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3 years ago

Calculate the measure of segment KL HELP PLS

Mathematics

1 answer:

iVinArrow [24]3 years ago

5 0

10x+3 = 3x+24

10x-3x= 24-3

7x=21

X= 21/7

Final answer:

X=3

I’m not so sure of my answer but wish you the best and good luck!!

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A square lot has sides that are 32 meters long. What is the perimeter of the lot

Hatshy [7]

To find this you would do 32+32+32+32= 128 meters.

4 0

3 years ago

Read 2 more answers

Shelby tossed 3 quarters into the air. What is the probability that all 3 quarters will land heads up

BaLLatris [955]

Answer:

For 3 quarters, the probability that all 3 quarters will land heads up is ;

P = 1/2 × 1/2 × 1/2 = 1/8

P = 1/8

Step-by-step explanation:

The probability that one quarter will land heads up is 1/2.

For 3 quarters, the probability that all 3 quarters will land heads up is ;

P = 1/2 × 1/2 × 1/2 = 1/8

P = 1/8

4 0

4 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

ABCD ~ WXYZ. AD = 6, DC = 3, and WZ = 59. Find YZ.

dusya [7]

You take alll the answers and just divide each one

5 0

4 years ago

A school team won 6 games this year against 4 games won last year . What is the percent increase .

Y_Kistochka [10]

Answer:

The increase in the number of wins (or amount of change) =6−4=2

Percentage increase

original amount or base

amount of change

×100

=50%

8 0

3 years ago