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3 years ago

Suppose you surveyed a group of students to see how many CDs they owned. The results are displayed in the

Mathematics

1 answer:

Alex17521 [72]3 years ago

7 0

You surveyed 30 students to see how many CDs they owned.

The number of students who owned 5 - 8 CDs was:

= 15

The number of students who owned 9 - 12 CDs was:

= 10

The number of students who owned 13 - 16 CDs was:

= 5

The number of students who owned 17 - 20 CDs was:

= 0

The total number of students surveyed is the sum of the above:

= 15 + 10 + 5 + 0

= 30 students

30 students were surveyed for this study.

<em>Find out more at brainly.com/question/16372388.</em>

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Equation: –4x – 5 + 2x = –11

BaLLatris [955]

<h2><u>ANSWER </u></h2>

x= 3

Step-by-step explanation:

=> -4x – 5 + 2x = -11

=> -2x = -11+5

=> -2x = -6

=> x= 6/2

=> x= 3

5 0

3 years ago

Read 2 more answers

Please help me! I am really struggling with this...

Alexeev081 [22]

Answer:

44°

Step-by-step explanation:

The secant- secant angle y is half the difference of the measure of its intercepted arcs, that is

$\frac{1}{2}$ (BHF - CGJ ) = y , that is

$\frac{1}{2}$ (156 - CGH) = 56° ( multiply both sides by 2 )

156 - CGH = 112° , thus

CGH = 156° - 112° = 44°

4 0

4 years ago

The table below shows two equations:

VMariaS [17]

| 3x - 1 | + 7 = 2
| 3x - 1 | = 2 - 7
| 3x - 1 | = -5.....no solution because an absolute value cannot equal a negative number.

| 2x + 1 | + 4 = 3
| 2x + 1 | = 3 - 4
|2x + 1 | = -1...same thing...no solution..cannot equal a neg. number

so Eq. 1 and Eq. 2 both have NO SOLUTIONS :)

8 0

4 years ago

Read 2 more answers

Tell whether the 2 rates form a proportion:

marshall27 [118]

A forms a proportion, 28x4=112 and 3x4=12.
B does not form a proportion.

3 0

3 years ago

Combine into a single logarithm.<br><br> 3log(x+y)+2log(x-y)-log(x^2 +y^2)

seropon [69]

Answer:

$3\log _{10}\left(x+y\right)+2\log _{10}\left(x-y\right)-\log _{10}\left(x^2+y^2\right)=\log _{10}\left(\frac{\left(x+y\right)^3\left(x-y\right)^2}{x^2+y^2}\right)$

Step-by-step explanation:

Given the expression

$3log\left(x+y\right)+2log\left(x-y\right)-log\left(x^2\:+y^2\right)$

solving to write into a single logarithm

$3log\left(x+y\right)+2log\left(x-y\right)-log\left(x^2\:+y^2\right)$

$\mathrm{Apply\:log\:rule}:\quad \:a\log _c\left(b\right)=\log _c\left(b^a\right)$

$3\log _{10}\left(x+y\right)=\log _{10}\left(\left(x+y\right)^3\right)$

$=\log _{10}\left(\left(x+y\right)^3\right)+2\log _{10}\left(x-y\right)-\log _{10}\left(x^2+y^2\right)$

$\mathrm{Apply\:log\:rule}:\quad \:a\log _c\left(b\right)=\log _c\left(b^a\right)$

$2\log _{10}\left(x-y\right)=\log _{10}\left(\left(x-y\right)^2\right)$

$=\log _{10}\left(\left(x+y\right)^3\right)+\log _{10}\left(\left(x-y\right)^2\right)-\log _{10}\left(x^2+y^2\right)$

$\mathrm{Apply\:log\:rule}:\quad \log _c\left(a\right)+\log _c\left(b\right)=\log _c\left(ab\right)$

$\log _{10}\left(\left(x+y\right)^3\right)+\log _{10}\left(\left(x-y\right)^2\right)=\log _{10}\left(\left(x+y\right)^3\left(x-y\right)^2\right)$

$=\log _{10}\left(\left(x+y\right)^3\left(x-y\right)^2\right)-\log _{10}\left(x^2+y^2\right)$

$\mathrm{Apply\:log\:rule}:\quad \log _c\left(a\right)-\log _c\left(b\right)=\log _c\left(\frac{a}{b}\right)$

$\log _{10}\left(\left(x+y\right)^3\left(x-y\right)^2\right)-\log _{10}\left(x^2+y^2\right)=\log _{10}\left(\frac{\left(x+y\right)^3\left(x-y\right)^2}{x^2+y^2}\right)$

$=\log _{10}\left(\frac{\left(x+y\right)^3\left(x-y\right)^2}{x^2+y^2}\right)$

Thus,

$3\log _{10}\left(x+y\right)+2\log _{10}\left(x-y\right)-\log _{10}\left(x^2+y^2\right)=\log _{10}\left(\frac{\left(x+y\right)^3\left(x-y\right)^2}{x^2+y^2}\right)$

6 0

3 years ago