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NISA [10]
3 years ago
6

A Pyramid is placed inside a prism as shown in the pyramid has the same base area B as a prison but half the height H of the pri

sm which expression gives the volume of the pyramid
Mathematics
2 answers:
guajiro [1.7K]3 years ago
8 0

Answer:

D) V= 1/6 bh

Step-by-step explanation:

The expression that provides the volume of the pyramid is as follows;

As mentioned in the question

It is given that the pyramid put inside a prism

Also the pyramid contains the similar base area but the height is half of the prism

so the base area of the pyramid is b

And, the Height of the pyramid = h ÷ 2

Now as we know that  

The volume of the pyramid is

V = 1 ÷3 × base ares × height

= 1 ÷ 3 × b × h ÷ 2

V = 1 ÷ 6 bh

So, the volume is 1 ÷6 bh cube unit.

LekaFEV [45]3 years ago
5 0

Answer:

V = 1/6 BH

Step-by-step explanation:

the volume of a pyramid is 1/3 BH , since the pyramid is only half the height of the prism you'll have to multiply 1/3 x 2 = 1/6

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Answer:

D. 91%

Step-by-step explanation:

Conditional Probability

We use the conditional probability formula to solve this question. It is

P(B|A) = \frac{P(A \cap B)}{P(A)}

In which

P(B|A) is the probability of event B happening, given that A happened.

P(A \cap B) is the probability of both A and B happening.

P(A) is the probability of A happening.

In this question:

Event A: Less than 15 minutes.

Event B: Less than 10 minutes.

We are given the following probability distribution:

f(T = t) = \frac{3}{5}(\frac{5}{t})^4, t \geq 5

Simplifying:

f(T = t) = \frac{3*5^4}{5t^4} = \frac{375}{t^4}

Probability of arriving in less than 15 minutes:

Integral of the distribution from 5 to 15. So

P(A) = \int_{5}^{15} = \frac{375}{t^4}

Integral of \frac{1}{t^4} = t^{-4} is \frac{t^{-3}}{-3} = -\frac{1}{3t^3}

Then

\int \frac{375}{t^4} dt = -\frac{125}{t^3}

Applying the limits, by the Fundamental Theorem of Calculus:

At t = 15, f(15) = -\frac{125}{15^3} = -\frac{1}{27}

At t = 5, f(5) = -\frac{125}{5^3} = -1

Then

P(A) = -\frac{1}{27} + 1 = -\frac{1}{27} + \frac{27}{27} = \frac{26}{27}

Probability of arriving in less than 15 minutes and less than 10 minutes.

The intersection of these events is less than 10 minutes, so:

P(B) = \int_{5}^{10} = \frac{375}{t^4}

We already have the integral, so just apply the limits:

At t = 10, f(10) = -\frac{125}{10^3} = -\frac{1}{8}

At t = 5, f(5) = -\frac{125}{5^3} = -1

Then

P(A \cap B) = -\frac{1}{8} + 1 = -\frac{1}{8} + \frac{8}{8} = \frac{7}{8}

If given the train arrived in less than 15 minutes, what is the probability it arrived in less than 10 minutes?

P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{\frac{7}{8}}{\frac{26}{27}} = 0.9087

Thus 90.87%, approximately 91%, and the correct answer is given by option D.

3 0
3 years ago
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