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3 years ago

Find the median. Round to the nearest tenth if necessary.

Mathematics

1 answer:

Rom4ik [11]3 years ago

5 0

Answer:

28.5

Step-by-step explanation:

(35+24+32+24+17+33+39+23)/8

(59+56+50+62)/8

(115+112)/8

227/8

=28.375

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What constant term should be added to both sides to complete the square on the left side?

grandymaker [24]

Answer:

9/64

Step-by-step explanation:

Given

x² - 3/4x + _____ = 10

Need to complete square on the left side of equation

The square should be in the form of:

(a + b)² = a² + 2ab + b²

We see a = x therefore

2b = -3/4, so
b = -3/8 and
b² = (-3/8)² = 9/64

So the missing constant term is 9/64

6 0

3 years ago

PLEASE HELP ME ROUND THIS TO NEAREST TENTH 251.327

aleksandr82 [10.1K]

251.327 = 251.33
Rounded to the nearest tenth is 251.3
Hope this helps!

5 0

3 years ago

Read 2 more answers

Help quick WILL GIVE BRAINLY

IceJOKER [234]

Answer:

The correct answer is 672.

Step-by-step explanation:

Plz mark as brainliest ! :D

5 0

3 years ago

Read 2 more answers

In the diagram, what is the value of x? A. 78° B. 108° C. 98° D. 82°

mamaluj [8]

the answer is 98 degrees. C:

5 0

3 years ago

Solve this algebra...

Mumz [18]

Step-by-step explanation:

$\text{Use}\ \dfrac{a^n}{a^m}=a^{n-m}\ \text{and}\ (a^n)^m=a^{nm}\\\\\left(\dfrac{a^x}{a^y}\right)^{x-y}=\left(a^{x-y}\right)^{x-y}=a^{(x-y)(x-y)}=a^{(x-y)^2}\\\\\left(\dfrac{a^y}{a^z}\right)^{y-z}=\left(a^{y-z}\right)^{y-z}=a^{(y-z)(y-z)}=a^{(y-z)^2}\\\\\left(\dfrac{a^z}{a^x}\right)^{z-x}=\left(a^{z-x}\right)^{z-x}=a^{(z-x)(z-x)}=a^{(z-x)^2}\\\\\text{Use}\ a^n\cdot a^m=a^{n+m}\\\\a^{(x-y)^2}\cdot a^{(y-z)^2}\cdot a^{(z-x)^2}=a^{(x-y)^2+(y-z)^2+(z-x)^2}$

$\text{use}\ (a-b)^2=a^2-2ab+b^2\\\\(x-y)^2=x^2-2xy+y^2\\(y-z)^2=y^2-2yz+z^2\\(z-x)^2=z^2-2zx+x^2\\\\=a^{x^2-2xy+y^2+y^2-2yz+z^2+z^2-2zx+x^2}\\\\\text{combine like terms}\\\\=a^{(x^2+x^2)+(y^2+y^2)+(z^2+z^2)-2xy-2yz-2zx}\\\\=a^{2x^2+2y^2+2z^2-2xy-2yz-2zx}\qquad\text{distributive}\\\\=a^{2(x^2+y^2+z^2)-2(xy-yz-zx)}\\\\\text{From the equastion we know:}\ x^2+y^2+z^2=xy+yx+zx.\\\text{Therefore}\\\\=a^{2(x^2+y^2+z^2)-2(x^2+y^2+z^2)}=a^0=1$

6 0

4 years ago