

[if a median placement is shared between two numbers, the mean/average of those two numbers is taken, and that is considered to be the median]

so, 7.5 is the current median

(this is an increase of 0.5)

--

the <em>mean</em> of a data set is what we commonly refer to as the "average"

[you find this value by adding all of the numbers in the data set together and dividing by the number of terms in the data set]

1 2 3 4 5 6 7 8 9 10 11 12 13

mean of original data set:

1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12 + 13 [= 91]

_______________________________________

                                    13

[91 ÷ 13 = 7]

because our number is <em>greater </em>than our original mean [12 > 7], we know that the mean must increase:

[91 + 12 = 103]

[103 ÷ 14 ≈ 9.36]

[we had an increase of 2.36]

so, median increased by 0.5, mean increased by 2.36

so, both values increased, whilst the mean increased by <em>more </em>than the median [as to be expected]

you could also express this as "The mean will increase more than the median, but both will increase." [third option listed]

hope this helps!! :)

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2 years ago
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