All categories

3 years ago

What is the equation of a horizontal line passing through the point (−3, 8)?

1 answer:

Studentka2010 [4]3 years ago

7 0

<span>horizontal line passing through the point (−3, 8) has x = -3 and y = 8
</span>horizontal line: slope is zero, no x-intercept, y-intercepts exists....

answer
y = 8

You might be interested in

What is the inverse of the function f(x) = 2x – 10?

VikaD [51]

y = 2x-10 switch x and y x = x = 2y-10 then solve for y 2y = x+10 y = (½)x+5 f⁻¹(x) = (½)x+5 Hope this helps and leave a rate, thanks and brainliest to help me reach expert ;)

8 0

3 years ago

Read 2 more answers

If 111 people attend a concert and tickets for adults cost $4 while tickets for children cost $3.25 and total receipts for the c

VikaD [51]

Answer:

57 children

54 adults

Step-by-step explanation:

Let's call x the number of children admitted and call z the number of adults admitted.

Then we know that:

$x + z = 111$

We also know that:

$3.25x + 4z = 401.25$

We want to find the value of x and z. Then we solve the system of equations:

-Multiplay the first equation by -4 and add it to the second equation:

$-4x - 4z = -444$

$3.25x + 4z = 401.25$

----------------------------------

$-0.75x = -42.75$

$x =\frac{-42.75}{-0.75}\\\\x=57$

Now we substitute the value of x in the first equation and solve for the variable z

$57 + z = 111$

$z = 111-57$

$z = 54$

5 0

3 years ago

Read 2 more answers

Calculate the sample standard deviation and sample variance for the following frequency distribution of hourly wages for a sampl

lidiya [134]

Answer:

(a) The sample variance is 16.51

(a) The sample standard deviation is 4.06

Step-by-step explanation:

Given

$\begin{array}{cc}{Class} & {Frequency} & 8.26 - 10.00 & 20 &10.01-11.75 & 38 &11.76 - 13.50& 36 & 13.51-15.25 &25&15.26-17.00 &27 &\ \end{array}$

Solving (a); The sample variance.

First, calculate the class midpoints.

This is the mean of the intervals.

i.e.

$x_1 = \frac{8.26+10.00}{2} = \frac{18.26}{2} = 9.13$

$x_2 = \frac{10.01+11.75}{2} = \frac{21.76}{2} = 10.88$

$x_3 = \frac{11.76+13.50}{2} = \frac{25.26}{2} = 12.63$

$x_4 = \frac{13.51+15.25}{2} = \frac{28.76}{2} = 14.38$

$x_5 = \frac{15.26+17.00}{2} = \frac{32.26}{2} = 16.13$

So, the table becomes:

$\begin{array}{ccc}{Class} & {Frequency} & {x} & 8.26 - 10.00 & 20&9.13 &10.01-11.75 & 38 &10.88&11.76 - 13.50& 36 &12.63& 13.51-15.25 &25&14.38&15.26-17.00 &27 &16.13\ \end{array}$