Answer:
Prat 9) 
Part 10) 
Part 11) 
Part 12) 
Step-by-step explanation:
we know that
If two figures are similar, then the ratio of its areas is equal to the scale factor squared
Let
SF-----> the scale factor
a----> area of the shaded figure
b----> area of the unshaded figure

Problem 9) we have


substitute in the formula


------> the scale factor
To find the value of x, multiply the length of the unshaded figure by the scale factor

Problem 10) we have


substitute in the formula


------> the scale factor
To find the value of x, multiply the length of the unshaded figure by the scale factor

Problem 11) we have


substitute in the formula


------> the scale factor
To find the value of x, multiply the length of the unshaded figure by the scale factor

Problem 12) we have


substitute in the formula


------> the scale factor
To find the value of x, divide the length of the shaded figure by the scale factor

1
Simplify n\times \frac{6}{9}n×96 to \frac{2n}{3}32n
\frac{2n}{3}=\frac{3}{12}32n=123
2
Simplify \frac{3}{12}123 to \frac{1}{4}41
\frac{2n}{3}=\frac{1}{4}32n=41
3
Multiply both sides by 33
2n=\frac{1}{4}\times 32n=41×3
4
Simplify \frac{1}{4}\times 341×3 to \frac{3}{4}43
2n=\frac{3}{4}2n=43
5
Divide both sides by 22
n=\frac{\frac{3}{4}}{2}n=243
6
Simplify \frac{\frac{3}{4}}{2}243 to \frac{3}{4\times 2}4×23
n=\frac{3}{4\times 2}n=4×23
7
Simplify 4\times 24×2 to 88
n=\frac{3}{8}n=83
Answer:
0.1433
Step-by-step explanation:
<h3><u>All possibilities:</u></h3>
The total number of different sequences in 40 rolls of a die is 
<h3><u>Exactly five 1s in 40 rolls:</u></h3>
The 1s may occur at any 5 rolls among 40. The number of ways of exactly five 1s occuring is therefore, equivalent to number of ways of selecting 5 fruits from 40 distinct fruits which is ⁴⁰C₅ ways. These 5 rolls have a fixed outcome <u>1</u>. Other 35 rolls each have 5 possible outcomes : <u>2</u> or <u>3</u> or <u>4</u> or <u>5</u> or <u>6</u>. So, the number of possible sequences of outcomes on other 35 rolls of the die is
.
⇒The total number of different sequences having exactly five 1s in 40 rolls of a die is (⁴⁰C₅)×(
)
∴The fraction of the total number of sequences having exactly five 1s is
)}{
}[/tex] ≈ <u>0.1433</u>