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Lesechka [4]
2 years ago
12

How to factorise 5h^2-12hg​

Mathematics
1 answer:
bonufazy [111]2 years ago
7 0

Answer:

h(5h-12g)

Step-by-step explanation:

5h^2-12hg

When we factor expressions, we look for factors within the terms that are alike, or in other words, we look for common factors. Here, 5h^2 and 12hg only have one common factor: h. Therefore, to factorize this expression, divide both terms by h.

= h(\frac{5h^2}{h}-\frac{12hg}{h}  )\\= h(5h-12g)

Now, we've "carried" h out of the expression and have therefore factored it.

I hope this helps!

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Y=-2x-40 will be equation
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There are three letter tiles—A, B, and C—in a bag; and there are three number tiles—1, 2, and 3—in another bag. Alexis picks a l
sdas [7]
If she picks the letter A, she can then pick either 1 , 2 or 3
If she picks B, she can pick 1, 2, or 3 and so on.

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Which data set is least Likely to resemble a normal distribution?<br> Look at picture
kicyunya [14]

Answer: B) The heights of girls who live on a certain street in the city of Buffalo

Every answer choice starts with "the heights of all 14-year-old girls who", so we can ignore that part. Choice A describes the largest population while choice B describes the smallest population. In other words, choice A is very general and broad, while choice B is very specific and narrow. The more specific you get and the smaller the population is, the less likely its going to be normally distributed.

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2 years ago
Let X denote the temperature (degree C) and let Y denote thetime in minutes that it takes for the diesel engine on anautomobile
BlackZzzverrR [31]

Answer:

Step-by-step explanation:

Given f_{XY} (x,y) = c(4x + 2y +1) ; 0 < x < 40\,and\, 0 < y

a)

we know that \int\limits^\infty_{-\infty}\int\limits^\infty_{-\infty} {f(x,y)} \, dxdy=1

therefore \int\limits^{40}_{-0}\int\limits^2_{0} {c(4x+2y+1)} \, dxdy=1

on integrating we get

c=(1/6640)

b)

P(X>20, Y>=1)=\int\limits^{40}_{20}\int\limits^2_{1} {\frca{1}{6640}(4x+2y+1)} \, dxdy

on doing the integration we get

                        =0.37349

c)

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f(x)=\int\limits^2_{0} {\frca{1}{6640}(4x+2y+1)} \, dy

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f(x)=(4x+3)/3320 ; 0<x<40

marginal density of Y is

f(y)=\int\limits^{40}_{0} {\frca{1}{6640}(4x+2y+1)} \, dx

on doing integration we get

f(y)=\frac{(y+40.5)}{83}

d)

P(01)=\int\limits^{40}_{0}\int\limits^2_{1} {\frca{1}{6640}(4x+2y+1)} \, dxdy

solve the above integration we get the answer

e)

P(X>20, 0

solve the above integration we get the answer

f)

Two variables are said to be independent if there jointprobability density function is equal to the product of theirmarginal density functions.

we know f(x,y)

In the (c) bit we got f(x) and f(y)

f(x,y)cramster-equation-2006112927536330036287f(x).f(y)

therefore X and Y are not independent

4 0
2 years ago
An insurance company reported that, on average, claims for a certain medical procedure are $942. An independent organization con
kirza4 [7]

Answer:

Step-by-step explanation:

The question is incomplete. The complete question is:

An insurance company reported that, on average claims for a certain medical procedure are $942. an independent organization constructed a 95% confidence interval of ($930, $950) for the average amount claimed for the particular medical procedure. what conclusion best evaluates the truthfulness of the number reported by the insurance company?

a) with 95% certainty, the average claim for this medical procedure is $942.

b) with 95% certainty, the average claim for this medical procedure is not $942.

c) the confidence interval is consistent with an average claim of $942 for this medical procedure

Solution:

Confidence interval is used to express how confident we are that the population parameter that we are looking for is contained in a range of given values. Looking at the given confident interval, the lower limit is $930 and the upper limit is $950. We can see that the population mean, $942 lies within these values. The correct option would be

c) the confidence interval is consistent with an average claim of $942 for this medical procedure

7 0
2 years ago
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