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Novay_Z [31]
3 years ago
8

If you double President Carter's age at the time of his first inauguration and subtract his age at the time he died, the result

is 22 years. How old was President Carter when he died?
Mathematics
1 answer:
Ratling [72]3 years ago
7 0

Answer:

82 years

Step-by-step explanation:

J. Carter age at first inauguration = 52 ---- (missing details)

Required

Determine his age when he died.

Represent his age when he died with x.

So, we have the following expression:

2 * Age\ at\ First\ Inauguration - x = 22

[Substitute 52 for age at first inauguration]

2 * 52 - x = 22

[Multiply 2 by 52]

104 - x = 22

[Collect Like Terms]

-x = 22 - 104

[Make x the subject]

x = 104 - 22

x = 82

<em>Hence, he died at 82 years old.</em>

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Solve for x.
Leto [7]

Answer:

x < 4

Step-by-step explanation:

Step 1: Subtract 10 from both sides.

  • -8x + 10 - 10 > -22 - 10
  • -8x > -32

Step 2: Divide both sides by -8 and flip the inequality sign.

  • \frac{-8x}{-8} > \frac{-32}{-8}
  • x < 4

Therefore, the answer is x < 4.

Have a lovely rest of your day/night, and good luck with your assignments! ♡

3 0
2 years ago
I need to find surface area of a cylinder. Formula being used: S.A= L x W +
Arturiano [62]

Answer:

90π yd²

Step-by-step explanation:

the surface area of a cylinder is the sum of the lateral area and twice the aera of one end of the cylinder:  π·d·l, where l represents the length of the cylinder.  Here, the lateral surface area is π·6 yd·12 yd, or 72π yd².

The two ends add the following to the total surface area:

2·π·(d/2)², or 2π·d²/4.

Thus, the total surface area of the cyl. is

A = 2π·(6 yd)²/4 + 72π yd², or

A = 18π yd² + 72π yd² = 90π yd²

Note:  Please check your source.  L x W +  2pi ·r ^2 is incorrect.

5 0
3 years ago
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Nutka1998 [239]

Answer:

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Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
For a certain river, suppose the drought length Y is the number of consecutive time intervals in which the water supply remains
AnnZ [28]

Answer:

a) There is a 9% probability that a drought lasts exactly 3 intervals.

There is an 85.5% probability that a drought lasts at most 3 intervals.

b)There is a 14.5% probability that the length of a drought exceeds its mean value by at least one standard deviation

Step-by-step explanation:

The geometric distribution is the number of failures expected before you get a success in a series of Bernoulli trials.

It has the following probability density formula:

f(x) = (1-p)^{x}p

In which p is the probability of a success.

The mean of the geometric distribution is given by the following formula:

\mu = \frac{1-p}{p}

The standard deviation of the geometric distribution is given by the following formula:

\sigma = \sqrt{\frac{1-p}{p^{2}}

In this problem, we have that:

p = 0.383

So

\mu = \frac{1-p}{p} = \frac{1-0.383}{0.383} = 1.61

\sigma = \sqrt{\frac{1-p}{p^{2}}} = \sqrt{\frac{1-0.383}{(0.383)^{2}}} = 2.05

(a) What is the probability that a drought lasts exactly 3 intervals?

This is f(3)

f(x) = (1-p)^{x}p

f(3) = (1-0.383)^{3}*(0.383)

f(3) = 0.09

There is a 9% probability that a drought lasts exactly 3 intervals.

At most 3 intervals?

This is P = f(0) + f(1) + f(2) + f(3)

f(x) = (1-p)^{x}p

f(0) = (1-0.383)^{0}*(0.383) = 0.383

f(1) = (1-0.383)^{1}*(0.383) = 0.236

f(2) = (1-0.383)^{2}*(0.383) = 0.146

Previously in this exercise, we found that f(3) = 0.09

So

P = f(0) + f(1) + f(2) + f(3) = 0.383 + 0.236 + 0.146 + 0.09 = 0.855

There is an 85.5% probability that a drought lasts at most 3 intervals.

(b) What is the probability that the length of a drought exceeds its mean value by at least one standard deviation?

This is P(X \geq \mu+\sigma) = P(X \geq 1.61 + 2.05) = P(X \geq 3.66) = P(X \geq 4).

We are working with discrete data, so 3.66 is rounded up to 4.

Either a drought lasts at least four months, or it lasts at most thee. In a), we found that the probability that it lasts at most 3 months is 0.855. The sum of these probabilities is decimal 1. So:

P(X \leq 3) + P(X \geq 4) = 1

0.855 + P(X \geq 4) = 1

P(X \geq 4) = 0.145

There is a 14.5% probability that the length of a drought exceeds its mean value by at least one standard deviation

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