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blsea [12.9K]
2 years ago
14

Which of the following is true of the points on the line y= 5/3x + 1/2

Mathematics
1 answer:
denpristay [2]2 years ago
7 0

The points on the line y= 5/3x + 1/2 is (0, 1/2)

<h3>What are linear equations?</h3>

Linear equations are equations that have constant average rates of change. Note that the constant average rates of change can also be regarded as the slope or the gradient

<h3>How to determine the points on the line y= 5/3x + 1/2?</h3>

A system of linear equations is a collection of at least two linear equations.

In this case, the system of equations is given as

y= 5/3x + 1/2

Set x = 0.

So, we have

y= 5/3 * 0 + 1/2

Evaluate the product

y = 0 + 1/2

Evaluate the sum

y = 1/2

This means that

x = 0 and y = 1/2

Rewrite as

(x, y) = (0, 1/2)

Hence, the points on the line y= 5/3x + 1/2 is (0, 1/2)

Read more about linear equations at:

brainly.com/question/1884491

#SPJ1

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What is 4.345 x 10^5 in standard form?
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Answer:

Choice D.)   434,500

Step-by-step explanation:

What is 4.345 x 10^5 in standard form?

4.345 x 10^5 = 4.345 *  100,000

multiply it out:

4.345 x 10^5 = 434,500

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Ben wants to estimate the cost of dinner and a movie for himself and his friend. He knows the following information
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B. $3100

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Estimate the value of 196 divided by 0.499
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Read 2 more answers
A manager at a local company asked his employees how many times they had given blood in the last year. The results of the survey
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Answer:

Var(X) = E(X^2) -[E(X)]^2 = 4.97 -(1.61)^2 =2.3779

And the deviation would be:

Sd(X) =\sqrt{2.3779}= 1.542 \approx 1.54

Step-by-step explanation:

For this case we have the following distribution given:

X        0         1       2       3       4         5        6

P(X)  0.3   0.25   0.2   0.12   0.07   0.04   0.02

For this case we need to find first the expected value given by:

E(X) = \sum_{i=1}^n X_i P(X_I)

And replacing we got:

E(X)= 0*0.3 +1*0.25 +2*0.2 +3*0.12 +4*0.07+ 5*0.04 +6*0.02=1.61

Now we can find the second moment given by:

E(X^2) =\sum_{i=1}^n X^2_i P(X_i)

And replacing we got:

E(X^2)= 0^2*0.3 +1^2*0.25 +2^2*0.2 +3^2*0.12 +4^2*0.07+ 5^2*0.04 +6^2*0.02=4.97

And the variance would be given by:

Var(X) = E(X^2) -[E(X)]^2 = 4.97 -(1.61)^2 =2.3779

And the deviation would be:

Sd(X) =\sqrt{2.3779}= 1.542 \approx 1.54

 

8 0
4 years ago
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