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2 years ago

Write an equation of the line that passes through the given points (7,9),m =4

Mathematics

1 answer:

Alla [95]2 years ago

8 0

Answer:

$y = 4x - 19$

You should also try solving it yourself to make sure I did not make any mistake.

Step-by-step explanation:

We already know the slope so, so far the equation should look like this.

$y = 4x + b$

Now we just need to insert the point into the equation. So point (7,9) 7 should replace the x and 9 will replace the y. It will look like this:

$9 = 4 \times 7 + b$

when you solve this equation you'll get b

$b = - 19$

So the equation is

$y = 4x - 19$

You might be interested in

Write an equation in slope-intercept form: (1,6) and (2,3) are on the line

s344n2d4d5 [400]

Step-by-step explanation:

eqn of line : y = mx+ c

m = slope or gradient

c = y-intercept.

Lets take (1,6) = (x1,y1) and (2,3) = (x2,y2)

$m \: = \frac{y1 - y2}{x1 - x2} \\ = \frac{6 - 3}{1 - 2} \\ = \frac{3}{ - 1} \\ = - 3$

Substitute x = 1 and y = 6 into equation,

$6 = ( - 3)(1) + c \\ 6 = - 3 + c \\ c = 6 + 3 \\ = 9$

Therefore substituting m and c value, equation of line is

$y = - 3x + 9$

5 0

2 years ago

Type the correct answer in the box

algol13

c = cost of food/med for cats

d = cost of food/med for dog

y = total cost/spent

What you know:

y = 6c + d [has 6 cats and 1 dog]

c = 2/3d - 5 [cost for cat is 5 less than 2/3 cost for dog]

y = $195

y = 6c + d

Substitute/plug in what you know, plug in (2/3d - 5) for c and 195 for y

195 = 6(2/3d - 5) + d Distribute/multiply 6 into (2/3d - 5)

195 = 4d - 30 + d Combine like terms

195 = 5d - 30 Add 30 on both sides of the equation

225 = 5d Divide 5 on both sides

$45 = d

5 0

3 years ago

(A) A small business ships homemade candies to anywhere in the world. Suppose a random sample of 16 orders is selected and each

Leviafan [203]

Following are the solution parts for the given question:

For question A:

$\to (n) = 16$

$\to (\bar{X}) = 410$

$\to (\sigma) = 40$

In the given question, we calculate $90\%$ of the confidence interval for the mean weight of shipped homemade candies that can be calculated as follows:

$\to \bar{X} \pm t_{\frac{\alpha}{2}} \times \frac{S}{\sqrt{n}}$

$\to C.I= 0.90\\\\\to (\alpha) = 1 - 0.90 = 0.10\\\\ \to \frac{\alpha}{2} = \frac{0.10}{2} = 0.05\\\\ \to (df) = n-1 = 16-1 = 15\\\\$

Using the t table we calculate $t_{ \frac{\alpha}{2}} = 1.753$ When $90\%$ of the confidence interval:

$\to 410 \pm 1.753 \times \frac{40}{\sqrt{16}}\\\\ \to 410 \pm 17.53\\\\ \to392.47 < \mu < 427.53$

So $90\%$ confidence interval for the mean weight of shipped homemade candies is between $392.47\ \ and\ \ 427.53$ .

For question B:

$\to (n) = 500$

$\to (X) = 155$

$\to (p') = \frac{X}{n} = \frac{155}{500} = 0.31$

Here we need to calculate $90\%$ confidence interval for the true proportion of all college students who own a car which can be calculated as

$\to p' \pm Z_{\frac{\alpha}{2}} \times \sqrt{\frac{p'(1-p')}{n}}$

$\to C.I= 0.90$

$\to (\alpha) = 0.10$

$\to \frac{\alpha}{2} = 0.05$

Using the Z-table we found $Z_{\frac{\alpha}{2}} = 1.645$

therefore $90\%$ the confidence interval for the genuine proportion of college students who possess a car is

$\to 0.31 \pm 1.645\times \sqrt{\frac{0.31\times (1-0.31)}{500}}\\\\ \to 0.31 \pm 0.034\\\\ \to 0.276 < p < 0.344$

So $90\%$ the confidence interval for the genuine proportion of college students who possess a car is between $0.28 \ and\ 0.34.$

For question C:

In question A, We are $90\%$ certain that the weight of supplied homemade candies is between 392.47 grams and 427.53 grams.
In question B, We are $90\%$ positive that the true percentage of college students who possess a car is between 0.28 and 0.34.

Learn more about confidence intervals:

brainly.in/question/16329412

7 0

2 years ago

If f(x)-79*), what is f(3)?

Anika [276]

Answer:

Step-by-step explanation:

Answer:

Step-by-step explanation:

you take the positive value for it

f(-3) = |2×(-3)+1|+5 ➡ |-6+1| + 5 ➡ |-5|+5 = 5+5

6 0

3 years ago

A painter charges $35 per hour for labor plus $40 for a ladder rental when he paints a house. The customer provides the paint. T

zhenek [66]

the answer is not there. first you take 950.00 and back out (subtract) the 40 dollar fee for a ladder rental which leaves 910.00 then divide the 910 by 35/hr and you get 26 hrs.

6 0

3 years ago