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3 years ago

12

PLEASE HELP ASAP!!!! I GIVE A BRAINLIEST TO THE RIGHT ANSWER!!!!

1 answer:

Mandarinka [93]3 years ago

4 0

Use cosine law to solve the problem. See image attached.
What we're looking for is the length of side A (a side which is opposite with angle A).

This is the formula of cosine law
A² = B² + C² - 2BC cos A

Input the numbers
A² = 75² + 90² - 2(75)(90) cos 85°
A = $\sqrt{75^{2}+90^{2}-2(75)(90)cos85^{0}}$
A = $\sqrt{90^{2}+75^{2}-2(90)(75)cos85^{0}}$

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The simple interest on a sum of money invested for 6 months at 5% per annum is $1680. Find the amount of money invested

olga55 [171]

9514 1404 393

Answer:

$67,200

Step-by-step explanation:

The amount if interest is given by the formula ...

I = Prt

where P is the amount invested, r is the annual rate, and t is the number of years. Here, we want to find P when r = 0.05 and t = 1/2 such that I = 1680.

P = I/(rt) = $1680/(0.05·0.5) = $1680/0.025

P = $67,200

The amount invested was $67,200.

3 0

3 years ago

A research study uses 800 men under the age of 55. Suppose that 30% carry a marker on the male chromosome that indicates an incr

ss7ja [257]

Answer:

a) There is a 12.11% probability that exactly 1 man has the marker.

b) There is a 85.07% probability that more than 1 has the marker.

Step-by-step explanation:

There are only two possible outcomes: Either the men has the chromosome, or he hasn't. So we use the binomial probability distribution.

Binomial probability

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}$

In which $C_{n,x}$ is the number of different combinatios of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And $\pi$ is the probability of X happening.

In this problem, we have that:

30% carry a marker on the male chromosome that indicates an increased risk for high blood pressure, so $\pi = 0.30$

(a) If 10 men are selected randomly and tested for the marker, what is the probability that exactly 1 man has the marker?

10 men, so $n = 10$

We want to find P(X = 1). So:

$P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}$

$P(X = 1) = C_{10,1}.(0.30)^{1}.(0.7)^{9} = 0.1211$

There is a 12.11% probability that exactly 1 man has the marker.

(b) If 10 men are selected randomly and tested for the marker, what is the probability that more than 1 has the marker?

That is $P(X > 1)$

We have that:

$P(X \leq 1) + P(X > 1) = 1$

$P(X > 1) = 1 - P(X \leq 1)$

We also have that:

$P(X \leq 1) = P(X = 0) + P(X = 1)$

$P(X = 0) = C_{10,0}.(0.30)^{0}.(0.7)^{10} = 0.0282$

So

$P(X \leq 1) = P(X = 0) + P(X = 1) = 0.0282 + 0.1211 = 0.1493$

Finally

$P(X > 1) = 1 - P(X \leq 1) = 1 - 0.1493 = 0.8507$

There is a 85.07% probability that more than 1 has the marker.

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mariarad [96]

Answer:

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Step-by-step explanation:

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Step-by-step explanation:

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Fittoniya [83]

This would be 8 feet

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