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Ganezh [65]
3 years ago
13

-5x-4y=-7 on a graph

Mathematics
1 answer:
Flauer [41]3 years ago
3 0

Answer:

Graph is attached below.

Step-by-step explanation:

If you want to solve for the variable x,

x = −4 / 5y + 7 / 5

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Find the slope of a line given the following two points: (4, 3) and (5, 2). 1 -1 2 -2
galben [10]
Given two points; (x₁,y₁) and (x₂,y₂) the slope is:

m=(y₁-y₂) / (x₁-x₂)      or      m=(y₂-y₁) / (x₂-x₁)

In this case:
(4,3)
(5,2)

m=(5-4)/(2-3)=1/-1=-1

Answer: the slope is -1
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3 years ago
Start time 6:57pm elasped time 47 min
Talja [164]
7:44 pm would be correct
3 0
3 years ago
PLEASE HELP ME!!
solniwko [45]

Answer:

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Step-by-step explanation:

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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?

So

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
Can someone help me with a b and c please
abruzzese [7]

Answer:

c.in this 2014,396.12million pets were owned

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