Line k is vertical and contains the point (8,3). Which of the following equations represents line k?

In London, the charge of a tour bus consists of a fixed rental price plus a per kilometre rate, that depends on the distance of

ella [17]

Using a linear function, we have that:

a) The fixed cost is of $10.

b) The equation is: C = 10 + 12d.

c) The cost of an 18 km trip is $226.

d) The distance traveled is of 20 km.

<h3>What is a linear function?</h3>

A linear function is modeled by:

y = mx + b

In which:

m 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, and can also be interpreted as the initial value of the function.

For this problem, the slope is of m = 12. When d = 5, C(d) = 70, hence we find the fixed cost as follows:

C(d) = 12d + b

70 = 60 + b

b = 10.

Hence the equation is:

C(d) = 12d + 10.

For a trip of 18 km, d = 18, hence the cost is:

C(12) = 12 x 18 + 10 = $226.

When the cost is of $250, the distance is found as follows:

250 = 12d + 10

12d = 240

d = 240/12

d = 20 km.

More can be learned about linear functions at brainly.com/question/24808124

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6 0

1 year ago

08.01) Which of the following statements shows a characteristic of a statistical question? (4 points) Select one: a. The questio

Alexeev081 [22]

Answer:

the distribution can be broken into categories

7 0

2 years ago

Read 2 more answers

Integrate this two questions

tankabanditka [31]

Simplify the integrands by polynomial division.

$\dfrac{t^2}{1 - 3t} = -\dfrac19 \left(3t + 1 - \dfrac1{1 - 3t}\right)$

$\dfrac t{1 + 4t} = \dfrac14 \left(1 - \dfrac1{1 + 4t}\right)$

Now computing the integrals is trivial.

5.

$\displaystyle \int \frac{t^2}{1 - 3t} \, dt = -\frac19 \int \left(3t + 1 - \frac1{1-3t}\right) \, dt \\\\ = -\frac19 \left(\frac32 t^2 + t + \frac13 \ln|1 - 3t|\right) + C \\\\ = \boxed{-\frac{t^2}6 - \frac t9 - \frac{\ln|1-3t|}{27} + C}$

where we use the power rule,

$\displaystyle \int x^n \, dx = \frac{x^{n+1}}{n+1} + C ~~~~ (n\neq-1)$

and a substitution to integrate the last term,

$\displaystyle \int \frac{dt}{1-3t} = -\frac13 \int \frac{du}u \\\\ = -\frac13 \ln|u| + C \\\\ = -\frac13 \ln|1-3t| + C ~~~ (u=1-3t \text{ and } du = -3\,dt)$

8.

$\displaystyle \int \frac t{1+4t} \, dt = \frac14 \int \left(1 - \frac1{1+4t}\right) \, dt \\\\ = \frac14 \left(t - \frac14 \ln|1 + 4t|\right) + C \\\\ = \boxed{\frac t4 - \frac{\ln|1+4t|}{16} + C}$

using the same approach as above.

5 0

1 year ago

A family has two cats named Gordo and Flaco. Gordo weighs 15 pounds and Flaco weighs 8 pounds. A cat’s weight is classified as u

Elena-2011 [213]

Answer:

Gordo's weight = 15 pounds

It is outside the healthy weight range and is in the top 5% of weights of cats.

Gordo's weight makes Gordo unhealthy.

Step-by-step explanation:

μ = mean weight = 9.5 pounds

σ = standard deviation = 1.5 pounds

This is a normal distribution problem

We first calculate the limit of the bottom 5% of weights

Let the z-score for this limit be z'

P(z < z') = 0.05

From the normal distribution table,

z' = -1.645

And the limit for the top 5% which is z" = 1.645.

The weight that corresponds to these scores are then calculated.

Standardized scores are given as

z = (x - μ)/σ

So,

z' = (limit for the bottom 5% - μ)/σ

-1.645 = (limit for the bottom 5% - 9.5)/1.5

limit of the bottom 5% = (-1.645)(1.5) + 9.5 = 7.033 pounds

z" = ( (limit for the top 5% - μ)/σ

1.645 = (limit for the top 5% - 9.5)/1.5

limit of the bottom 5% = (1.645)(1.5) + 9.5 = 11.968 pounds

Therefore the healthy weight range for cats is (7.033 < x < 9.968)

Gordo's weight = 15 pounds

It is outside the healthy weight range and is in the top 5% of weights of cats.

Gordo's weight makes the cat unhealthy.

Flaco's weight = 8 pounds

Flaco's weight lies in the healthy weight range for cats. Hence, Flaco is a healthy cat.

8 0

2 years ago

True or False Atomic theory has changed over time

jolli1 [7]

Answer:

True

Step-by-step explanation:

As modern science has advanced we're able to learn more about what makes up matter. Hope this helps :)

5 0

2 years ago