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Margaret [11]
2 years ago
12

Rita is starting a running program. the table shows the total number of miles she runs in different weeks. what is the equation

of the line of best fit for the data? state each number to the thousandths place. y ≈ x
Mathematics
2 answers:
Stolb23 [73]2 years ago
7 0

Based on the table, the equation of the line of best fit to the nearest thousandths is equal to y = 1.671x + 4.699.

<h3>How to determine the equation of the line of best fit?</h3>

Mathematically, the equation of the line of best fit is calculated by using this formula:

y = mx + c

<u>Where:</u>

  • m is the slope.
  • c is the intercept.

Based on the table, the slope and intercept to the nearest thousandths are equal to 1.671 and 4.699 respectively.

Therefore, the equation of the line of best fit is equal to y = 1.671x + 4.699.

Read more on line of best fit here: brainly.com/question/4674926

#SPJ4

Schach [20]2 years ago
4 0

Answer:

Y=1.671 and X=4.699

Step-by-step explanation:

it is right

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d. A five-hour bus trip is divided into six equal legs.

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(This question is related to the picture question)
UNO [17]

Answer:

5. f(x) = 10,000 (1.5)^x

Step-by-step explanation:

==>>We would have to multiply the original amount by 1.50^x because the initial amount would be 1, and 50% increase would be .5 so 1.5 and you raise it to the number of years to show the total increase.

  • Let's test it

Initial: 10,000

After 1 year

10,000 + (.5*10000)

10,000 + 5000 = 15,000

--

After 2 years

15,000 + (.5*15000)

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Let's try our equation.

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10,000(1.5)^2

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6 0
3 years ago
A random sample of 100 automobile owners in thestate of Virginia shows that an automobile is driven onaverage 23,500 kilometers
Anastaziya [24]

Answer:

a) 22497.7 < μ< 24502.3

b)  With 99% confidence the possible error will not exceed 1002.3

Step-by-step explanation:

Given that:

Mean (μ) = 23500 kilometers per year

Standard deviation (σ) = 3900 kilometers

Confidence level (c) = 99% = 0.99

number of samples (n) = 100

a) α = 1 - c = 1 - 0.99 = 0.01

\frac{\alpha }{2} =\frac{0.01}{2}=0.005\\ z_{\frac{\alpha }{2}}=z_{0.005}=2.57

Using normal distribution table, z_{0.005 is the z value of 1 - 0.005 = 0.995 of the area to the right which is 2.57.

The margin of error (e) is given as:

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Confidence interval = 22497.7 < μ< 24502.3

b) With 99% confidence the possible error will not exceed 1002.3

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