The midpoint of AB is M(4,-2). If the coordinates of A are (3, -6), what are the

Mathematics

1 answer:

satela [25.4K]2 years ago

5 0

Answer:

okay here we go

Step-by-step explanation:

$in \: vectos \: ab \: means \: ob - oa \\ but \: the \: midpoint \: m \: is \: equal \: to \: a \: half \: of \: ab \\ \\ therefore \: m = \frac{1}{2} ab \\ = \frac{1}{2} (ob - oa) \\ = \frac{1}{2} (ob - (3. - 6)) \\ (4. - 2) = \frac{1}{2} (ob - (3. - 6)) \\ (8. - 4) = (ob - (3. - 6)) \\ ob = (11. - 2)$

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Explain to me how to do this please

nataly862011 [7]

You must first attach the problems with your question.

6 0

3 years ago

HELPPPP look at image

posledela

Your answer would be 24 and 168
take the 48 and divide by 4 to get the 1st one then just multiply 24 by 7 and you get 168

5 0

3 years ago

Read 2 more answers

A cookie factory monitored the number of broken cookies per pack yesterday.

trapecia [35]

Answer:

Confidence Interval - 2.290 < S < 2.965

Step-by-step explanation:

Complete question

A chocolate chip cookie manufacturing company recorded the number of chocolate chips in a sample of 50 cookies. The mean is 23.33 and the standard deviation is 2.6. Construct a 80% confidence interval estimate of the standard deviation of the numbers of chocolate chips in all such cookies.

Solution

Given

n=50

x=23.33

s=2.6

Alpha = 1-0.80 = 0.20

X^2(a/2,n-1) = X^2(0.10, 49) = 63.17

sqrt(63.17) = 7.948

X^2(1 - a/2,n-1) = X^2(0.90, 49) = 37.69

sqrt(37.69) = 6.139

s*sqrt(n-1) = 18.2

$s\sqrt{\frac{n-1}{X^2 _{(n-1), \frac{\alpha }{2} } }$ $\leq \sigma \leq s\sqrt{\frac{n-1}{X^2 _{(n-1), 1-\frac{\alpha }{2} } }$