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3 years ago

WILL GIVE BRAINLIEST

Mathematics

2 answers:

ludmilkaskok [199]3 years ago

8 0

Answer:

Air that is moving primarily in a horizontal direction.

Alex777 [14]3 years ago

8 0

Answer:

Wind is the movement of air horizontally.

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In ΔWXY, W Y ‾ WY is extended through point Y to point Z, m ∠ Y W X = ( 2 x + 12 ) ∘ m∠YWX=(2x+12) ∘ , m ∠ W X Y = ( 2 x + 5 ) ∘

oksian1 [2.3K]

Answer:

∠XVW = 36

Step-by-step explanation:

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2 years ago

15 more than twice a number

coldgirl [10]

Answer:

15×2x

Step-by-step explanation:

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4 years ago

Read 2 more answers

Solve the inequality<br>-6p<12 <br><br><br>Ah this is due by tomorrow

777dan777 [17]

-6p < 12 Divide -6 on both sides to get "p" by itself

$\frac{-6p}{-6} < \frac{12}{-6}$

p > -2

When you divide or multiply a negative number to both sides of the inequality, the sign (< , > , ≤ , ≥) is flipped.

3 0

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.

3 0

3 years ago

Help plzzzzzzzzzzzzzzzzzzzz

bulgar [2K]

Answer:

y = -4

Step-by-step explanation:

First, let's find the slope. Since the line is horizontal, it's slope has to be 0 (this is because slope = rise/run, there is no "rise" in a horizontal line therefore the slope is 0)

Now, we need to find the y-intercept. This is where the line intercepts the y-axis. Looking at the graph, the line intercepts the axis at (0,-4), therefore the y-intercept is -4.

The equation would be:

y=0x-4

y=-4

5 0

2 years ago