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3 years ago

What is the equation of a line that passes through the points (–3, 4) and (2, 8)?

Mathematics

2 answers:

N76 [4]3 years ago

4 0

Answer:

Step-by-step explanation:

(–3, 4) and (2, 8)

y2 = 8, y1 = 4, x2 = 2 and x1 = -3

Slope m = {y2-y1}/{x2-x1}

m = {8 - 4}/{2 - (-3)}

m = 4/(2+3)

m = 4/5

And

y - y1 = m(x - x1)

y - 4 = 4/5(x - (-3))

y - 4 = 4/5(x + 3)

Multiply each term by 5

5y - 20 = 4(x + 3)

5y - 20 = 4x + 12

5y = 4x + 20 + 12

5y = 4x + 32

Dividing by 5

y = 4x/5 + 32/5

y = 0.8x + 6.4

GREYUIT [131]3 years ago

3 0

Answer:

Step-by-step explanation:

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Jay bought 6 bags of maize at shs 2700 each.He sold the bags at shs.3300 each. what was his profit?

NISA [10]

Answer:

3300-2700= 600 x 6 = 3600

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3 years ago

Evaluate 9/m+4 when m = 3.

irinina [24]

Answer:

9/3+4=

4+3=

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

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

What is the absolute value of -250?

Andrei [34K]

Answer:

250

Step-by-step explanation:

Pretty much every time it asks for the abasloute value it just asks how much to get the zero on the number line. It will ALWAYS be positive. For example -5 will be 5 to get to zero on the numberline. Same thing with a positive 5.

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3 years ago

PLEASE HELP ME 100 POINTS

Ghella [55]

Answer:

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b b b b b b b b bb b b b b bb b b bb b bb bb bb b bb b b b b b b b b bb b b b b bb b bb b b b b bb b b b b b b b bb bb

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3 years ago