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3 years ago

Write the equation of the line represented by the following table

Mathematics

1 answer:

s344n2d4d5 [400]3 years ago

4 0

Answer:

y = 100x + 400

Step-by-step explanation:

The equation of a line in slope- intercept form is

y = mx + c ( m is the slope and c the y- intercept )

Calculate m using the slope formula

m = (y₂ - y₁ ) / (x₂ - x₁ )

with (x₁, y₁ ) = (0, 400) and (x₂, y₂ ) = (1, 500) ← 2 points from the table

m = $\frac{500-400}{1-0}$ = 100

note the line crosses the y- axis at (0, 400) ⇒ c = 400

y = 100x + 400 ← equation of line

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Makovka662 [10]

Let $u=x^2-4$ and $v=4x-5$ . By the product rule,

$\dfrac{\mathrm d(u^5v^4)}{\mathrm dx}=\dfrac{\mathrm d(u^5)}{\mathrm dx}v^4+u^5\dfrac{\mathrm d(v^4)}{\mathrm dx}$

By the power rule, we have $(u^5)'=5u^4$ and $(v^4)'=4v^3$ , but $u,v$ are functions of $x$ , so we also need to apply the chain rule:

$\dfrac{\mathrm d(u^5)}{\mathrm dx}=5u^4\dfrac{\mathrm du}{\mathrm dx}$

$\dfrac{\mathrm d(v^4)}{\mathrm dx}=4v^3\dfrac{\mathrm dv}{\mathrm dx}$

and we have

$\dfrac{\mathrm du}{\mathrm dx}=2x$

$\dfrac{\mathrm dv}{\mathrm dx}=4$

So we end up with

$\dfrac{\mathrm d(u^5v^4)}{\mathrm dx}=10xu^4v^4+16u^5v^3$

Replace $u,v$ to get everything in terms of $x$ :

$\dfrac{\mathrm d((x^2-4)^5(4x-5)^4)}{\mathrm dx}=10x(x^2-4)^4(4x-5)^4+16(x^2-4)^5(4x-5)^3$

We can simplify this by factoring:

$10x(x^2-4)^4(4x-5)^4+16(x^2-4)^5(4x-5)^3=2(x^2-4)^4(4x-5)^3\bigg(5x(4x-5)+8(x^2-4)\bigg)$

$=2(x^2-4)^4(4x-5)^3(28x^2-57)$

7 0

3 years ago

Which equation is correctly rewritten to solve for x?<br><br> x/r - s = t

wolverine [178]

Answer:

x = r(t + s) (second option)

Step-by-step explanation:

First add s to both sides and get x/r = t + s.

Then mutiply both sides by r to separate x and this is the final equation:

x = r(t + s)

5 0

2 years ago

Plz I need the awnser to this

Scilla [17]

Answer:A

Step-by-step explanation: because it’s an even number

8 0

3 years ago

The prostate-specific antigen (PSA) test is a simple blood test to screen for prostate cancer. It has been used in men over 50 a

Ugo [173]

The question is incomplete. I am writing the complete question below:

The prostate-specific antigen (PSA) test is a simple blood test to screen for prostate cancer. It has been used in men over 50 as a routine part of a physical exam, with levels above 4 ng/mL indicating possible prostate cancer. The test result is not always correct, sometimes indicating prostate cancer when it is not present and often missing prostate cancer that is present. Suppose that these are the approximate conditional probabilities of a positive (above 4 ng/ml) and negative test result given cancer is present or absent.

Positive Result Negative Result

Cancer Present 0.21 0.79

Cancer Absent 0.06 0.94

In a large study of prostate cancer screening, it was found that about 6.6% of the population has prostate cancer.

What is the probability that the test is positive for a randomly chosen person from this population? (Enter your answer to five decimal places.)

P(Positive test) =

Answer:

P(Positive Result) = 0.0699

Step-by-step explanation:

We are given that 6.6% of the population has prostate cancer. So,

P(Cancer Present) = 0.066

P(Cancer Absent) = 1 - 0.066 = 0.934

From the given conditional probability table, we have:

P(Positive Result | Cancer Present) = 0.21

P(Positive Result | Cancer Absent) = 0.06

We need to find the probability that the result is positive. It can be either that the result is positive and cancer is present or the result is positive and the cancer is absent. So,

P(Positive Result) = P(Positive Result∩Cancer Present) + P(Positive Result∩Cancer Absent)

= P(Positive Result | Cancer Present)*P(Cancer Present) + P(Positive Result | Cancer Absent)*P(Cancer Absent)

= (0.21)*(0.066) + (0.06)*(0.934)

= 0.01386 + 0.05604

P(Positive Result) = 0.0699

5 0

3 years ago

The population in thousands of people of a city is approximated by the function P(t) = 1200(2)0.1038twhere t is the number of ye

asambeis [7]

The population of the group in 2019 is 2,134,000 people

Let P represent the population in thousands of people in the year t after 2011

Given the equation:

$P=1200(2)^{0.1038t}$

a) The population of the group in 2019, that is t = 8(2019 - 2011):

$P=1200(2)^{0.1038*8}=2134$

b) The population of the group in 2029, that is t = 18(2029 - 2011):

$P=1200(2)^{0.1038*18}=4382$

The population of the group in 2019 is 2,134,000 people

Find out more on equation at: brainly.com/question/2972832

7 0

2 years ago