I need to find the median ,mean,mode,range,and

Mathematics

1 answer:

Julli [10]2 years ago

7 0

So it looks like the numbers are already in numerical order, that's good. The median is simply the middle number. I shouldn't have to help you find that. If there are 2 middle numbers, take the average and that's the median.

To find mean, you need a calculator. Add up all the numbers, and then divide by 10, since there's 10 numbers total. That's the mean.

The mode is 99, because that's the most frequent number. It appears twice.

The range is the span of the data set. To find this, take the largest number (99) and minus the smallest number (1). This is the range.

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Evaluate the integral. (Assume a ≠ b. Remember to use absolute values where appropriate. Use C for the constant of integration.)

Bess [88]

Answer:

F(x)= 5*[ $\frac{x^{3} }{3}$ + (a*b)* $\frac{x^{2} }{2}$ + a*b*x + C.

Step-by-step explanation:

First step we aplicate distributive property to the function.

5*(x+a)*(x+b)= 5*[ $x^{2}$ +x*b+a*x+a*b]

5*[ $x^{2}$ +x*(b+a)+a*b]= f(x), where a, b are constant and a≠b

integrating we find ⇒∫f(x)*dx= F(x) + C, where C= integration´s constant

∫^5*[ $x^{2}$ +x*(a+b)+a*b]*dx, apply integral´s property

5*[∫ $x^{2}$ dx+∫(a*b)*x*dx + ∫a*b*dx], resolving the integrals

5*[ $\frac{x^{3} }{3}$ + (a*b)* $\frac{x^{2} }{2}$ + a*b*x

Finally we can write the function F(x)

F(x)= 5*[ $\frac{x^{3} }{3}$ + (a*b)* $\frac{x^{2} }{2}$ + a*b*x ]+ C.

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3 years ago

Find the product of (√3-5) and (√3+2)

beks73 [17]

Step-by-step explanation:

check it.. i have solved it for you.

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X²+8x-16 by factorising method what is the ans pls I need it fast ASAP

ch4aika [34]

Answer:

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Step-by-step explanation:

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1 year ago

Need help with this ignore what a put in the box it isn’t right

Zarrin [17]

$f=6cm\\g=8cm$

Why?

The first thing we need to do is find the area of the triangle, we can to that by subtracting the area of ABCD from ACBE, then, we can use the formulas to calculate the area for both triangle and rectangle to find "f" and "g".

Calculating we have:

$TriangleArea=ABCE-ABCD\\\\TriangleArea=60cm^{2}-48cm^{2}=12cm^{2}$