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3 years ago

What is the mean of the following set of numbers rounded to the nearest whole number?

1 answer:

4 0

The mean is when you add up all the numbers and divide by the amount of numbers there are.
143+312+41+28+308+619+321+352+465=2589
2589/9=287.666666667
round to the whole number you get
288

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Please help me with this question.

Gelneren [198K]

Answer 1/4

Step-by-step explanation:

3/12 divided by 3 is 1/4

7 0

3 years ago

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Write an equation for three-fourth of m is 20

alisha [4.7K]

Answer:

Since, three fourth of $t$ is $15$. Hence, the required equation is $\frac{3t}{4}=15$

Step-by-step explanation:

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2 years ago

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Fine the product:3 x 5 (2)

Delvig [45]

Answer:

30 because 3 x 5=15 then x2 it's 30

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3 years ago

Kaitlin, Tony, and Austin sent a total of 103 text messages over their cell phones during the weekend. Austin sent 4 times as ma

emmainna [20.7K]

The number of messages Kaitlyn, Tony and Austin sent are 13, 18 and 72 messages respectively.

<h3>How to write and solve equation?</h3>

let

Number of messages Kaitlyn sent = x
Number of messages Tony sent = x + 5
Number of messages Austin sent = 4(x + 5)
Total messages = 103

103 = x + (x + 5) + 4(x + 5)

103 = x + x + 5 + 4x + 20

103 = 6x + 25

103 - 25 = 6x

78 = 6x

x = 78/6

x = 13

Number of messages each sent:

Number of messages Kaitlyn sent = x

= 13 messages

Number of messages Tony sent = x + 5

= 13 + 5

= 18 messages

Number of messages Austin sent = 4(x + 5)

= 4(13 + 5)

= 4(18)

= 72 messages

Learn more about equation:

brainly.com/question/2972832

#SPJ1

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1 year ago

A wire b units long is cut into two pieces. One piece is bent into an equilateral triangle and the other is bent into a circle.

mezya [45]

1. Divide wire b in parts x and b-x.

2. Bend the b-x piece to form a triangle with side (b-x)/3

There are many ways to find the area of the equilateral triangle. One is by the formula A= $\frac{1}{2}sin60^{o}side*side= \frac{1}{2} \frac{ \sqrt{3} }{2} (\frac{b-x}{3}) ^{2}= \frac{ \sqrt{3} }{36}(b-x)^{2}$
A= $\frac{ \sqrt{3} }{36}(b-x)^{2}=\frac{ \sqrt{3} }{36}( b^{2}-2bx+ x^{2} )=\frac{ \sqrt{3} }{36}b^{2}-\frac{ \sqrt{3} }{18}bx+ \frac{ \sqrt{3} }{36}x^{2}$

Another way is apply the formula A=1/2*base*altitude,
where the altitude can be found by applying the pythagorean theorem on the triangle with hypothenuse (b-x)/3 and side (b-x)/6

3. Let x be the circumference of the circle.

$2 \pi r=x$

so $r= \frac{x}{2 \pi }$

Area of circle = $\pi r^{2}= \pi ( \frac{x}{2 \pi } )^{2} = \frac{ \pi }{ 4 \pi ^{2} }* x^{2} = \frac{1}{4 \pi } x^{2}$

4. Let f(x)= $\frac{ \sqrt{3} }{36}b^{2}-\frac{ \sqrt{3} }{18}bx+ \frac{ \sqrt{3} }{36}x^{2}+\frac{1}{4 \pi } x^{2}$

be the function of the sum of the areas of the triangle and circle.

5. f(x) is a minimum means f'(x)=0

f'(x)= $\frac{ -\sqrt{3} }{18}b+ \frac{ \sqrt{3} }{18}x+\frac{1}{2 \pi } x$ =0

$\frac{ -\sqrt{3} }{18}b+ \frac{ \sqrt{3} }{18}x+\frac{1}{2 \pi } x=0$

$(\frac{ \sqrt{3} }{18}+\frac{1}{2 \pi }) x=\frac{ \sqrt{3} }{18}b$

$x= \frac{\frac{ \sqrt{3} }{18}b}{(\frac{ \sqrt{3} }{18}+\frac{1}{2 \pi }) }$

6. So one part is $\frac{\frac{ \sqrt{3} }{18}b}{(\frac{ \sqrt{3} }{18}+\frac{1}{2 \pi }) }$ and the other part is b- $\frac{\frac{ \sqrt{3} }{18}b}{(\frac{ \sqrt{3} }{18}+\frac{1}{2 \pi }) }$

4 0

3 years ago

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