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3 years ago

What is the mean of this set 18,21,20,14,17

Mathematics

2 answers:

konstantin123 [22]3 years ago

6 0

Answer: 18

Step-by-step explanation:

Add them all up

18+21+20+14+17=90

Divide by 5 since that is how many numbers there are

90/5=18

Darya [45]3 years ago

5 0

Answer:

the mean is 18

Step-by-step explanation:

you add them all up and then divide by the number of numbers

18+21+20+14+17=90

90/5=18

hey diddle diddle the median is in the middle, you add and divide for the mean the mode is the one that appears the most and the range is the difference between.

Hope this helps!!

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Pls help if you actually can

Setler79 [48]

The triangles that are similar would be ΔGCB and ΔPEB due to Angle, Angle, Angle similarity theorem.

<h3>How to identify similar triangles?</h3>

From the image attached, we see that we are given the Parallelogram GRPC. Thus;

A. The triangles that are similar would be ΔGCB and ΔPEB due to Angle, Angle, Angle similarity theorem.

B. The proof of the fact that ΔGCB and ΔPEB are similar pairs of triangles is as follow;

∠CGB ≅ ∠PEB (Alternate Interior Angles)

∠BPE ≅ ∠BCG (Alternate Interior Angles)

∠GBC ≅ ∠EBP (Vertical Angles)

C. To find the distance from B to E and from P to E, we will first find PE and then BE by proportion;

225/325 = PE/375

PE = 260 ft

BE/425 = 225/325

BE = 294 ft

Read more about Similar Triangles at; brainly.com/question/14285697

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2 years ago

Find three positive numbers whose sum is 140 and whose product is a maximum. (Enter your answers as a comma-separated list.)

Paladinen [302]

Answer:

$\frac{140}{3}, \frac{140}{3}, \frac{140}{3}$

Step-by-step explanation:

Let the numbers be $x, y\ and\ z.$

Such that:

$x + y + z = 140$

Make z the subject

$z = 140 -x - y$

For their product to be maximum, we have:

$f(x,y,z) = xyz$

Substitute $z = 140 -x - y$ in $f(x,y,z) = xyz$

$f(x,y) = xy(140 - x - y)$

Open bracket

$f(x,y) = 140xy - x^2y - xy^2$

Differentiate w.r.t x and y

$f_x=140y - 2xy - y^2$

$f_y=140x - x^2 - 2xy$

Since the products are maximum, then $f_x = f_y = 0$

For $f_x=140y - 2xy - y^2$

$140y - 2xy - y^2 = 0$

Factorize:

$y(140 - 2x - y) = 0$

Split

$y = 0\ or\ 140 - 2x - y = 0$

Make y the subject

$y = 0\ or\ y = 140 - 2x$

For $f_y=140x - x^2 - 2xy$

$140x - x^2 - 2xy = 0$

---------------------------------------------------

Substitute y = 0

$140x - x^2 -2x*0 = 0$

$140x - x^2 = 0$

Factorize

$x(140 - x)= 0$

$x = 0\ or\ 140-x = 0$

$x = 0\ or\ x = 140$

---------------------------------------------------

Substitute $y = 140 - 2x$

$140x - x^2 - 2xy = 0$

$140x - x^2 - 2x(140 - 2x) = 0$

$140x - x^2 - 280x + 4x^2 = 0$

Re-arrange

$4x^2 -x^2 +140x - 280x = 0$

$3x^2 -140x = 0$

Factor x out

$x(3x - 140) = 0$

Divide through by x

$3x - 140 = 0$

$3x = 140$

$x = \frac{140}{3}$

Recall that: $y = 140 - 2x$

$y = 140 - 2 * \frac{140}{3}$

$y = 140 - \frac{280}{3}$

Take LCM

$y = \frac{140*3-280}{3}$

$y = \frac{140}{3}$

Recall that:

$z = 140 -x - y$

$z = 140 - \frac{140}{3} - \frac{140}{3}$

Take LCM

$z = \frac{3 * 140- 140 - 140}{3}$

$z = \frac{140}{3}$

Hence, the numbers are:

$\frac{140}{3}, \frac{140}{3}, \frac{140}{3}$

8 0

3 years ago