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3 years ago

This is my very first trig problem and I'd like to know how to tackle this. Please explain how I can do problems like these in t

he future!

Mathematics

1 answer:

Paha777 [63]3 years ago

5 0

$\tan(50°) = \frac{QR}{PQ} \Leftrightarrow QR = PQ \tan(50°) = 8.1 \tan(50°) \approx9.65 \\ \cos(50°) = \frac{PQ}{PR} \Leftrightarrow PR = \frac{PQ}{\cos(50°)} = \frac{8.1}{\cos(50°)} \approx12.6$

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It’s B because it’s smaller than -3 but slightly greater than -4

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3 years ago

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Which of the following fractions can be represented by a terminating decimal?

Karo-lina-s [1.5K]

Answer:

B) 13/8

Step-by-step explanation:

First we need to understand what a terminating decimal is: It is a decimal that has a finite number of decimal values that are not zero, meaning that its decimal values end at some point

Let's go through our possible answers with trial and error:

A) 8/9

8/9 = 0.888888888888...9 (Incorrect)

B) 13/8

13/8 = 1.625 (Correct)

C) 4/3

4/3 = 1.3333333333...4 (Incorrect)

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2 years ago

George and Davon begin their hike at 9 a.m. one morning. They plan to hike from the 2 1/5 -mile marker to the 9 1/15 -mile marke

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Yes because when your multiply every thing together you’ll have 9/15 miles

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3 years ago

Suppose a particular type of cancer has a 0.9% incidence rate. Let D be the event that a person has this type of cancer, therefo

natita [175]

Answer:

There is a 12.13% probability that the person actually does have cancer.

Step-by-step explanation:

We have these following probabilities.

A 0.9% probability of a person having cancer

A 99.1% probability of a person not having cancer.

If a person has cancer, she has a 91% probability of being diagnosticated.

If a person does not have cancer, she has a 6% probability of being diagnosticated.

The question can be formulated as the following problem:

What is the probability of B happening, knowing that A has happened.

It can be calculated by the following formula

$P = \frac{P(B).P(A/B)}{P(A)}$

Where P(B) is the probability of B happening, P(A/B) is the probability of A happening knowing that B happened and P(A) is the probability of A happening.

In this problem we have the following question

What is the probability that the person has cancer, given that she was diagnosticated?

P(B) is the probability of the person having cancer, so $P(B) = 0.009$

P(A/B) is the probability that the person being diagnosticated, given that she has cancer. So $P(A/B) = 0.91$

P(A) is the probability of the person being diagnosticated. If she has cancer, there is a 91% probability that she was diagnosticard. There is also a 6% probability of a person without cancer being diagnosticated. So

$P(A) = 0.009*0.91 + 0.06*0.991 = 0.06765$

What is the probability that the person actually does have cancer?

$P = \frac{P(B).P(A/B)}{P(A)} = \frac{0.91*0.009}{0.0675} = 0.1213$

There is a 12.13% probability that the person actually does have cancer.

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3 years ago

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