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4 years ago

Through:(5,0), parallel to y=4/5x-3

2 answers:

7 0

If you are looking for a line: remember point-slope form.
In this question you have a point (5,0) and you can get the slope from the equation. Since parallel lines have the same slope you can just use 4/5.
Just plug the numbers into the equation: y-y1=m(x-x1)
y-0=4/5(x-5)
Therefore y=4/5x-4

OverLord2011 [107]4 years ago

7 0

To be parallel lines both must have the same slope. Our reference line is:

y=0.8x-3 so the slope is 0.8, given the point (5,0) we can solve for the y-intercept, or "b" of y=mx+b for our parallel line.

0=0.8(5)+b

0=4+b

b=-4 so

y=0.8x-4 is the parallel line to y=0.8x-3 that passes through (5,0)

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What is the volume of one rubber ball? round to the nearest hundredth of a centimeter. cm3 if the price of the rubber needed to

sertanlavr [38]

The cost of producing one ball will be $ 0.09261. Then the profit will be if each ball is $ 0.41.

<h3>What is Geometry?</h3>

It deals with the size of geometry, region, and density of the different forms both 2D and 3D.

A toy company produces rubber balls that have a radius of 1. 7 cm.

A sphere has a radius of 1. 7 centimeters.

Then the volume of the ball will be

Volume = 4/3 x π x r³

Volume = 4/3 x π x 1.7³

Volume = 20.58 cubic cm

If the price of the rubber needed to produce a ball is $0.0045/cm³.

Then the cost of producing one ball will be

→ 20.58 x 0.0045

→ $ 0.09261

If the company sells a ball for $0.50. Then the profit will be

Profit = 0.50 - 0.09261

Profit = $ 0.41

More about the geometry link is given below.

brainly.com/question/7558603

#SPJ1

3 0

2 years ago

Read 2 more answers

A photoconductor film is manufactured at a nominal thickness of 25 mils. The product engineer wishes to increase the mean speed

AURORKA [14]

Answer:

A 98% confidence interval estimate for the difference in mean speed of the films is [-0.042, 0.222].

Step-by-step explanation:

We are given that Eight samples of each film thickness are manufactured in a pilot production process, and the film speed (in microjoules per square inch) is measured.

For the 25-mil film, the sample data result is: Mean Standard deviation 1.15 0.11 and For the 20-mil film the data yield: Mean Standard deviation 1.06 0.09.

Firstly, the pivotal quantity for finding the confidence interval for the difference in population mean is given by;

P.Q. = $\frac{(\bar X_1 -\bar X_2)-(\mu_1- \mu_2)}{s_p \times \sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }$ ~ $t__n_1_+_n_2_-_2$

where, $\bar X_1$ = sample mean speed for the 25-mil film = 1.15

$\bar X_1$ = sample mean speed for the 20-mil film = 1.06

$s_1$ = sample standard deviation for the 25-mil film = 0.11

$s_2$ = sample standard deviation for the 20-mil film = 0.09

$n_1$ = sample of 25-mil film = 8

$n_2$ = sample of 20-mil film = 8

$\mu_1$ = population mean speed for the 25-mil film

$\mu_2$ = population mean speed for the 20-mil film

Also, $s_p =\sqrt{\frac{(n_1-1)s_1^{2}+ (n_2-1)s_2^{2}}{n_1+n_2-2} }$ = $\sqrt{\frac{(8-1)\times 0.11^{2}+ (8-1)\times 0.09^{2}}{8+8-2} }$ = 0.1005

Here for constructing a 98% confidence interval we have used a Two-sample t-test statistics because we don't know about population standard deviations.

So, 98% confidence interval for the difference in population means, ( $\mu_1-\mu_2$ ) is;

P(-2.624 < $t_1_4$ < 2.624) = 0.98 {As the critical value of t at 14 degrees of

freedom are -2.624 & 2.624 with P = 1%}

P(-2.624 < $\frac{(\bar X_1 -\bar X_2)-(\mu_1- \mu_2)}{s_p \times \sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }$ < 2.624) = 0.98

P( $-2.624 \times {s_p \times \sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }$ < $2.624 \times {s_p \times \sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }$ < ) = 0.98

P( $(\bar X_1-\bar X_2)-2.624 \times {s_p \times \sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }$ < ( $\mu_1-\mu_2$ ) < $(\bar X_1-\bar X_2)+2.624 \times {s_p \times \sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }$ ) = 0.98

98% confidence interval for ( $\mu_1-\mu_2$ ) = [ $(\bar X_1-\bar X_2)-2.624 \times {s_p \times \sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }$ , $(\bar X_1-\bar X_2)+2.624 \times {s_p \times \sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }$ ]

= [ $(1.15-1.06)-2.624 \times {0.1005 \times \sqrt{\frac{1}{8}+\frac{1}{8} } }$ , $(1.15-1.06)+2.624 \times {0.1005 \times \sqrt{\frac{1}{8}+\frac{1}{8} } }$ ]

= [-0.042, 0.222]

Therefore, a 98% confidence interval estimate for the difference in mean speed of the films is [-0.042, 0.222].

Since the above interval contains 0; this means that decreasing the thickness of the film doesn't increase the speed of the film.

7 0

3 years ago

The figure in the picture is made up of 5 congruent squares. If the perimeter of the entire figure is 21.6 cm, find its area.

vovikov84 [41]

Perimeter is length plus width so use that to help you

3 0

3 years ago

How to find the radius of square that tells you length, width, and height.

Anna [14]

Well diameter will be the diagnol of that square..

let side of square be x, then diagonal will be root2 * x = diameter

so, raidus will be = dia/ 2
= (root2 * x ) /2

5 0

3 years ago

What is the bases, height, perimeter, and the area from the trapezoid? please help me

Sergio039 [100]

Answer:

base : 30 cm

height : 20 cm

perimeter : 92 cm

Area : 400 cm

Step-by-step explanation:

base : 30 cm

height : 20 cm

perimeter : 27 + 30 + 25 + 10

: 92 cm

Area : (1/2 ( a + b) ) x h

: (1/2 ( 10 + 30) ) x 20

: (20) x 20

: 400 cm

3 0

3 years ago