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Alenkinab [10]
2 years ago
5

What is the decimal expansion of the following fraction? 1/8 A. 0.125 B. 0.125 C. 1.8 D. 0.18​

Mathematics
1 answer:
qaws [65]2 years ago
3 0

Answer:

0.125....................

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(3 marks) A certain type of storage battery lasts, on average, 3.0 years with a standard deviation of 0.5 year. The battery live
Ksenya-84 [330]

Answer:

0.8041 = 80.41% probability that a given battery will last between 2.3 and 3.6 years

Step-by-step explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the z-score of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

A certain type of storage battery lasts, on average, 3.0 years with a standard deviation of 0.5 year

This means that \mu = 3, \sigma = 0.5

What is the probability that a given battery will last between 2.3 and 3.6 years?

This is the p-value of Z when X = 3.6 subtracted by the p-value of Z when X = 2.3. So

X = 3.6

Z = \frac{X - \mu}{\sigma}

Z = \frac{3.6 - 3}{0.5}

Z = 1.2

Z = 1.2 has a p-value of 0.8849

X = 2.3

Z = \frac{X - \mu}{\sigma}

Z = \frac{2.3 - 3}{0.5}

Z = -1.4

Z = -1.4 has a p-value of 0.0808

0.8849 - 0.0808 = 0.8041

0.8041 = 80.41% probability that a given battery will last between 2.3 and 3.6 years

4 0
2 years ago
2x +3y =6 Standard form to slope intercept form
makvit [3.9K]
2x+3y=6
3y=-2x+6
3y/3=-2x+6/3
Y=-2/3x+2
3 0
3 years ago
Find the value of x 50 and please explain how?
nikklg [1K]

Answer:

180-50=130

So the answer is 130

7 0
2 years ago
Read 2 more answers
a slitter assembly contains 48 blades five blades are selected at random and evaluated each day for sharpness if any dull blade
12345 [234]

Answer:

P(at least 1 dull blade)=0.7068

Step-by-step explanation:

I hope this helps.

This is what it's called dependent event probability, with the added condition that at least 1 out of 5 blades picked is dull, because from your selection of 5, you only need one defective to decide on replacing all.

So if you look at this from another perspective, you have only one event that makes it so you don't change the blades: that 5 out 5 blades picked are sharp. You also know that the probability of changing the blades plus the probability of not changing them is equal to 100%, because that involves all the events possible.

P(at least 1 dull blade out of 5)+Probability(no dull blades out of 5)=1

P(at least 1 dull blade)=1-P(no dull blades)

But the event of picking one blade is dependent of the previous picking, meaning there is no chance of picking the same blade twice.

So you have 38/48 on getting a sharp one on your first pick, then 37/47 (since you remove 1 sharp from the possibilities, and 1 from the whole lot), and so on.

Also since are consecutive events, you need to multiply the events.

The probability that the assembly is replaced the first day is:

P(at least 1 dull blade)=1-P(no dull blades)

P(at least 1 dull blade)=1-(\frac{38}{48}* \frac{37}{47} *\frac{36}{46}*\frac{35}{45}*\frac{34}{44})

P(at least 1 dull blade)=1-0.2931

P(at least 1 dull blade)=0.7068

6 0
2 years ago
Solve the equation 20.5 = x + 4
devlian [24]

Answer:

16.5

Step-by-step explanation:

20.5 = x + 4

16.5 = x

x = 16.5

4 0
3 years ago
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