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ololo11 [35]
3 years ago
10

Rowan reads at a constant rate of 1.5 pages per minute and has already

Mathematics
1 answer:
saul85 [17]3 years ago
3 0

Using linear functions, it is found that Rowan and Kennedy will have read the same number of pages in 10 minutes.

What is a linear function?

  • A linear function is modeled by:

y = ax + b

In which:

  • a is the slope, which is the rate of change, that is, by how much y changes when x changes by 1.
  • b is the y-intercept, which is the value of y when x = 0.

For Rowan, we have that:

  • He has already read 20 pages, hence the y-intercept is b = 20.
  • He reads 1.5 pages per minute, hence the slope is a = 1.5

Then:

y_R = 1.5x + 20

For Kennedy, we have that:

  • He has already read 25 pages, hence the y-intercept is b = 25.
  • He reads 1 page per minute, hence the slope is a = 1

Then:

y_K = x + 25

They will have read the same number of pages after x minutes, for which:

y_R = y_K

Then:

1.5x + 20 = x + 25

0.5x = 5

x = \frac{5}{0.5}

x = 10

To learn more about linear function, you can take a look at brainly.com/question/25823744

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jeyben [28]
3427 + 65.75s = 5000 + 35.5s
65.75s - 35.5s = 5000 - 3427
30.25s = 1573
s = 1573 / 30.25
s = 52 seconds <== they are the same at 52 seconds

3427 + 65.75(52) = 6846
5000 + 35.5(52) = 6846

and their altitude is 6846 <===
6 0
4 years ago
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Does Anyone know this?
lukranit [14]

Answer:

∠AEC = 139°

Step-by-step explanation:

Since EC bisects ∠BED then ∠BEC = ∠CED = 4x + 1

∠AED = ∠AEB + ∠BEC + ∠CED = 180 ← straight angle

Substitute values into the equation

11x - 12 + 4x + 1 + 4x + 1 = 180, that is

19x - 10 = 180 ( add 10 to both sides )

19x = 190 ( divide both sides by 19 )

x = 10

Hence

∠AEC = ∠AEB + ∠BEC = 11x - 12 + 4x + 1 = 15x - 11, hence

∠AEC = (15 × 10) - 11 = 150 - 11 = 139°

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3 years ago
The scores on a standardized test are normally distributed with the mean of 750 and standard deviation 70. Find the probability
mafiozo [28]
Hey the answer is C.47.7 
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Suppose that the length of a side of a cube X is uniformly distributed in the interval 9
Nastasia [14]

Answer:

f(v) = \left \{ {{\frac{1}{3}v^{-\frac{2}{3}}\ 9^3 \le v \le 10^3} \atop {0, elsewhere}} \right.

Step-by-step explanation:

Given

9 < x < 10 --- interval

Required

The probability density of the volume of the cube

The volume of a cube is:

v = x^3

For a uniform distribution, we have:

x \to U(a,b)

and

f(x) = \left \{ {{\frac{1}{b-a}\ a \le x \le b} \atop {0\ elsewhere}} \right.

9 < x < 10 implies that:

(a,b) = (9,10)

So, we have:

f(x) = \left \{ {{\frac{1}{10-9}\ 9 \le x \le 10} \atop {0\ elsewhere}} \right.

Solve

f(x) = \left \{ {{\frac{1}{1}\ 9 \le x \le 10} \atop {0\ elsewhere}} \right.

f(x) = \left \{ {{1\ 9 \le x \le 10} \atop {0\ elsewhere}} \right.

Recall that:

v = x^3

Make x the subject

x = v^\frac{1}{3}

So, the cumulative density is:

F(x) = P(x < v^\frac{1}{3})

f(x) = \left \{ {{1\ 9 \le x \le 10} \atop {0\ elsewhere}} \right. becomes

f(x) = \left \{ {{1\ 9 \le x \le v^\frac{1}{3} - 9} \atop {0\ elsewhere}} \right.

The CDF is:

F(x) = \int\limits^{v^\frac{1}{3}}_9 1\  dx

Integrate

F(x) = [v]\limits^{v^\frac{1}{3}}_9

Expand

F(x) = v^\frac{1}{3} - 9

The density function of the volume F(v) is:

F(v) = F'(x)

Differentiate F(x) to give:

F(x) = v^\frac{1}{3} - 9

F'(x) = \frac{1}{3}v^{\frac{1}{3}-1}

F'(x) = \frac{1}{3}v^{-\frac{2}{3}}

F(v) = \frac{1}{3}v^{-\frac{2}{3}}

So:

f(v) = \left \{ {{\frac{1}{3}v^{-\frac{2}{3}}\ 9^3 \le v \le 10^3} \atop {0, elsewhere}} \right.

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