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gregori [183]
2 years ago
10

To check a solution, you can ____________the solutions into the equations and verify that both equations are true.

Mathematics
1 answer:
kondor19780726 [428]2 years ago
7 0

Answer: To check a solution, you can substitute the solutions into the equations and verify that both equations are true.

Step-by-step explanation:

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Please Help!! A septic tank has the shape shown in the figure. How many gallons does it hold? (1 cu ft = 7.48 gallons.) (Round t
Hoochie [10]

The overall volume is the sum of the volume of a cylinder of height 5'9" and diameter 3'6", and a sphere of diameter 3'6" (two hemispheres = full sphere).

Volume of the cylinder = (area of the base) x (height) = pi * (diameter/2)^2 * 5.75ft = 3.1415 * (3.5ft/2)^2 * 5.75 ft = 55.32 ft^3

Volume of the sphere = 4/3 * pi * (3.5ft)^3 / 8 = 22.45 ft^3

Total volume = (Volume of cylinder) + (Volume of sphere) = (55.32 + 22.45) ft^3 = 77.77 ft^3

3 0
3 years ago
Which phrases represent the expression k/2 - 5
Contact [7]

Answer:

K over two minus five

Step-by-step explanation:

I think?

3 0
3 years ago
The probability that a randomly selected 3-year-old male chipmunk will live to be 4 years old is 0.96516.
mezya [45]

Using the binomial distribution, it is found that there is a:

a) The probability that two randomly selected 3-year-old male chipmunks will live to be 4 years old is 0.93153 = 93.153%.

b) The probability that six randomly selected 3-year-old male chipmunks will live to be 4 years old is 0.80834 = 80.834%.

c) The probability that at least one of six randomly selected 3-year-old male chipmunks will not live to be 4 years old is 0.19166 = 19.166%. This probability is not unusual, as it is greater than 5%.

-----------

For each chipmunk, there are only two possible outcomes. Either they will live to be 4 years old, or they will not. The probability of a chipmunk living is independent of any other chipmunk, which means that the binomial distribution is used to solve this question.

Binomial probability distribution  

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

C_{n,x} = \frac{n!}{x!(n-x)!}

The parameters are:

  • x is the number of successes.
  • n is the number of trials.
  • p is the probability of a success on a single trial.

In this problem:

  • 0.96516 probability of a chipmunk living through the year, thus p = 0.96516

Item a:

  • Two is P(X = 2) when n = 2, thus:

P(X = 2) = C_{2,2}(0.96516)^2(1-0.96516)^{0} = 0.9315

The probability that two randomly selected 3-year-old male chipmunks will live to be 4 years old is 0.93153 = 93.153%.

Item b:

  • Six is P(X = 6) when n = 6, then:

P(X = 6) = C_{6,6}(0.96516)^6(1-0.96516)^{0} = 0.80834

The probability that six randomly selected 3-year-old male chipmunks will live to be 4 years old is 0.80834 = 80.834%.

Item c:

  • At least one not living is:

P(X < 6) = 1 - P(X = 6) = 1 - 0.80834 = 0.19166

The probability that at least one of six randomly selected 3-year-old male chipmunks will not live to be 4 years old is 0.19166 = 19.166%. This probability is not unusual, as it is greater than 5%.

A similar problem is given at brainly.com/question/24756209

6 0
3 years ago
Circle the problems below that you could use mental math to solve.
SashulF [63]
I think this is more of an opinion question. What someone could solve mentally may not be the same as someone else's.If it were my worksheet, I'd circle all of them but it may be different for you.
At your level, I'd probably circle:
54 - 10
93 - 20
and 39 - 2
4 0
3 years ago
How do you do this? Yeah
Aliun [14]

Answer:

17

Step-by-step explanation:

You have to plug -2 into every x in the function, so:

(-2)^2 - 3 (-2) + 7

4 + 6 + 7

17

7 0
3 years ago
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