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Aleonysh [2.5K]
2 years ago
5

3. Bob the plumber charges $15 per hour plus a service fee of $40 for coming out to your

Mathematics
1 answer:
Tju [1.3M]2 years ago
5 0
15 + 40 = 55
55 x 5 = 275
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Which statement is true about the ranges for the box plots?
agasfer [191]

Answer:

The range of the Morning box plot is the same as the range of the Afternoon box plot.

Step-by-step explanation:

Data provided in the question

Number line = 0 to 16

Morning whiskers range = 3 to 15

Box range = 5 to 12

Line divided = 8

Afternoon whiskers range = 4 to 16

Box range = 8 to 15

Line divided = 14

Based on the above information,

The morning whiskers range difference is

= 12 - 5

= 7

And, the afternoon whiskers range difference is

= 15 - 8

= 7

Therefore the first option is correct

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3 years ago
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zloy xaker [14]
D is the correct answer
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2 years ago
Put the numbers in order from least to greatest. ITEM BANK: Move to Bottom 0.00025 0.025 0.025 x 10 0.5 x 10-5 1.25 x 10-1 1.45
LuckyWell [14K]

Answer:

Least to Greatest:

0.5*10^-5, 0.00025, 3.5 x 10^-4, 1.45 x 10^-3, 25.4 x 10^-4, 0.025, 0.025 x 10, 1.25 x 10^-1

4 0
3 years ago
What is the answer to 12x23
Triss [41]
The answer to 12x23 is 276
8 0
3 years ago
A simple random sample of size nequals10 is obtained from a population with muequals68 and sigmaequals15. ​(a) What must be true
valentina_108 [34]

Answer:

(a) The distribution of the sample mean (\bar x) is <em>N</em> (68, 4.74²).

(b) The value of P(\bar X is 0.7642.

(c) The value of P(\bar X\geq 69.1) is 0.3670.

Step-by-step explanation:

A random sample of size <em>n</em> = 10 is selected from a population.

Let the population be made up of the random variable <em>X</em>.

The mean and standard deviation of <em>X</em> are:

\mu=68\\\sigma=15

(a)

According to the Central Limit Theorem if we have a population with mean <em>μ</em> and standard deviation <em>σ</em> and we take appropriately huge random samples (<em>n</em> ≥ 30) from the population with replacement, then the distribution of the sample mean will be approximately normally distributed.

Since the sample selected is not large, i.e. <em>n</em> = 10 < 30, for the distribution of the sample mean will be approximately normally distributed, the population from which the sample is selected must be normally distributed.

Then, the mean of the distribution of the sample mean is given by,

\mu_{\bar x}=\mu=68

And the standard deviation of the distribution of the sample mean is given by,

\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}=\frac{15}{\sqrt{10}}=4.74

Thus, the distribution of the sample mean (\bar x) is <em>N</em> (68, 4.74²).

(b)

Compute the value of P(\bar X as follows:

P(\bar X

                    =P(Z

*Use a <em>z</em>-table for the probability.

Thus, the value of P(\bar X is 0.7642.

(c)

Compute the value of P(\bar X\geq 69.1) as follows:

Apply continuity correction as follows:

P(\bar X\geq 69.1)=P(\bar X> 69.1+0.5)

                    =P(\bar X>69.6)

                    =P(\frac{\bar X-\mu_{\bar x}}{\sigma_{\bar x}}>\frac{69.6-68}{4.74})

                    =P(Z>0.34)\\=1-P(Z

Thus, the value of P(\bar X\geq 69.1) is 0.3670.

7 0
3 years ago
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