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saveliy_v [14]
4 years ago
10

When a data set is normally distributed, about how much of the data fall within one standard deviation of the mean?

Mathematics
1 answer:
Bezzdna [24]4 years ago
6 0

That would be 68%   answer

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What is -1/3 plus 4/5
Doss [256]

-1/3 + 4/5

-5/15 + 12/15

7/15

Hope this helped!

-TTL

8 0
3 years ago
Read 2 more answers
The width of a rectangle is 1m less than half of its length, and the perimeter is 46m. Find the dimensions.
VLD [36.1K]
The width is 15.
The length is 31.
6 0
3 years ago
A chemical plant has an emergency alarm system. When an emergency situation exists, the alarm sounds with probability 0.95. When
Lapatulllka [165]

Answer:

6.56% probability that a real emergency situation exists.

Step-by-step explanation:

We have these following probabilities:

A 0.4% probability that a real emergency situation exists.

A 99.6% probability that a real emergency situation does not exist.

If an emergency situation exists, a 95% probability that the alarm sounds.

If an emergency situation does not exist, a 2% probability that the alarm sounds.

The problem can be formulated as the following question:

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:

What is the probability of a real emergency situation existing, given that the alarm has sounded.

P(B) is the probability of there being a real emergency situation. So P(B) = 0.004.

P(A/B) is the probability of the alarm sounding when there is a real emergency situation. So P(A/B) = 0.95.

P(A) is the probability of the alarm sounding. This is 95% of 0.4%(real emergency situation) and 2% of 99.6%(no real emergency situation). So

P(A) = 0.95*0.04 + 0.02*0.996 = 0.05792

Given that the alarm has just sounded, what is the probability that a real emergency situation exists?

P = \frac{P(B).P(A/B)}{P(A)} = \frac{0.004*0.95}{0.05792} = 0.0656

6.56% probability that a real emergency situation exists.

6 0
3 years ago
Ratio equivalent to 2:1
Nataliya [291]

Answer:

4:2

8:4

12:6

16:8

6 0
3 years ago
9.W
gogolik [260]
The correct answer is A
5 0
4 years ago
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