All categories

3 years ago

Graph y = 2x + 5. Need help ASAP plz and thank you

Mathematics

2 answers:

soldi70 [24.7K]3 years ago

8 0

Answer:

*see graph below*

Step-by-step explanation:

y = mx + b

m = slope (rise over run)

b = y-intercept

Corresponding with the given equation, the graph below has a y-intercept of 5, and a slope of $\frac{2}{1}$ or 2.

andrezito [222]3 years ago

6 0

Answer:

Below

Step-by-step explanation:

i'll explain how to get the x and y intercepts and then u can graph it :). Also it'll help you with future questions like this rather than just looking at a graph

To find the x intercept:

Sub y for 0 : 0 = 2x + 5

Move variables : -2x = 5

Divide both sides by -2 : x = -5/2

The x intercept (which is the horizontal line) is -5/2 or -2.5

To find the y intercept:

Sub x for 0 : y = (2)(0) + 5

Solve for y : y = 5

The y intercept (which is the vertical line) is 5

now to find the slope so you can graph it

y = mx + b

y = 2x + 5

Therefore the slope (m) is 2.

hope this help

You might be interested in

How many ways can a baseball manager arrange a batting order of 9 players?

ELEN [110]

9! or 9×8×7×6×5×4×3×2×1 which is 362,880

6 0

4 years ago

Read 2 more answers

How much weight did the kitten gain in that month?

Aliun [14]

Answer:

21 ounces

Step-by-step explanation:

brainly.com/question/15457567

8 0

3 years ago

Read 2 more answers

Can I have help with the question

sleet_krkn [62]

Answer:

Midpoint = (3.5, 4.5)

Perpendicular bisector = y = $\frac{7}{9}$ x + $\frac{16}{9}$

Step-by-step explanation:

[] We can solve this using the midpoint formula:

-> See attached

[] Plug-in our coordinates and solve:

$(\frac{7+0}{2} ,\frac{0+9}{2} )=(\frac{7}{2} ,\frac{9}{2} )=(3.5,4.5 )$

[] Now we will find the slope to solve for the perpendicular bisector.

-> We will use slope-intercept form, see attached

$\frac{9-0}{0-7}=\frac{9}{-7}$

-> The slopes of two perpendicular lines are negative reciprocals of each other, so $\frac{7}{9}$ will be the slope of or perpendicular bisector

-> Now we can solve for the equation by using y – y1 = m ( x – x1), were y1 and x1 are the coordinates of our midpoint

y - 4.5 = $\frac{7}{9}$ (x-3.5)

y - 4.5 = $\frac{7}{9}$ x- $\frac{49}{18}$

y = $\frac{7}{9}$ x- $\frac{49}{18}$ + 4.5

y = $\frac{7}{9}$ x + $\frac{16}{9}$

Have a nice day!

I hope this is what you are looking for, but if not - comment! I will edit and update my answer accordingly. (ノ^∇^)

- Heather

6 0

2 years ago

An article contained the following observations on degree of polymerization for paper specimens for which viscosity times concen

devlian [24]

Answer:

(a) A 95% confidence interval for the population mean is [433.36 , 448.64].

(b) A 95% upper confidence bound for the population mean is 448.64.

Step-by-step explanation:

We are given that article contained the following observations on degrees of polymerization for paper specimens for which viscosity times concentration fell in a certain middle range:

420, 425, 427, 427, 432, 433, 434, 437, 439, 446, 447, 448, 453, 454, 465, 469.

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

P.Q. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean = $\frac{\sum X}{n}$ = 441

s = sample standard deviation = $\sqrt{\frac{\sum (X-\bar X)^{2} }{n-1} }$ = 14.34

n = sample size = 16

$\mu$ = population mean

Here for constructing a 95% confidence interval we have used One-sample t-test statistics as we don't know about population standard deviation.

So, 95% confidence interval for the population mean, $\mu$ is ;

P(-2.131 < $t_1_5$ < 2.131) = 0.95 {As the critical value of t at 15 degrees of

freedom are -2.131 & 2.131 with P = 2.5%}

P(-2.131 < $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ < 2.131) = 0.95

P( $-2.131 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $2.131 \times {\frac{s}{\sqrt{n} } }$ ) = 0.95

P( $\bar X-2.131 \times {\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+2.131 \times {\frac{s}{\sqrt{n} } }$ ) = 0.95

95% confidence interval for $\mu$ = [ $\bar X -2.131 \times {\frac{s}{\sqrt{n} } }$ , $\bar X +2.131 \times {\frac{s}{\sqrt{n} } }$ ]

= [ $441-2.131 \times {\frac{14.34}{\sqrt{16} } }$ , $441+2.131 \times {\frac{14.34}{\sqrt{16} } }$ ]

= [433.36 , 448.64]

(a) Therefore, a 95% confidence interval for the population mean is [433.36 , 448.64].

The interpretation of the above interval is that we are 95% confident that the population mean will lie between 433.36 and 448.64.

(b) A 95% upper confidence bound for the population mean is 448.64 which means that we are 95% confident that the population mean will not be more than 448.64.

6 0

3 years ago

Write two expressions that have a GCF of 8xy

Svetllana [295]

32x^2y
8xy^6
the gcf is 8xy

6 0

3 years ago

Read 2 more answers