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3 years ago

Help me l,ll give you brainest

Mathematics

2 answers:

natima [27]3 years ago

7 0

Answer:

Step-by-step explanation:

he did the question wrong completely, so, check #4, #2, and #3

stepan [7]3 years ago

7 0

Answer:

the last three choices

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A rectangle measures 2 1/2 inches by 1 5/16 inches. What is its area?

Norma-Jean [14]

Answer:

it would be 3.28125

Step-by-step explanation:

do base times height to get the area

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3 years ago

The amount of money A accrued at the end of

Rudiy27

B is the answer no capppppp

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2 years ago

A kite string is 35m long. The angle the string makes with the ground is 50o

kipiarov [429]

Answer:

22.50m

Step-by-step explanation:

With the ground, the horizontal ground distance is the adjacent side, so by SOHCAHTOA, the ratio used is cosine, i.e.

Ground distance

= hypotenuse * cos(50°)

= 35 cos(50°)

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7 0

3 years ago

8g - 7 = 5g + 6. What is g?

Whitepunk [10]

In fraction form g would be 13/3, decimal form is 4.3

4 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