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3 years ago

The is the mean score ? A.25.5 B.26 C.

Mathematics

1 answer:

pentagon [3]3 years ago

5 0

So to find the mean you add up all the numbers (or the exam scores) which will get you 212 then you take the number and divide it by the total numbers in the set (how many numbers did you add up?)

212 divided by 8 is 26.5 which would be C. I hope this helps!

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Given the original number n. Multiply the number by 9. Add 99. Divide this sum by 9. Subtract the original number, n, from the q

tangare [24]

Answer:

Step-by-step explanation:

Hello,

Given the original number n.

$\large \boxed{n}$

Multiply the number by 9.

$\large \boxed{9n}$

Add 99.

$\large \boxed{9n+99}$

Divide this sum by 9.

$\large \boxed{\dfrac{9n+99}{9}=n+11}$

Subtract the original number, n, from the quotient.

$\large \boxed{n+11-n=11}$

Thank you.

4 0

3 years ago

Help Me Help Me Help!!!!

lions [1.4K]

X=6 because
9-6=3

X=3 because
2•3=6

X=2 because
1.5 or 1 1/2 • 2 = 3

Not sure if this was what you needed or not!

3 0

3 years ago

Read 2 more answers

Triangle ABC has area 150 cm^2 find the value of x BA= 17 cm BC= x cm sin68

Romashka-Z-Leto [24]

Here's a picture of the problem

8 0

3 years ago

Explain how to get that answer!!

ra1l [238]

We need to simplify $\frac{ \sqrt{14x^3} }{ \sqrt{18x} }$

First lets factor $\sqrt{14x^3}$

$\sqrt{14x^3}$ = $\sqrt{14} \sqrt{x^3}$
$\sqrt{14} = \sqrt{2} \sqrt{7}$ by applying the radical rule $\sqrt[n]{ab} = \sqrt[n]{a} \sqrt[n]{b}$
$\sqrt{x^3} = x^{3/2}$ By applying the radical rule $\sqrt[n]{x^m} = x^{m/n}$

So
$\sqrt{14x^3}$ = $\sqrt{14} \sqrt{x^3}$ = $\sqrt{2} \sqrt{7}x^{3/2}$

Now let's factor $\sqrt{18x}$
By applying the radical rule $\sqrt[n]{ab} = \sqrt[n]{a} \sqrt[n]{b}$ ,
$\sqrt{18x} = \sqrt{18} \sqrt{x}$
$\sqrt{18} = \sqrt{2} * 3$

So $\sqrt{18x}$ = $\sqrt{2}*3 \sqrt{x}$

So $\frac{ \sqrt{14x^3} }{ \sqrt{18x} }$ = $\frac{ \sqrt{2} \sqrt{7} x^{3/2} }{ \sqrt{2}*3 \sqrt{x} }$

We know that $\sqrt[n]{x} = x^{1/n}$ so $\sqrt{x} = x^{1/2}$

We now have $\frac{ \sqrt{2} \sqrt{7} x^{3/2} }{ \sqrt{2}*3 \sqrt{x}} = \frac{ \sqrt{2} \sqrt{7} x^{3/2} }{ \sqrt{2}*3x^{1/2}}$

We know that $\frac{x^a}{x^b} = x^{a-b}$
So $\frac{x^{3/2}}{x^{1/2}} = x^{3/2 - 1/2} = x$

We now got $\frac{ \sqrt{2} \sqrt{7} x^{3/2} }{ \sqrt{2}*3x^{1/2}} = \frac{ \sqrt{2} \sqrt{7} x }{ \sqrt{2}*3}
$

We can notice that the numerator and the denominator both got √2 in a multiplication, so we can simplify them, and we get:
$\frac{ \sqrt{2} \sqrt{7} x }{ \sqrt{2}*3} = \frac{ \sqrt{7}x }{3}$

All in All, we get $\frac{ \sqrt{14x^3} }{ \sqrt{18x} }$ = $\frac{ \sqrt{7}x }{3}$

Hope this helps! :D

6 0

3 years ago

Please Help!

borishaifa [10]

Answer:

Both triangles have acute angles

8 0

3 years ago