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AveGali [126]
2 years ago
11

The box plots below show the distribution of salaries, in thousands, among employees of two small companies.

Mathematics
2 answers:
Shtirlitz [24]2 years ago
7 0

Answer:

Mean and IQR

Step-by-step explanation:

The measure of centre gives the central or the measure which gives the best mid term of a distribution. Based in the details of the box plot, the median is the value which divides the box in the box plot.

For company A:

Range = 25 to 80 with a median value at 30 ; this means the median does not give a good centre measure of the distribution ad it is very close to the minimum value. This goes for the Company B plot too; with values ranging from (35 to 90) and the median designated at 40.

Hence, the mean will be the best measure of centre rather Than the median in this case.

For the variability, the interquartile range would best suit the distribution. With the lower quartile and upper quartile both having reasonable width to the minimum and maximum value of the distribution.

jeyben [28]2 years ago
3 0

Answer:

B

Step-by-step explanation:

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A vcr and tv were bought for 8000 each. The shopkeeper made a loss of 4 per cent on vcr and 8 per cent prophit on the tv. Find t
aleksley [76]

Answer:

  2% gain

Step-by-step explanation:

We assume the shopkeeper bought the appliances at the indicated prices, and that gain is computed on the basis of that cost price.

Since the base cost is the same for each appliance, the percentages can be added directly to find the percentage gain on 8000. However, the shopkeeper's total outlay was 16000, not 8000, so the final gain percentage is half of that total.

  gain percent = (-4% + 8%)/2 = 2%

_____

If you want to see the actual numbers:

  Loss on VCR = 4% × 8000 = 320.

  Gain on TV = 8% × 8000 = 640.

Total gain on 16000 is -320 +640 = 320. As a percentage, that is ...

  320/16000 × 100% = 2%

7 0
3 years ago
Please help math experts need your help 15 points !
Alinara [238K]

Answer:

12 = -3 - 3x

Step-by-step explanation:

Same concept as the last question I answered. First, let's derive a formula from the given model. So we have 3 negative x and 3 negative 1 which equal to 12 positive ones, this can be written as:

-3x - 3 = 12

which can be rewritten as

12 = -3 - 3x

7 0
3 years ago
5 . 9 × ____ = 73.16 which is the missing factor?
ladessa [460]

Answer:

12.4

Step-by-step explanation:

73.16/5.9

3 0
3 years ago
Read 2 more answers
The taxi and takeoff time for commercial jets is a random variable x with a mean of 8.3 minutes and a standard deviation of 3.3
In-s [12.5K]

Answer:

a) There is a 74.22% probability that for 37 jets on a given runway, total taxi and takeoff time will be less than 320 minutes.

b) There is a 1-0.0548 = 0.9452 = 94.52% probability that for 37 jets on a given runway, total taxi and takeoff time will be more than 275 minutes.

c) There is a 68.74% probability that for 37 jets on a given runway, total taxi and takeoff time will be between 275 and 320 minutes.

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a random variable X, with mean \mu and standard deviation \sigma, a large sample size can be approximated to a normal distribution with mean \mu and standard deviation \frac{\sigma}{\sqrt{n}}.

Problems of normally distributed samples can be solved using 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 problem, we have that:

The taxi and takeoff time for commercial jets is a random variable x with a mean of 8.3 minutes and a standard deviation of 3.3 minutes. This means that \mu = 8.3, \sigma = 3.3.

(a) What is the probability that for 37 jets on a given runway, total taxi and takeoff time will be less than 320 minutes?

We are working with a sample mean of 37 jets. So we have that:

s = \frac{3.3}{\sqrt{37}} = 0.5425

Total time of 320 minutes for 37 jets, so

X = \frac{320}{37} = 8.65

This probability is the pvalue of Z when X = 8.65. So

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

Z = \frac{8.65 - 8.3}{0.5425}

Z = 0.65

Z = 0.65 has a pvalue of 0.7422. This means that there is a 74.22% probability that for 37 jets on a given runway, total taxi and takeoff time will be less than 320 minutes.

(b) What is the probability that for 37 jets on a given runway, total taxi and takeoff time will be more than 275 minutes?

Total time of 275 minutes for 37 jets, so

X = \frac{275}{37} = 7.43

This probability is subtracted by the pvalue of Z when X = 7.43

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

Z = \frac{7.43 - 8.3}{0.5425}

Z = -1.60

Z = -1.60 has a pvalue of 0.0548.

There is a 1-0.0548 = 0.9452 = 94.52% probability that for 37 jets on a given runway, total taxi and takeoff time will be more than 275 minutes.

(c) What is the probability that for 37 jets on a given runway, total taxi and takeoff time will be between 275 and 320 minutes?

Total time of 320 minutes for 37 jets, so

X = \frac{320}{37} = 8.65

Total time of 275 minutes for 37 jets, so

X = \frac{275}{37} = 7.43

This probability is the pvalue of Z when X = 8.65 subtracted by the pvalue of Z when X = 7.43.

So:

From a), we have that for X = 8.65, we have Z = 0.65, that has a pvalue of 0.7422.

From b), we have that for X = 7.43, we have Z = -1.60, that has a pvalue of 0.0548.

So there is a 0.7422 - 0.0548 = 0.6874 = 68.74% probability that for 37 jets on a given runway, total taxi and takeoff time will be between 275 and 320 minutes.

7 0
3 years ago
F(x) = 2x - 1, g(x) = 3x, h(x) = x2 + 1<br> Compute the following:<br> h(f(9))
Tamiku [17]
H(2)=5-F+g(2)=(f • h)(2)=f
6 0
3 years ago
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