ANSWER ASAP, FIND LENGTH OF MISSING SIDE USING PYTHAGOREOON THEOREM

Mathematics

1 answer:

miss Akunina [59]3 years ago

5 0

Answer:

51?

Step-by-step explanation:

I am sorry if it's wrong

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How can you write 2,000 in a exponent

Mnenie [13.5K]

1,000 = 10^3 (3 because there are 3 digits after the first one)

2,000 = 2 * 10^3

3 0

3 years ago

Each side of the square below is 8 inches. A triangle inside of a square. The top of the triangle divides a side of the square i

dmitriy555 [2]

The probability that a point is chosen at random in the square is in the blue region is 0.8.

<h3>What is Probability?</h3>

The probability helps us to know the chances of an event occurring.

$\rm{Probability=\dfrac{Desired\ Outcomes}{Total\ Number\ of\ outcomes\ possible}$

As we know that the area of the shaded region is the sum of the area of triangle A and the area of triangle B. Therefore, the area of the blue shaded region is,

$\text{Area of shaded region} = (\dfrac{1}{2} \times 8 \times 8) + (\dfrac{1}{2} \times 4 \times 8)$

$= (32) + (16)\\\\= 48\rm\ in^2$

The area of the square can be written as,

$\text{Area of Square} = \rm(Side)^2 = 8^2 = 64\ in^2$

Now, the probability that a point is chosen at random in the square is in the blue region can be written as,

$\rm{Probability=\dfrac{Desired\ Outcomes}{Total\ Number\ of\ outcomes\ possible}\\\\$

$\rm Probability = \dfrac{\text{Area of blue region}}{\text{Area of square}}\\\\Probability = \dfrac{48}{60} = 0.8$

Hence, the probability that a point is chosen at random in the square is in the blue region is 0.8.

Learn more about Probability:

brainly.com/question/795909

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2 years ago

In the upcoming election for governor, the most recent poll, based on 900 respondents, predicts that the incumbent will be reele

Paul [167]

Answer:

.0166

Step-by-step explanation:

Central Limit Theorem for proportions:

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean $\mu = p$ and standard deviation $\sigma_{p} = \sqrt{\frac{p(1-p)}{n}}$