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IRINA_888 [86]
3 years ago
8

Tom is building a pool. He plans to dig a hole 20 feet wide and 50 feet long. He wants the pool to hold 7500 cubic feet of water

. How deep must he dig the hole?
Mathematics
1 answer:
9966 [12]3 years ago
4 0

Answer: 7.5 feet

Step-by-step explanation:

.

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You need 12 gallons of paint to paint the exterior of your house. At one store the paint sells for $19.95 a gallon. At a second
nikklg [1K]

Answer:

it is $239.40 for $19.95 and 12 gallons and for $37.98 for 6 gallons it is $227.88. you would save $11.52

Step-by-step explanation:

5 0
3 years ago
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Solve for x: 3 < x + 3 < 6 6 < x < 9 6 > x > 9 0 < x < 3 0 > x > 3
kifflom [539]

Given:

The compound inequality is:

3

To find:

The solution for x.

Solution:

We have,

3

To solve the inequality, we need to isolate the variable x.

Subtracting 3 from each side, we get

3-3

0

The solution for x is 0. Therefore, the correct option is C.

8 0
2 years ago
4x + 7y - 3x - y show work please
Marina86 [1]

Answer:

Step-by-step explanation:

4x + 7y - 3x - y

Move like terms next to each other.

4x - 3x + 7y - y

Subtract them

4x-3x= 1x or x

7y-y = 6y

So we have:

x + 6y

5 0
2 years ago
In a sample of 1200 U.S.​ adults, 191 dine out at a resaurant more than once per week. Two U.S. adults are selected at random
allochka39001 [22]

Answer:

a) The probability that both adults dine out more than once per week = 0.0253

b) The probability that neither adult dines out more than once per week = 0.7069

c) The probability that at least one of the two adults dines out more than once per week = 0.2931

d) Of the three events described, the event that can be considered unusual because of its low probability of occurring, 0.0253 (2.53%), is the event that the two randomly selected adults both dine out more than once per week.

Step-by-step explanation:

In a sample of 1200 U.S. adults, 191 dine out at a restaurant more than once per week.

Assuming this sample.is a random sample and is representative of the proportion of all U.S. adults, the probability of a randomly picked U.S. adult dining out at a restaurant more than once per week = (191/1200) = 0.1591666667 = 0.1592

Now, assuming this probability per person is independent of each other.

Two adults are picked at random from the entire population of U.S. adults, with no replacement, thereby making sure these two are picked at absolute random.

a) The probability that both adults dine out more than once per week.

Probability that adult A dines out more than once per week = P(A) = 0.1592

Probability that adult B dines out more than once per week = P(B) = 0.1592

Probability that adult A and adult B dine out more than once per week = P(A n B)

= P(A) × P(B) (since the probability for each person is independent of the other person)

= 0.1592 × 0.1592

= 0.02534464 = 0.0253 to 4 d.p.

b) The probability that neither adult dines out more than once per week.

Probability that adult A dines out more than once per week = P(A) = 0.1592

Probability that adult A does NOT dine out more than once per week = P(A') = 1 - P(A) = 1 - 0.1592 = 0.8408

Probability that adult B dines out more than once per week = P(B) = 0.1592

Probability that adult B does NOT dine out more than once per week = P(B') = 1 - P(B) = 1 - 0.1592 = 0.8408

Probability that neither adult dines out more than once per week = P(A' n B')

= P(A') × P(B')

= 0.8408 × 0.8408

= 0.70694464 = 0.7069 to 4 d.p.

c) The probability that at least one of the two adults dines out more than once per week.

Probability that adult A dines out more than once per week = P(A) = 0.1592

Probability that adult A does NOT dine out more than once per week = P(A') = 1 - P(A) = 1 - 0.1592 = 0.8408

Probability that adult B dines out more than once per week = P(B) = 0.1592

Probability that adult B does NOT dine out more than once per week = P(B') = 1 - P(B) = 1 - 0.1592 = 0.8408

The probability that at least one of the two adults dines out more than once per week

= P(A n B') + P(A' n B) + P(A n B)

= [P(A) × P(B')] + [P(A') × P(B)] + [P(A) × P(B)]

= (0.1592 × 0.8408) + (0.8408 × 0.1592) + (0.1592 × 0.1592)

= 0.13385536 + 0.13385536 + 0.02534464

= 0.29305536 = 0.2931 to 4 d.p.

d) Which of the events can be considered unusual? Explain.

The event that can be considered as unusual is the event that has very low probabilities of occurring, probabilities of values less than 5% (0.05).

And of the three events described, the event that can be considered unusual because of its low probability of occurring, 0.0253 (2.53%), is the event that the two randomly selected adults both dine out more than once per week.

Hope this Helps!!!

6 0
3 years ago
Someone help its due
Sphinxa [80]

Answer:

15

Step-by-step explanation:

PQR = 150.

RPQ = PRQ

RPQ + PRQ = 30

RPQ = 15

7 0
2 years ago
Read 2 more answers
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