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Neko [114]
3 years ago
15

Find the surface area of this cube

Mathematics
2 answers:
alukav5142 [94]3 years ago
7 0
Okay so I will not be doing an explanation cause I do them to long so I will just give you the answer 512^2 in.Hope this helps :)
Mkey [24]3 years ago
3 0

Answer:

512 in^3

Step-by-step explanation:

It would be length x height x width = AREA

8 x 8 x 8= 512 and since its 3d it will be 512^3

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2x + 3/ x - 4 - 2x - 8/ 2× + 1 = 1
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6 0
3 years ago
20 Points:
belka [17]

Answer:

D. 105 students and 150 guests

Step-by-step explanation:

105 students and $1 for each ticket is $105.(105 x 1 = 105)

150 guests and $5 for each ticket is $750.(150 x 5 =750)

$105 + $750 = $855

6 0
3 years ago
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Katie played 5 consecutive games of soccer without being taken off the field. Then, after a single game on the sidelines, she pl
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3 years ago
Waiting on the platform, a commuter hears an announcement that the train is running five minutes late. He assumes the arrival ti
natima [27]

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.

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Step-by-step explanation:

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