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4 years ago

Write an equation of a line that is parallel to the line whose equation is 3y=x+6 and that passes through the point (-3,4).

1 answer:

3 0

The standard form of this equation is

$y=\frac{1}{3}x+2$

The slope of two parallel lines i the same, so the slope of the line we're looking for is also $\frac{1}{3}$

Since the line passes through $(-3,4)$ , the equation is:

$(y-4)=\frac{1}{3}(x+3)$

$\boxed{y=\frac{1}{3}x+5}$

You might be interested in

What is 17- 6 11/12<br> Can someone please solve in steps?

Airida [17]

17- 6 11/12

borrow from the 17 and use a common denominator

16 12/12 - 6 11/12

10 1/12

6 0

3 years ago

For what value of constant c is the function k(x) continuous at x = 0 if k =

nlexa [21]

The value of constant c for which the function k(x) is continuous is zero.

<h3>What is the limit of a function?</h3>

The limit of a function at a point k in its field is the value that the function approaches as its parameter approaches k.

To determine the value of constant c for which the function of k(x) is continuous, we take the limit of the parameter as follows:

$\mathbf{ \lim_{x \to 0^-} k(x) = \lim_{x \to 0^+} k(x) = 0 }$

$\mathbf{\implies \lim_{x \to 0 } \ \ \dfrac{sec \ x - 1}{x}= c }$

Provided that:

$\mathbf{\implies \lim_{x \to 0 } \ \ \dfrac{sec \ x - 1}{x}= \dfrac{0}{0} \ (form) }$

Using l'Hospital's rule:

$\mathbf{\implies \lim_{x \to 0} \ \ \dfrac{\dfrac{d}{dx}(sec \ x - 1)}{\dfrac{d}{dx}(x)}= \lim_{x \to 0} sec \ x \ tan \ x = 0}$

Therefore:

$\mathbf{\implies \lim_{x \to 0 } \ \ \dfrac{sec \ x - 1}{x}=0 }$

Hence; c = 0

Learn more about the limit of a function x here:

brainly.com/question/8131777

#SPJ1

5 0

2 years ago

Juan went to Walmart to buy notebooks and

9966 [12]

Answer:

Juan went to Walmart to buy notebooks and

highlighters for his classes. Notebooks cost $3

while highlighters cost $4. Juan bought 12 items

and he spent a total of $42. Write a system of

equations that describes this situation

Step-by-step explanation:

3+4÷42

8 0

3 years ago

A mathematics teacher wanted to see the correlation between test scores and

SpyIntel [72]

Answer:

The homework grade, to the nearest integer, for a student with a test grade of 68 is 69.

Step-by-step explanation:

The general form of the linear regression equation is:

$y=a+bx$

Here,

<em>y</em> = dependent variable = test grade

<em>x</em> = independent variable = homework grade

<em>a</em> = intercept

<em>b</em> = slope

Compute the value of <em>a</em> and <em>b</em> as follows:

$a=\frac{\sum Y\cdot \sum X^{2}-\sum X\cdot\sum XY}{n\cdot \sum X^{2}-(\sum X)^{2}}\\\\=\frac{(592\times 44909)-(591\times45227)}{(8\times44909)-(591)^{2}}\\\\=-14.316$ $b=\frac{n\cdot \sum XY-\sum X\cdot\sum Y}{n\cdot \sum X^{2}-(\sum X)^{2}}\\\\=\frac{(8\times 45227)-(591\592)}{(8\times44909)-(591)^{2}}\\\\=1.195$

The linear regression equation that represents the set of data is:

$y=-14.316+1.195x$

Compute the value of <em>x</em> for <em>y</em> = 68 as follows:

$y=-14.316+1.195x$

$68=-14.316+1.195x\\1.195x=68+14.316\\1.195x=82.316\\x=68.884\\x\approx 69$

Thus, the homework grade, to the nearest integer, for a student with a test grade of 68 is 69.

8 0

3 years ago

c. Assume that the manufacturer's claim is true, what is the probability of such a tire lasting more than 60,000 miles?

KATRIN_1 [288]

Answer:

The probability of such a tire lasting more than 60,000 miles is 0.0228, for the complete question provided in explanation.

Step-by-step explanation:

<h2>Q. This is the question: </h2>

The lifetime of a certain type of car tire are normally distributed. The mean lifetime of a car tire is 50,000 miles with a standard deviation of 5,000 miles. Assume that the manufacturer's claim is true, what is the probability of such a tire lasting more than 60,000 miles?

<h2>Answer:</h2>

This is the question of normal distribution:

First w calculate the value of Z corresponding to X = 60,000 miles

We, have; Mean = μ = 50,000 miles, and Standard Deviation = σ = 5,000 miles

Now, for Z, we know that:

Z = (x-μ)/σ

Z = (60,000 - 50,000)/5,000

Now, we have standard tables, for normal distribution in terms of values of Z. One is attached in this answer.

P(X > 60,000) = P(Z > 2) = 1 - P(Z = 2)

P(X > 60,000) = 1 - 0.9772

4 0

3 years ago