If A is 5 and B is 12, what is C?

A 2-column table with 5 rows. The first column is labeled hours spent exercising with entries 2, 4, 3, 4, 5. The second column i

slamgirl [31]

Answer:

second answer or B

Step-by-step explanation:

5 0

3 years ago

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Two angles are complementary. One angle is labeled 2p+6°. The other angles is labeled 4p°. Find the value of p.

makkiz [27]

Answer:

Value of p is 14

Step-by-step explanation:

$given \: complementary \: angles \: add \: up \: to \: 90 \: degrees \\ then \\ 2p + 6 + 4p = 90 \\ 6p + 6 = 90 \\ 6(p + 1) = 90 \\ p + 1 = \frac{90}{6} \\ p + 1 = 15 \\ p = 15 - 1 \\ p = 14$

HAVE A NICE DAY!

THANKS FOR GIVING ME THE OPPORTUNITY TO ANSWER YOUR QUESTION. 

5 0

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.

3 0

3 years ago

GUYS PLEASE HELP ME PLEASE HELP HELP PLEASE HELP ASAP GUYS HELP!!!!!

slega [8]

Answer:

$SA=5\ yd^2$

Step-by-step explanation:

we know that

The surface area of the square pyramid is equal to the area of the square base plus the area of its four triangular faces

so

$SA=b^2+4[\frac{1}{2}(b)(h)]$

we have

$b=1\ yd\\h=2\ yd$

substitute

$SA=1^2+4[\frac{1}{2}(1)(2)]$

$SA=5\ yd^2$

5 0

3 years ago

What is the value of 4^-2? Leave your answer in fraction form of necessary.

valkas [14]

Answer:

1/16

Step-by-step explanation:

We know that a^ -b is equal to 1 / a^b

That means

4^-2 = 1/ 4^2 = 1/16

8 0

4 years ago

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