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gayaneshka [121]

2 years ago

15

The accompanying table gives the population of 6 regions in 2008 and the area of these regions. Use the

1 answer:

Natali [406]2 years ago

6 0

The answer is b click the icon to view the data table

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A bag contains 5 white, 3 black, and 2 green balls. Balls are picked at random.

Soloha48 [4]

Answer:

look down there

Step-by-step explanation:

First ball:

Probability of drawing a white ball is 5/8

Probability of drawing a black ball is 3/8

Second ball:

This depends on the first ball drawn, lets say you drew a white ball initially, 4 white balls are left out of 7 balls in total. The probability of a white ball in the second pick is 4/7.

Total probability of drawing two white balls is 5/8*4/7 (since they are independent events).

If you picked a black ball initially, picking another black ball would have a probability of 2/7, on similar grounds , total prob for 2 blacks would be 3/8*2/7.

The probability that you pick 2 balls of same color is (5/14 + 3/28) = 13/28. (Since they are mutually exclusive events)

6 0

3 years ago

Please help need answers

bogdanovich [222]

see explanation below

(1) $\frac{1}{5}$ × $\frac{2}{2}$ = $\frac{2}{10}$ = 0.2

(2) $\frac{6}{25}$ × $\frac{4}{4}$ = $\frac{24}{100}$ = 0.24

(3) 2 $\frac{3}{4}$ = 2 + $\frac{75}{100}$ = 2.75

(4) 3 $\frac{9}{10}$ = 3 + 0.9 = 3.9

(5) 1.25 = 1 $\frac{1}{4}$ = $\frac{5}{4}$

(6) 3.29 = 3 $\frac{29}{100}$ = $\frac{329}{100}$

(7) 0.65 = $\frac{65}{100}$ = $\frac{13}{20}$ in simplest form

(8) 5.6 = 5 $\frac{6}{10}$ = 5 $\frac{3}{5}$ = $\frac{28}{5}$

(9) he is incorrect

$\frac{3}{5}$ × $\frac{20}{20}$ = $\frac{60}{100}$ = 0.6 ≠ 3.5

7 0

3 years ago

Which us the simplified form of <br><img src="https://tex.z-dn.net/?f=%20%7Bp%7D%5E%7B0%7D%20" id="TexFormula1" title=" {p}^{0}

Mice21 [21]

1
anything with the power of 0=1

3 0

2 years ago

(1+cos2x)/(1-cos2x) = cot^2x

sesenic [268]

We will turn the left side into the right side.

$\dfrac{1 + \cos 2x}{1 - \cos2x} = \cot^2 x$

Use the identity:

$\cos 2x = \cos^2 x - \sin^2 x$

$\dfrac{1 + \cos^2 x - \sin^2 x}{1 - ( \cos^2 x - \sin^2 x)} = \cot^2 x$

$\dfrac{1 - \sin^2 x + \cos^2 x }{1 - \cos^2 x + \sin^2 x} = \cot^2 x$

Now use the identity

$\sin^2 x + \cos^2 x = 1$ solved for sin^2 x and for cos^2 x.

$\dfrac{\cos^2 x + \cos^2 x }{\sin^2 x + \sin^2 x} = \cot^2 x$

$\dfrac{2\cos^2 x}{2\sin^2 x} = \cot^2 x$

$\dfrac{\cos^2 x}{\sin^2 x} = \cot^2 x$

$\cot^2 x = \cot^2 x$

8 0

3 years ago

Rodrick and Valentina drove to the coast. Rodrick drove 38 9−− 10miles. Then Valentina drove the last 51 3− 5miles. How far did

ivolga24 [154]

I believe the answer would be 69 1/2 and/or 69.5

3 0

2 years ago

Read 2 more answers