Based on the scatterplot below, what is the most likely value for “Pounds of Apples” when “Number of Apples” is 80?

Mathematics

2 answers:

kodGreya [7K]3 years ago

5 0

The answer you are looking for is 42.

ollegr [7]3 years ago

3 0

Find the approx. slope of the regression line through the given 5 points. As x increases from 9 to 45 (a change of +36), y increases from 5 to 25, or 20.
20 5
Thus, the slope of the regression line is approx. m = ------ = -----
36 9
We need to continue, to find the equation of this regression line.

Since the slope is 5/9 and one point on the line is (25,12), the equation of the line is:

y-12 = (5/9)(x-25).

Now let x = 80 and calculate y: y = 12 + (5/9)(55) = 12 = 42.6

When the # of apples is 80, the weight of the apples is approx. 43 lb.

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valkas [14]

I believe it’s 7/10

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

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