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Lisa [10]
3 years ago
12

What is the are of a triangle (picture provided)

Mathematics
1 answer:
77julia77 [94]3 years ago
3 0

Answer:

Area Δ = 102.3 units² ⇒ The answer is (d)

Step-by-step explanation:

* Use the formula of the area:

∵ Area of the triangle = 1/2 (a)(b) sin(C)

∵ We have the length of the 3 sides

∴ Use cos Rule to find the angle C

∵ cos(C) = (a² + b² - c²)/2ab

∵ a = 25 , b = 13 , c = 17

∴ cos(C) = (25² + 13² - 17²)/2(25)(13) = 625 + 169 - 289/650 = 505/650

∴ m∠C = 39°

∴ Area Δ = (1/2)(25)(13)sin(39) = 102.3 units²

∴ The answer is (d)

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What is 10b +19 when b= -2
otez555 [7]

Answer:

-1    i hope this helps!    :)

Step-by-step explanation:

you are given both 10b + 19    and b = -2

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4 0
3 years ago
In the library on a university campus, there is a sign in the elevator that indicates a limit of 16 persons. In addition, there
lesya692 [45]

Answer:

15.87% probability that a random sample of 16 people will exceed the weight limit

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For sums, the theorem can be applied, with mean n*\mu and standard deviation s = \sqrt{n}*\sigma

In this problem, we have that:

n = 16, \mu = 16*150 = 2400, s = \sqrt{16}*27 = 108

If a random sample of 16 persons from the campus is to be taken, what is the chance that a random sample of 16 people will exceed the weight limit

This is 1 subtracted by the pvalue of Z when X = 2508. So

Z = \frac{X - \mu}{\sigma}

By the Central Limit Theorem

Z = \frac{X - \mu}{s}

Z = \frac{2508 - 2400}{108}

Z = 1

Z = 1 has a pvalue of 0.8413

1 - 0.8413 = 0.1587

15.87% probability that a random sample of 16 people will exceed the weight limit

3 0
3 years ago
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