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1 year ago

Mia’s work to find the slope of a trend line through the points (3, 10) and (35, 91) is shown below.

2 answers:

5 0

For this case the first thing you should know is that by definition the slope of the line is given by:
m = (y2-y1) / (x2-x1)
Where we have two ordered pairs:
P1 = (x1, y1)
P2 = (x2, y2)
We observe then that according to the formula, the error of mine was to invert the numerator with the denominator.
Answer:
C: Mia switched the numerator and the denominator.

Mrrafil [7]1 year ago

5 0

Answer:

Answer:

C: Mia switched the numerator and the denominator.

Step-by-step explanation:

just took the test

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What are the solutions of the quadratic equation 40-x^2=0

Nataliya [291]

Use the quadratic formula to find the solutions:−b±√b2−4(ac)2a-b±b2-4(ac)2a Substitute the values:
a=−1a=-1, b=0b=0, and c=40c=40 into the quadratic formula and solve for x 0±√02−4⋅(−1⋅40)2⋅−10±02-4⋅(-1⋅40)2⋅-1 Simplify:
x=±2√10x=±210 The result can be shown in both exact and approximate form: x=±2√10x=±210x≈6.324555,−6.324555
I hope this helps. :)

3 0

1 year ago

Read 2 more answers

Two ocean beaches are being affected by erosion. The table shows the width, in feet, of each beach measured at high tide where 1

TiliK225 [7]

Answer:

1. Western erodes 2 ft/yr; Dunes builds up at 5 ft/yr

2. Sometime in 2006

3. Solve simultaneous equations

Step-by-step explanation:

1. Erosion patterns

(a) Western Beach

In 15 yr, Western Beach erodes from 100 ft to 70 ft.

The rate of erosion is 30 ft/15 yr = 2 ft/yr.

(b) Dunes Beach

In 15 yr, Dunes Beach builds up from 20 ft to 95 ft.

The rate of buildup is 75 ft/15 yr = 5 ft/yr.

2. Beaches with equal width

From the table, it appears that the beaches will have the same width sometime in year 11 (2006).

3. Best approximation

The graph below also shows that it happens part way through year 11 (2006).

We could get an even better solution by calculating the equations of the two lines and solving the simultaneous equations.

5 0

2 years ago

Suppose an automobile manufacturer designed a radically new lightweight engine and wants to recommend the grade of gasoline that

Deffense [45]

Answer:

The F-statistic used to test the hypothesis that the miles per gallon for each fuel are the same is 4.07.

Step-by-step explanation:

There are four treatments in the data given, i.e. k = 4.

Total number of observations, n = 12.

Note: degrees of freedom is denoted as df.

For treatment, the degrees of freedom = k-1 = 4-1 =3 df.

The total degrees of freedom = n-1 = 12-1 = 11 df.

The error in degrees of freedom = df (total) - df(treatment)

The error in degrees of freedom = 11 - 3 = 8 df

At α = 0.05 level,from the F table, the F-statistic with (3 , 8)df is 4.07.

Therefore, the F-statistic used to test the hypothesis that the miles per gallon for each fuel are the same is 4.07.

8 0

1 year ago

The profitability, P, of a popular restaurant franchise can be modeled by the function P (t) = t4 – 10t3 + 28t2 – 24t, where t i

vekshin1

In this question, the profit of the restaurant after t months is given by a polynomial function. To find when it begins to show a profit, we find the numerical values of the function for t, and it shows a profit when $t > 0$

Profit after t months:

$P(t) = t^4 - 10t^3 + 28t^2 - 24t$

0 months:

This is P(0). So

$P(0) = 0^4 - 10*0^3 + 28*0^2 - 24*0 = 0$

1 month:

This is P(1). So

$P(1) = 1^4 - 10*1^3 + 28*1^2 - 24*1 = -5$

2 months:

This is P(2). So

$P(2) = 2^4 - 10*2^3 + 28*2^2 - 24*2 = 0$

3 months:

This is P(3). So

$P(3) = 3^4 - 10*3^3 + 28*3^2 - 24*3 = -9$

4 months:

This is P(4). So

$P(4) = 4^4 - 10*4^3 + 28*4^2 - 24*4 = -32$

5 months:

This is P(5). So

$P(5) = 5^4 - 10*5^3 + 28*5^2 - 24*5 = -45$

6 months:

This is P(6). So

$P(6) = 6^4 - 10*6^3 + 28*6^2 - 24*6 = 0$

7 months:

This is `P(7). So

$P(7) = 7^4 - 10*7^3 + 28*7^2 - 24*7 = 175$

After 7 months it shows profit, so it starts showing profit on the 6th month, and thus, the correct answer is given by option D.

For another example of a function involving numeric value, you can check brainly.com/question/24231879

8 0

2 years ago

Using the standard normal table, the probability that Seth’s light bulb will last no more than 14,500 (P(z ≤ 1.25)) hours is abo

lana66690 [7]

Answer:

How many standard deviations above the mean is 14,500 hours? 1.25 1.5 2.5 Using the standard normal table, the probability that Seth's light bulb will last no more than 14,500 (P(z ≤ 1.25)) hours is about ✔ 89% .

5 0

1 year ago

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