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Aliun [14]
2 years ago
7

A cylindrical pressure vessel has a height of 9 feet and a diameter of 6 feet. How much gas can the pressure vessel hold when fu

ll?
Mathematics
2 answers:
stellarik [79]2 years ago
4 0
Height of the cylindrical pressure vessel = 9 feet
Diameter of the cylindrical pressure vessel = 6 feet
Then
Radius of the cylindrical pressure vessel = (6/2) feet
                                                                   = 3 feet
Then
Volume of the cylindrical pressure vessel = <span>π<span>r^2</span></span><span>h</span>
                                                                   = 3.14 * (3)^2 * 9 cubic feet
                                                                   = 3.14 * 9 * 9 cubic feet
                                                                   = 254.34 cubic feet
So the amount of gas that the cylindrical pressure vessel can hold is 254.34 cubic feet. I hope the procedure is clear enough for you to understand.
lianna [129]2 years ago
4 0

Answer:

81The majority f math problems do not multiply pi

Step-by-step explanation:

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Two cones are similar. The surface area of the larger cone is 65π square inches. The surface area of the smaller cone is 41.6π s
devlian [24]

Answer:

8 inches

Step-by-step explanation:

The surface area of a cone (without the base) is given by:

Surface area = pi*r*s

Where r is the base radius and s is the slant height.

The smaller cone has a surface area of 41.6pi in2 and a radius of 6.4 inches, so the slant height is:

41.6pi = pi*r*s

41.6 = 6.4s

s = 6.5 in

If the cones are similar, the radius and the slant height increase in the same proportion, so we have that:

r'/s' = r/s = 6.4/6.5

s' = r'*6.5/6.4

So for the larger cone, we have:

65pi = pi*r'*s'

65 = r'*r'*6.5/6.4

r'^2 = 64

r' = 8 inches

5 0
2 years ago
Read 2 more answers
A scale drawing of a house addition shows a scale factor of 1 in. = 3.3 ft. Josh decides to make the house addition smaller, and
emmasim [6.3K]
It is decreased by a factor of 3. You can determine this by realizing that one inch original equals 3.3 feet, but the equals 1.1 feet, and 3.3/1.1 = 3.
7 0
1 year ago
Read 2 more answers
The service time distribution describes the probability P that the service time of the customer will be no more than T hours. If
Molodets [167]

Answer: option d.

Step-by-step explanation:

You have the following formul given in the problem:

p=1-e^{-mt}

You know that:

The number of  customers serviced in an hour by the technical support representative is 6 costumbers, therefore:

m=6

As the problem asked for the probability that  a costumber will be on hold less than 30 minutes, we know that:

t=0.5

Substitute the values above into the formula.

Then, you obtain:

p=1-e^{-(6)(0.5)}=0.95 or 95%

7 0
1 year ago
Which point shows the location of 5 – 2i on the complex plane below? On a coordinate plane, points A, B, C, and D are shown. Poi
Romashka-Z-Leto [24]

Answer:

The Point C shows the location of 5-2i in the complex plane: 5 points to the right of the origin and 2 points down from the origin.

Step-by-step explanation:

We have the complex number 5-2i and we have to show the location of the point that represents that number in the complex plane

In the complex plane the real numbers are located in the horizontal axis, increasing to the right. The positives real numbers are at the right of the origin and the negatives to the left.

The complex numbers are located in the vertical axis, with the positives over the origin and the negatives below the origin.

This complex number 5-2i is the sum of a real part (5) and a imaginary part (-2i), so the point will be 5 units rigth on the horizontal axis (for the real part) and 2 units down in the vertical axis (for the imaginary part).

8 0
1 year ago
Read 2 more answers
Suppose that 4% of the 2 million high school students who take the SAT each year receive special accommodations because of docum
dangina [55]

Answer:

a. 0.0122

b. 0.294

c. 0.2818

d. 30.671%

e. 2.01 hours

Step-by-step explanation:

Given

Let X represents the number of students that receive special accommodation

P(X) = 4%

P(X) = 0.04

Let S = Sample Size = 30

Let Y be a selected numbers of Sample Size

Y ≈ Bin (30,0.04)

a. The probability that 1 candidate received special accommodation

P(Y = 1) = (30,1)

= (0.04)¹ * (1 - 0.04)^(30 - 1)

= 0.04 * 0.96^29

= 0.012244068467946074580191760542164986632531806368667873050624

P(Y=1) = 0.0122 --- Approximated

b. The probability that at least 1 received a special accommodation is given by:

This means P(Y≥1)

But P(Y=0) + P(Y≥1) = 1

P(Y≥1) = 1 - P(Y=0)

Calculating P(Y=0)

P(Y=0) = (0.04)° * (1 - 0.04)^(30 - 0)

= 1 * 0.96^36

= 0.293857643230705789924602253011959679180763352848028953214976

= 0.294 --- Approximated

c.

The probability that at least 2 received a special accommodation is given by:

P (Y≥2) = 1 -P(Y=0) - P(Y=1)

= 0.294 - 0.0122

= 0.2818

d. The probability that the number among the 15 who received a special accommodation is within 2 standard deviations of the number you would expect to be accommodated?

First, we calculate the standard deviation

SD = √npq

n = 15

p = 0.04

q = 1 - 0.04 = 0.96

SD = √(15 * 0.04 * 0.96)

SD = 0.758946638440411

SD = 0.759

Mean =np = 15 * 0.04 = 0.6

The interval that is two standard deviations away from .6 is [0, 2.55] which means that we want the probability that either 0, 1 , or 2 students among the 20 students received a special accommodation.

P(Y≤2)

P(0) + P(1) + P(2)

=.

P(0) + P(1) = 0.0122 + 0.294

Calculating P(2)

P(2) = (0.04)² * (1 - 0.04)^(30 - 2(

P(2) = 0.00051

So,

P(0) +P(1) + P(2). = 0.0122 + 0.294 + 0.00051

= 0.30671

Thus it 30.671% probable that 0, 1, or 2 students received accommodation.

e.

The expected value from d) is .6

The average time is [.6(4.5) + 19.2(3)]/30 = 2.01 hours

8 0
2 years ago
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