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3 years ago

Write an equation for an equation with a slope of 1/2 and goes through the point C(−4,3).

Mathematics

2 answers:

Degger [83]3 years ago

6 0

I think the answer should be
C9-3,4

mamaluj [8]3 years ago

4 0

Answer:

c9-3,4)

Step-by-step explanation:

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Sindrei [870]

11. 0.03, 3/10,1 1/8,1 5/6,2

7 0

3 years ago

Point R lies on the directed line segment from L (-8,-10) to M (4,-2) and partitions the segment in the ratio 3 to 5. What are t

Dvinal [7]

Answer:

R = (- 3.5, - 7 )

Step-by-step explanation:

Using the Section formula

$x_{R}$ = $\frac{3(4)+5(-8)}{3+5}$ = $\frac{12-40}{8}$ = $\frac{-28}{8}$ = - 3.5

$y_{R}$ = $\frac{3(-2)+5(-10)}{3+5}$ = $\frac{-6-50}{8}$ = $\frac{-56}{8}$ = - 7

Thus coordinates of R = (- 3.5, - 7 )

7 0

3 years ago

Read 2 more answers

A laboratory scale is known to have a standard deviation (sigma) or 0.001 g in repeated weighings. Scale readings in repeated we

weqwewe [10]

Answer:

99% confidence interval for the given specimen is [3.4125 , 3.4155].

Step-by-step explanation:

We are given that a laboratory scale is known to have a standard deviation (sigma) or 0.001 g in repeated weighing. Scale readings in repeated weighing are Normally distributed with mean equal to the true weight of the specimen.

Three weighing of a specimen on this scale give 3.412, 3.416, and 3.414 g.

Firstly, the pivotal quantity for 99% confidence interval for the true mean specimen is given by;

P.Q. = $\frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\bar X$ = sample mean weighing of specimen = $\frac{3.412+3.416+3.414}{3}$ = 3.414 g

$\sigma$ = population standard deviation = 0.001 g

n = sample of specimen = 3

$\mu$ = population mean

<em>Here for constructing 99% confidence interval we have used z statistics because we know about population standard deviation (sigma).</em>

So, 99% confidence interval for the population mean, $\mu$ is ;

P(-2.5758 < N(0,1) < 2.5758) = 0.99 {As the critical value of z at 0.5% level

of significance are -2.5758 & 2.5758}

P(-2.5758 < $\frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } }$ < 2.5758) = 0.99

P( $-2.5758 \times {\frac{\sigma}{\sqrt{n} } }$ < ${\bar X - \mu}$ < $2.5758 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.99

P( $\bar X-2.5758 \times {\frac{\sigma}{\sqrt{n} } }$ < $\mu$ < $\bar X+2.5758 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.99

<u>99% confidence interval for</u> $\mu$ = [ $\bar X-2.5758 \times {\frac{\sigma}{\sqrt{n} } }$ , $\bar X+2.5758 \times {\frac{\sigma}{\sqrt{n} } }$ ]

= [ $3.414-2.5758 \times {\frac{0.001}{\sqrt{3} } }$ , $3.414+2.5758 \times {\frac{0.001}{\sqrt{3} } }$ ]