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3 years ago

12.what is the probability that a boy and a girl chosen randsomly will be seniors?

Mathematics

1 answer:

zhenek [66]3 years ago

3 0

$\frac{\textbf{1}}{\textbf{20}}$

Step-by-step explanation:

Probability= $\frac{\text{number of favourable outcomes}}{\text{total number of outcomes}}$

Probability for a randomly chosen girl to be senior= $\frac{\text{number of senior girls}}{\text{total number of girls}}$

Probability for a randomly chosen girl to be senior= $\frac{7}{8+11+9+7}=\frac{7}{35}=\frac{1}{5}$

Probability for a randomly chosen boy to be senior= $\frac{\text{number of senior boys}}{\text{total number of boys}}$

Probability for a randomly chosen girl to be senior= $\frac{9}{10+7+10+9}=\frac{9}{36}=\frac{1}{4}$

For two independent events,

Probability for both event 1 and event 2 to take place= $\text{probability of event 1} \times \text{probability of event 2}$

Since choosing boys and girls is independent,

Probability for both boy an girl chosen to be senior= $\text{probability for boy to be senior}\times\text{probability for girl to be senior}$

Probability for both boy and girl chosen to be senior= $\frac{1}{5} \times \frac{1}{4} = \frac{1}{20}$

So,required probability is $\frac{1}{20}$

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A dartboard consists of two concentric circles. The probability of hitting the inner circle is

stepan [7]

Given:
Diameter of outer circle = 20 inches.

We need to find the Area of the outer circle to get the radius of the inner circle.

Area = πr²

Outer circle Area = 3.14 * (10in)² = 314 in²

314 in² * 64% probability = 200.96 in² Area of the inner circle.

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64 in² = r²
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radius of inner circle is 8 inches.

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3 years ago

In a certain sequence of numbers, each term after the first is found by doubling and then adding $3$ to the previous term. If th

eimsori [14]

Given $a_7=125$ you can find a_6:

$a_6= \frac{125-3}{2} =61$
$a_5= \frac{61-3}{2} =29 \\ a_4= \frac{29-3}{2}=13 \\ a_3= \frac{13-3}{2} =5 \\ a_2= \frac{5-3}{2} =1 \\ a_1= \frac{1-3}{2}=-1$
Answer: -1

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3 years ago

For what value of a should you solve the system of elimination?

SIZIF [17.4K]

$\begin{bmatrix}3x+5y=10\\ 2x+ay=4\end{bmatrix}$

$\mathrm{Multiply\:}3x+5y=10\mathrm{\:by\:}2: 6x+10y=20$
$\mathrm{Multiply\:}2x+ay=4\mathrm{\:by\:}3: 3ay+6x=12$

$\begin{bmatrix}6x+10y=20\\ 6x+3ay=12\end{bmatrix}$

6x + 3ay = 12
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6x + 10y = 20
/
3a - 10y = -8

$\begin{bmatrix}6x+10y=20\\ 3a-10y=-8\end{bmatrix}$

$3a-10y=-8 \ \textgreater \ \mathrm{Subtract\:}3a\mathrm{\:from\:both\:sides}$
$3a-10y-3a=-8-3a$

$\mathrm{Simplify} \ \textgreater \ -10y=-8-3a \ \textgreater \ \mathrm{Divide\:both\:sides\:by\:}-10$
$\frac{-10y}{-10}=-\frac{8}{-10}-\frac{3a}{-10}$

Simplify more.

$\frac{-10y}{-10} \ \textgreater \ \mathrm{Apply\:the\:fraction\:rule}: \frac{-a}{-b}=\frac{a}{b} \ \textgreater \ \frac{10y}{10}$

$\mathrm{Divide\:the\:numbers:}\:\frac{10}{10}=1 \ \textgreater \ y$

$-\frac{8}{-10}-\frac{3a}{-10} \ \textgreater \ \mathrm{Apply\:rule}\:\frac{a}{c}\pm \frac{b}{c}=\frac{a\pm \:b}{c} \ \textgreater \ \frac{-8-3a}{-10}$

$\mathrm{Apply\:the\:fraction\:rule}: \frac{a}{-b}=-\frac{a}{b} \ \textgreater \ -\frac{-3a-8}{10} \ \textgreater \ y=-\frac{-8-3a}{10}$

$\mathrm{For\:}6x+10y=20\mathrm{\:plug\:in\:}\ \:y=\frac{8}{10-3a} \ \textgreater \ 6x+10\cdot \frac{8}{10-3a}=20$

$10\cdot \frac{8}{10-3a} \ \textgreater \ \mathrm{Multiply\:fractions}: \:a\cdot \frac{b}{c}=\frac{a\:\cdot \:b}{c} \ \textgreater \ \frac{8\cdot \:10}{10-3a}$
$\mathrm{Multiply\:the\:numbers:}\:8\cdot \:10=80 \ \textgreater \ \frac{80}{10-3a}$

$6x+\frac{80}{10-3a}=20 \ \textgreater \ \mathrm{Subtract\:}\frac{80}{10-3a}\mathrm{\:from\:both\:sides}$
$6x+\frac{80}{10-3a}-\frac{80}{10-3a}=20-\frac{80}{10-3a}$

$\mathrm{Simplify} \ \textgreater \ 6x=20-\frac{80}{10-3a} \ \textgreater \ \mathrm{Divide\:both\:sides\:by\:}6 \ \textgreater \ \frac{6x}{6}=\frac{20}{6}-\frac{\frac{80}{10-3a}}{6}$

$\frac{6x}{6} \ \textgreater \ \mathrm{Divide\:the\:numbers:}\:\frac{6}{6}=1 \ \textgreater \ x$

$\frac{20}{6}-\frac{\frac{80}{10-3a}}{6} \ \textgreater \ \mathrm{Apply\:rule}\:\frac{a}{c}\pm \frac{b}{c}=\frac{a\pm \:b}{c} \ \textgreater \ \frac{20-\frac{80}{-3a+10}}{6}$

$20-\frac{80}{10-3a} \ \textgreater \ \mathrm{Convert\:element\:to\:fraction}: \:20=\frac{20}{1} \ \textgreater \ \frac{20}{1}-\frac{80}{-3a+10}$

$\mathrm{Find\:the\:least\:common\:denominator\:}1\cdot \left(-3a+10\right)=-3a+10$

$Adjust\:Fractions\:based\:on\:the\:LCD \ \textgreater \ \frac{20\left(-3a+10\right)}{-3a+10}-\frac{80}{-3a+10}$

$\mathrm{Since\:the\:denominators\:are\:equal,\:combine\:the\:fractions}: \frac{a}{c}\pm \frac{b}{c}=\frac{a\pm \:b}{c}$
$\frac{20\left(-3a+10\right)-80}{-3a+10} \ \textgreater \ \frac{\frac{20\left(-3a+10\right)-80}{-3a+10}}{6} \ \textgreater \ \mathrm{Apply\:the\:fraction\:rule}: \frac{\frac{b}{c}}{a}=\frac{b}{c\:\cdot \:a}$

$20\left(-3a+10\right)-80 \ \textgreater \ Rewrite \ \textgreater \ 20+10-3a-4\cdot \:20$

$\mathrm{Factor\:out\:common\:term\:}20 \ \textgreater \ 20\left(-3a+10-4\right) \ \textgreater \ Factor\;more$

$10-3a-4 \ \textgreater \ \mathrm{Subtract\:the\:numbers:}\:10-4=6 \ \textgreater \ -3a+6 \ \textgreater \ Rewrite$
$-3a+2\cdot \:3$

$\mathrm{Factor\:out\:common\:term\:}3 \ \textgreater \ 3\left(-a+2\right) \ \textgreater \ 3\cdot \:20\left(-a+2\right) \ \textgreater \ Refine$
$60\left(-a+2\right)$

$\frac{60\left(-a+2\right)}{6\left(-3a+10\right)} \ \textgreater \ \mathrm{Divide\:the\:numbers:}\:\frac{60}{6}=10 \ \textgreater \ \frac{10\left(-a+2\right)}{\left(-3a+10\right)}$

$\mathrm{Remove\:parentheses}: \left(-a\right)=-a \ \textgreater \ \frac{10\left(-a+2\right)}{-3a+10}$

$Therefore\;our\;solutions\;are\; y=\frac{8}{10-3a},\:x=\frac{10\left(-a+2\right)}{-3a+10}$

Hope this helps!

7 0

3 years ago

Read 2 more answers

Solve the inequality x + 2 1/2 > -1/2. Graph the solution set on a number line.show your work.

antiseptic1488 [7]

Answer:

x > - 3

Step-by-step explanation:

x + 2 $\frac{1}{2}$ > - $\frac{1}{2}$ First turn 2 $\frac{1}{2}$ into an fraction since its easier. So it becomes $\frac{5}{2}$

x + $\frac{5}{2}$ > - $\frac{1}{2}$

- $\frac{5}{2}$ - $\frac{5}{2}$ Do inverse operations, so subtract $\frac{5}{2}$ from both sides.

x > - $\frac{6}{2}$ Combine - $\frac{5}{2}$ with - $\frac{1}{2}$ , which gives us - $\frac{6}{2}$

x > - 3 Since we know that 6 divided by 2 is 3.

This would not allow me to attach an image so I am going to type a visual representation f what the graph would look like below)

<-------------O--------------------------------------------------->

-5 -4 -3 -2 -1 0 1 2 3 4 5

So, the number line would be a solid line rather than dashed (but I was not able to attach an image so I kind of had to improvise). When using ≥ ≤ sings the circle of the point is closed. When using < and > signs, the circle of the point is open. Thus, in this scenario since he sign is >, then the circle of the point is open. The direction of the arrow and ray for this equality is going to the right because the answer is that x is greater than - 3, Therefore any values greater than - 3 are a possibility which is why the arrow goes on indefinitely.

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3 years ago

Susan put the following numbers together.27 thousands,41 hundreds,15 tens and 8 ones. write this number in standard form.

MArishka [77]

27258 tell me if i wrong plz

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3 years ago