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3 years ago

Write an equation in slope-intercept form from the points (0, 3) and (-4,-1)

Mathematics

1 answer:

Bezzdna [24]3 years ago

8 0

Answer:

y = x + 3

Step-by-step explanation:

Slope-intercept form is represented by the formula $y = mx + b$ . We can write an equation in point-slope form first, then convert it to that form.

1) First, find the slope of the line. Use the slope formula $m = \frac{y_2-y_1}{x_2-x_1}$ and substitute the x and y values of the given points into it. Then, simplify to find the slope, or $m$ :

$m = \frac{(-1)-(3)}{(-4)-(0)} \\m = \frac{-1-3}{-4-0} \\m = \frac{-4}{-4} \\m = 1$

Thus, the slope of the line must be 1.

2) Now, since we know a point the line intersects and its slope, use the point-slope formula $y-y_1=m(x-x_1)$ and substitute values for $m$ , $x_1$ , and $y_1$ . From there, we can convert the equation into slope-intercept form.

Since $m$ represents the slope, substitute 1 in its place. Since $x_1$ and $y_1$ represent the x and y values of a point the line intersects, choose any one of the given points (either one is fine) and substitute its x and y values into the equation, too. (I chose (0,3).) Finally, isolate y to find the answer:

$y-3=1(x-0)\\y-3 = x\\y = x + 3$

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3 years ago

A study of long-distance phone calls made from General Electric Corporate Headquarters in Fairfield, Connecticut, revealed the l

Katena32 [7]

Answer:

(a) The fraction of the calls last between 4.50 and 5.30 minutes is 0.3729.

(b) The fraction of the calls last more than 5.30 minutes is 0.1271.

(c) The fraction of the calls last between 5.30 and 6.00 minutes is 0.1109.

(d) The fraction of the calls last between 4.00 and 6.00 minutes is 0.745.

(e) The time is 5.65 minutes.

Step-by-step explanation:

We are given that the mean length of time per call was 4.5 minutes and the standard deviation was 0.70 minutes.

Let X = the length of the calls, in minutes.

So, X ~ Normal( $\mu=4.5,\sigma^{2} =0.70^{2}$ )

The z-score probability distribution for the normal distribution is given by;

Z = $\frac{X-\mu}{\sigma}$ ~ N(0,1)

where, $\mu$ = population mean time = 4.5 minutes

$\sigma$ = standard deviation = 0.7 minutes

(a) The fraction of the calls last between 4.50 and 5.30 minutes is given by = P(4.50 min < X < 5.30 min) = P(X < 5.30 min) - P(X $\leq$ 4.50 min)

P(X < 5.30 min) = P( $\frac{X-\mu}{\sigma}$ < $\frac{5.30-4.5}{0.7}$ ) = P(Z < 1.14) = 0.8729

P(X $\leq$ 4.50 min) = P( $\frac{X-\mu}{\sigma}$ $\leq$ $\frac{4.5-4.5}{0.7}$ ) = P(Z $\leq$ 0) = 0.50

The above probability is calculated by looking at the value of x = 1.14 and x = 0 in the z table which has an area of 0.8729 and 0.50 respectively.

Therefore, P(4.50 min < X < 5.30 min) = 0.8729 - 0.50 = 0.3729.

(b) The fraction of the calls last more than 5.30 minutes is given by = P(X > 5.30 minutes)

P(X > 5.30 min) = P( $\frac{X-\mu}{\sigma}$ > $\frac{5.30-4.5}{0.7}$ ) = P(Z > 1.14) = 1 - P(Z $\leq$ 1.14)

= 1 - 0.8729 = 0.1271

The above probability is calculated by looking at the value of x = 1.14 in the z table which has an area of 0.8729.

(c) The fraction of the calls last between 5.30 and 6.00 minutes is given by = P(5.30 min < X < 6.00 min) = P(X < 6.00 min) - P(X $\leq$ 5.30 min)

P(X < 6.00 min) = P( $\frac{X-\mu}{\sigma}$ < $\frac{6-4.5}{0.7}$ ) = P(Z < 2.14) = 0.9838

P(X $\leq$ 5.30 min) = P( $\frac{X-\mu}{\sigma}$ $\leq$ $\frac{5.30-4.5}{0.7}$ ) = P(Z $\leq$ 1.14) = 0.8729

The above probability is calculated by looking at the value of x = 2.14 and x = 1.14 in the z table which has an area of 0.9838 and 0.8729 respectively.

Therefore, P(4.50 min < X < 5.30 min) = 0.9838 - 0.8729 = 0.1109.

(d) The fraction of the calls last between 4.00 and 6.00 minutes is given by = P(4.00 min < X < 6.00 min) = P(X < 6.00 min) - P(X $\leq$ 4.00 min)

P(X < 6.00 min) = P( $\frac{X-\mu}{\sigma}$ < $\frac{6-4.5}{0.7}$ ) = P(Z < 2.14) = 0.9838

P(X $\leq$ 4.00 min) = P( $\frac{X-\mu}{\sigma}$ $\leq$ $\frac{4.0-4.5}{0.7}$ ) = P(Z $\leq$ -0.71) = 1 - P(Z < 0.71)

= 1 - 0.7612 = 0.2388

The above probability is calculated by looking at the value of x = 2.14 and x = 0.71 in the z table which has an area of 0.9838 and 0.7612 respectively.

Therefore, P(4.50 min < X < 5.30 min) = 0.9838 - 0.2388 = 0.745.

(e) We have to find the time that represents the length of the longest (in duration) 5 percent of the calls, that means;

P(X > x) = 0.05 {where x is the required time}

P( $\frac{X-\mu}{\sigma}$ > $\frac{x-4.5}{0.7}$ ) = 0.05

P(Z > $\frac{x-4.5}{0.7}$ ) = 0.05

Now, in the z table the critical value of x which represents the top 5% of the area is given as 1.645, that is;

$\frac{x-4.5}{0.7}=1.645$

${x-4.5}{}=1.645 \times 0.7$

x = 4.5 + 1.15 = 5.65 minutes.

SO, the time is 5.65 minutes.

7 0

3 years ago

What are the next three terms in the sequence? -3, 6, 15, 24

bearhunter [10]

Its adding 9 every time.
-3, 6, 15, 24, 33, 42, 51, 60

7 0

3 years ago

Compare the decimals 16.30 and 16.3

pshichka [43]

16.30 and 16.3 equal. If you ever have a problem like that then you just add on a zero.

Example:
19.4500 = 19.45 Just add two zeros on the end
19.45 + two zeros = 1.4500

8 0

3 years ago

Pls help I’ll give brainliest

DanielleElmas [232]

Answer:

B) 7 gallons

Step-by-step explanation:

car begins with 12 gallons and, at a rate of 29 mpg, would allow the car to travel for 348 miles

car only traveled 140 miles so 208 miles more can be driven

208÷29 is approx 7 gallons

7 0

3 years ago