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3 years ago

Which number line correctly illustrates 0.97 ?

Mathematics

1 answer:

Verdich [7]3 years ago

8 0

Answer:

Number Line C shows 0.97 exactly.

Step-by-step explanation:

Look the number line starts from 0.95.

So the 3rd line shows 0.97.

I hope it helps.

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The time taken to assemble a laptop computer in a certain plant is a random variable having a normal distribution of 20 hours an

ludmilkaskok [199]

Answer:

a) 40.13% probability that a laptop computer can be assembled at this plant in a period of time of less than 19.5 hours.

b) 34.13% probability that a laptop computer can be assembled at this plant in a period of time between 20 hours and 22 hours.

Step-by-step explanation:

When the distribution is normal, we use the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore 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 pvalue, we get the probability that the value of the measure is greater than X.

In this question:

$\mu = 20, \sigma = 2$

a)Less than 19.5 hours?

This is the pvalue of Z when X = 19.5. So

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

$Z = \frac{19.5 - 20}{2}$

$Z = -0.25$

$Z = -0.25$ has a pvalue of 0.4013.

40.13% probability that a laptop computer can be assembled at this plant in a period of time of less than 19.5 hours.

b)Between 20 hours and 22 hours?

This is the pvalue of Z when X = 22 subtracted by the pvalue of Z when X = 20. So

X = 22

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

$Z = \frac{22 - 20}{2}$

$Z = 1$

$Z = 1$ has a pvalue of 0.8413

X = 20

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

$Z = \frac{20 - 20}{2}$

$Z = 0$

$Z = 0$ has a pvalue of 0.5

0.8413 - 0.5 = 0.3413

34.13% probability that a laptop computer can be assembled at this plant in a period of time between 20 hours and 22 hours.

4 0

3 years ago

What is the value of x

KIM [24]

Answer is the letter A

5 0

3 years ago

1. 8x-2y = -24

mel-nik [20]

Answer:

8x-2y = -24

Step-by-step explanation:

Slope = 34

Y-intercept = 7

7 0

3 years ago

What is the solution of 64 x 64 to the power of 2?

Andrei [34K]

Answer:

16777216

Step-by-step explanation:

64 times 64 equals 4096. 4096 to the power of 2 is 16777216.

8 0

3 years ago

Suppose that you had the following data set. 500 200 250 275 300 Suppose that the value 500 was a typo, and it was suppose to be

hodyreva [135]

Answer:

$\bar X_B = \frac{\sum_{i=1}^5 X_i}{5} =\frac{500+200+250+275+300}{5}=\frac{1525}{5}=305$

$s_B = \sqrt{\frac{\sum_{i=1}^5 (X_i-\bar X)^2}{n-1}}=\sqrt{\frac{(500-305)^2 +(200-305)^2 +(250-305)^2 +(275-305)^2 +(300-305)^2)}{5-1}} = 115.108$

$\bar X_A = \frac{\sum_{i=1}^5 X_i}{5} =\frac{-500+200+250+275+300}{5}=\frac{525}{5}=105$

$s_A = \sqrt{\frac{\sum_{i=1}^5 (X_i-\bar X)^2}{n-1}}=\sqrt{\frac{(-500-105)^2 +(200-105)^2 +(250-105)^2 +(275-105)^2 +(300-105)^2)}{5-1}} = 340.221$

The absolute difference is:

$Abs = |340.221-115.108|= 225.113$

If we find the % of change respect the before case we have this:

$\% Change = \frac{|340.221-115.108|}{115.108} *100 = 195.57\%$

So then is a big change.

Step-by-step explanation:

The subindex B is for the before case and the subindex A is for the after case

Before case (with 500)

For this case we have the following dataset:

500 200 250 275 300

We can calculate the mean with the following formula:

$\bar X_B = \frac{\sum_{i=1}^5 X_i}{5} =\frac{500+200+250+275+300}{5}=\frac{1525}{5}=305$

And the sample deviation with the following formula:

$s_B = \sqrt{\frac{\sum_{i=1}^5 (X_i-\bar X)^2}{n-1}}=\sqrt{\frac{(500-305)^2 +(200-305)^2 +(250-305)^2 +(275-305)^2 +(300-305)^2)}{5-1}} = 115.108$

After case (With -500 instead of 500)

For this case we have the following dataset:

-500 200 250 275 300

We can calculate the mean with the following formula:

$\bar X_A = \frac{\sum_{i=1}^5 X_i}{5} =\frac{-500+200+250+275+300}{5}=\frac{525}{5}=105$

And the sample deviation with the following formula:

$s_A = \sqrt{\frac{\sum_{i=1}^5 (X_i-\bar X)^2}{n-1}}=\sqrt{\frac{(-500-105)^2 +(200-105)^2 +(250-105)^2 +(275-105)^2 +(300-105)^2)}{5-1}} = 340.221$

And as we can see we have a significant change between the two values for the two cases.

The absolute difference is:

$Abs = |340.221-115.108|= 225.113$

If we find the % of change respect the before case we have this:

$\% Change = \frac{|340.221-115.108|}{115.108} *100 = 195.57\%$

So then is a big change.

8 0

3 years ago