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Alex73 [517]
2 years ago
6

A 4-column table with 5 rows. The first column is labeled age (years) with entries 1, 2, 3, 4, 5. The second column is labeled g

iven value with entries 15, 12, 9, 5, 4. The third column is labeled predicted with entries 14.8, 11.9, 9, 6.1, 3.2. The fourth column is labeled residual with entries 0.2, 0.1, 0, negative 1.1, 0.8. Which point would be on a residual plot of the data?
Mathematics
2 answers:
Rashid [163]2 years ago
8 0

Answer:

3,0

Step-by-step explanation:

sweet [91]2 years ago
5 0

Answer:

any other info?

Step-by-step explanation:

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PLZ HELP<br><br> Factor:<br><br> x^9 - 9x + 20
Blizzard [7]

Answer:

The problem can't be factor

Step-by-step explanation:


8 0
3 years ago
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someone please help, i don’t understand this and i’m stuck. this is my last homework i need to do for today. if you can help tha
Ad libitum [116K]

Answer:

0.38 and 0.7 maybe it's the answer

Step-by-step explanation:

power of 10 in above questions are 2 and 3 respectively

5 0
2 years ago
THIS IS THE LAST ONE THX
Nataly_w [17]
Letter D it’s incorrect, not sure tho
7 0
3 years ago
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Sharon is twice as old as Brian, and in 5 years the sum of their ages will be 25 years. Find their present ages.
Vinil7 [7]
Sharon's age= 2x
Brian's age= x

(2x-5)+(x-5)= 25
3x-10=25
3x=15
x=5
Brian is 5 years old and Sharon is 10

3 0
3 years ago
Suppose that an airline uses a seat width of 16.5 in. Assume men have hip breadths that are normally distributed with a mean of
Alexxx [7]

Answer:

a) 0.018

b) 0            

Step-by-step explanation:

We are given the following information in the question:

Mean, μ =  14.4 in

Standard Deviation, σ = 1 in

We are given that the distribution of breadths is a bell shaped distribution that is a normal distribution.

Formula:

z_{score} = \displaystyle\frac{x-\mu}{\sigma}

a) P(breadth will be greater than 16.5 in)

P(x > 16.5)

P( x > 16.5) = P( z > \displaystyle\frac{16.5 - 14.4}{1}) = P(z > 2.1)

= 1 - P(z \leq 2.1)

Calculation the value from standard normal z table, we have,  

P(x > 16.5) = 1 - 0.982 = 0.018 = 1.8\%

0.018 is the probability that if an individual man is randomly​ selected, his hip breadth will be greater than 16.5 in.

b) P( with 123 randomly selected​ men, these men have a mean hip breadth greater than 16.5 in)

Formula:

z_{score} = \displaystyle\frac{x-\mu}{\frac{\sigma}{\sqrt{n}}}  

P(x > 16.5)  

P( x > 16.5) = P( z > \displaystyle\frac{16.5-14.4}{\frac{1}{\sqrt{123}}}) = P(z > 23.29)  

= 1 - P(z \leq 23.29)

Calculation the value from standard normal z table, we have,  

P(x > 16.5) = 1 - 1 = 0

There is 0 probability that 123 randomly selected men have a mean hip breadth greater than 16.5 in

4 0
3 years ago
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