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DENIUS [597]
3 years ago
7

24, 34, 32, 16, 45, 38, 28MediaMean

Mathematics
2 answers:
Mekhanik [1.2K]3 years ago
5 0
Do u mean media as in median ? Or is it media ?
The mean is super easy add all them up then divide by how many numbers there are(For mean)
KengaRu [80]3 years ago
4 0
Mean is the average so add

24+34+32+16+45+38+28=217

Then divide by 7 ( because there are 7 numbers

Mean=31
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Answer: The numbers are 1 and 3.

Step-by-step explanation:

Let x = smaller number , y= larger number.

As per given,

x=\dfrac{y}{3}   ...(i)

x+y=y^2-5       ...(ii)

Put value of x from (i) in (ii)

\dfrac{y}{3}+y=y^2-5\\\\\Rightarrow\ \dfrac43y=y^2-5\\\\\Rightarrow\ 3y^2-4y-15=0\\\\\Rightarrow\ 3y^2-9y+5y-15=0\\\\\Rightarrow\ 3y(y-3)+5(y-3)=0\\\\\Rightarrow\ (y-3)(3y+5)=0\\\\\Rightarrow\ y=3 \ \ or\ y=\dfrac{-5}{3}

Since numbers are positive , so y=3 is correct.

And x will be 1  [from (i)]

Hence, the numbers are 1 and 3.

7 0
3 years ago
Find the surface area <br> 3 in<br> 4 in<br> 4 in<br> 10 in<br> PLZ HELP
Artyom0805 [142]

2(½ x 3 x 4) + 3(10 x 4)

=2(½ x 12) + 3(40)

=2(6) +120

=12+120

=132

4 0
3 years ago
A circle is inscribed with quadrilateral A B C D.
Vlada [557]

Answer:

Its incribed angle theorem

Step-by-step explanation:

I took the test

3 0
3 years ago
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Would it be 3.4, 4.3, 5.2, 6.1, 7
6 0
3 years ago
A cylinder shaped can needs to be constructed to hold 400 cubic centimeters of soup. The material for the sides of the can costs
LenKa [72]

Answer:

The dimensions of the can that will minimize the cost are a Radius of 3.17cm and a Height of 12.67cm.

Step-by-step explanation:

Volume of the Cylinder=400 cm³

Volume of a Cylinder=πr²h

Therefore: πr²h=400

h=\frac{400}{\pi r^2}

Total Surface Area of a Cylinder=2πr²+2πrh

Cost of the materials for the Top and Bottom=0.06 cents per square centimeter

Cost of the materials for the sides=0.03 cents per square centimeter

Cost of the Cylinder=0.06(2πr²)+0.03(2πrh)

C=0.12πr²+0.06πrh

Recall: h=\frac{400}{\pi r^2}

Therefore:

C(r)=0.12\pi r^2+0.06 \pi r(\frac{400}{\pi r^2})

C(r)=0.12\pi r^2+\frac{24}{r}

C(r)=\frac{0.12\pi r^3+24}{r}

The minimum cost occurs when the derivative of the Cost =0.

C^{'}(r)=\frac{6\pi r^3-600}{25r^2}

6\pi r^3-600=0

6\pi r^3=600

\pi r^3=100

r^3=\frac{100}{\pi}

r^3=31.83

r=3.17 cm

Recall that:

h=\frac{400}{\pi r^2}

h=\frac{400}{\pi *3.17^2}

h=12.67cm

The dimensions of the can that will minimize the cost are a Radius of 3.17cm and a Height of 12.67cm.

3 0
3 years ago
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