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Yuri [45]
3 years ago
5

Find the volume of the prism.

Mathematics
1 answer:
puteri [66]3 years ago
7 0

Answer:

Volume of prism = 6.3 cm³

Step-by-step explanation:

Given:

Dimensions of parallels line = 2.3 and 1.2 cm

Height of prism = 1.8 cm

Width of prism = 2 cm

Find:

Volume of prism

Computation:

Volume of prism = Area of base x Width

Volume of prism = [(1/2)(sum of parallels line)(Height)]Width

Volume of prism = [(1/2)(2.3 + 1.2)(1.8)]2

Volume of prism = [(1/2)(3.5)(1.8)]2

Volume of prism = [(3.5)(1.8)]

Volume of prism = 6.3 cm³

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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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3 years ago
Ali’s dog weighs 8 times as much as her cat. Together, the two pets weigh 54 pounds. How much does Ali’s dog weigh?
baherus [9]
Ali's dog weighs 48 pounds.
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3 years ago
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Plz help me I really need it
IrinaVladis [17]

Answer:

i think [after solving] that it is option B and C

Step-by-step explanation:

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Answer:

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Step-by-step explanation:

Those are the factors of 10

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Question 1 of 40<br> If f(x) = 4x² - 6 and g(x) = x² - 4x-8, find (f- g)(x).
Aleksandr-060686 [28]

Answer:

\huge\boxed{\sf (f-g)(x) = 3x\² + 4x - 14}

Step-by-step explanation:

<h3><u>Given functions:</u></h3>
  • f(x) = 4x² - 6
  • g(x) = x² - 4x - 8
<h3><u>Solution:</u></h3>

Subtract both functions

(f-g)(x) = 4x² - 6 - (x² - 4x - 8)

(f-g)(x) = 4x² - 6 - x² + 4x - 8

Combine like terms

(f-g)(x) = 4x² - x² + 4x - 6 - 8

(f-g)(x) = 3x² + 4x - 14

\rule[225]{225}{2}

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