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statuscvo [17]
3 years ago
10

Can someone please answer this question please answer it correctly and please show work please help me I need it

Mathematics
1 answer:
Dennis_Churaev [7]3 years ago
5 0

Answer:

$ -400

Step-by-step explanation:

Let's find each year's profits:

2010 = (-400)

2011 = (-200)

2012 = (200)

To find the total profits from 2010 - 2012, we will add all the values.

(-400) + (-200) + 200

(-600) + 200

200 - 600 = (-400)

$ -400

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Find the midpoint of the line segment JK if j (4,6) and k(0,-4)
Snowcat [4.5K]

Answer:

(2,1)

General Formulas and Concepts:

Order of Operations: BPEMDAS

Midpoint Formula: (\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})

Step-by-step explanation:

<u>Step 1: Define points</u>

J (4, 6)

K (0, -4)

<u>Step 2: Find midpoint</u>

  1. Substitute:                         (\frac{4+0}{2},\frac{6-4}{2})
  2. Add/Subtract:                    (\frac{4}{2},\frac{2}{2})
  3. Divide:                               (2,1)
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Dominik [7]

Answer:

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What is the correct answer to 42r - 18
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Answer:

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Simplify the Square root of 72a6b3c2
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5 0
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The overhead reach distances of adult females are normally distributed with a mean of 205 cm and a standard deviation of 7.8 cm.
devlian [24]

Answer:

Given the mean = 205 cm and standard deviation as 7.8cm

a. To calculate the probability that an individual distance is greater than 218.4 cm, we subtract the probability of the distance given (i.e 218.4 cm) from the mean (i.e 205 cm) divided by the standard deviation (i.e 7.8cm) from 1. Therefore, we have 1- P(Z\leq 1.72). Using the Z distribution table we have 1-0.9573. Therefore P(X >218.4)= 0.0427.

b. To calculate the probability that mean of 15 (i.e n=15) randomly selected distances is greater than 202.8, we subtract the probability of the distance given (i.e 202.8cm) from the mean (i.e 205 cm) divided by the standard deviation (i.e 7.8cm) divided by the square root of mean (i.e n= 15)  from 1. Therefore, we have 1- P(Z\leq -1.09). Using the Z distribution table we have 1-0.1378. Therefore P(X >202.8)= 0.8622.

c. This will also apply to a normally distributed data even if it is not up to the sample size of 30 since the sample distribution is not a skewed one.

Step-by-step explanation:

Given the mean = 205 cm and standard deviation as 7.8cm

a. To calculate the probability that an individual distance is greater than 218.4 cm, we subtract the probability of the distance given (i.e 218.4 cm) from the mean (i.e 205 cm) divided by the standard deviation (i.e 7.8cm) from 1. Therefore, we have 1- P(Z\leq 1.72). Using the Z distribution table we have 1-0.9573. Therefore P(X >218.4)= 0.0427.

b. To calculate the probability that mean of 15 (i.e n=15) randomly selected distances is greater than 202.8, we subtract the probability of the distance given (i.e 202.8cm) from the mean (i.e 205 cm) divided by the standard deviation (i.e 7.8cm) divided by the square root of mean (i.e n= 15)  from 1. Therefore, we have 1- P(Z\leq -1.09). Using the Z distribution table we have 1-0.1378. Therefore P(X >202.8)= 0.8622.

c. This will also apply to a normally distributed data even if it is not up to the sample size of 30 since the sample distribution is not a skewed one.

4 0
3 years ago
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