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1 year ago

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Jacob has a bag of his favorite marbles. It has 3 red marbles, 4 blue, and 10 of his most favorite color, neon orange. What is t

he probability that, if he removes 2 marbles without looking and without replacement, he will get two orange marbles

1 answer:

drek231 [11]1 year ago

7 0

Answer:

$P(1\ and\ 2) = 0.3309$

Step-by-step explanation:

Given

$Red = 3$

$Blue = 4$

$Orange=10$

Required

Probability of selecting 2 orange marbles

The total number of marbles is:

$Total = Red + Blue + Orange$

$Total = 3 + 4 + 10$

$Total = 17$

The probability that the first selection is orange is:

$P(1) = \frac{10}{17}$

Because it is a selection without replacement, the number of orange marbles and the total number of marbles would decrease by 1, respectively.

So, the probability that the selection is orange is:

$P(2) = \frac{10-1}{17-1}$

$P(2) = \frac{9}{16}$

The required probability is:

$P(1\ and\ 2) = P(1) * P(2)$

$P(1\ and\ 2) = \frac{10}{17} * \frac{9}{16}$

$P(1\ and\ 2) = \frac{10*9}{17*16}$

$P(1\ and\ 2) = \frac{90}{272}$

$P(1\ and\ 2) = 0.3309$

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Answer:

Sadie's error is " she made error in step 2 $=\sqrt{3^2\times 6\times a^2\times a^5\times b^2\times b}$ where $a\geqslant 0$ "

Because she made error in splitting the powers to simplify the square root

<h3>Therefore the correct answer for Sadie's expression is $3ab\sqrt{6ab}$ where $a\geqslant 0$ </h3>

Step-by-step explanation:

Given that " Sadie simplified the expression StartRoot 54 a Superscript 7 b cubed EndRoot, where a greater-than-or-equal-to 0, "

It can be written as $\sqrt{54a^7b^3}$ where $a\geqslant 0$

The given expression is $\sqrt{54a^7b^3}$ where $a\geqslant 0$

To find Sadie's error and explain the correct answer :

Sadie's steps are

$\sqrt{54a^7b^3}$ where $a\geqslant 0$

$=\sqrt{3^2\times 6\times a^2\times a^5\times b^2\times b}$

$=3ab\sqrt{6a^5b}$

<h3> $\sqrt{54a^7b^3}=3ab\sqrt{6a^5b}$ where $a\geqslant 0$ </h3><h3><u>Now corrected steps are</u></h3>

$\sqrt{54a^7b^3}$ where $a≥0$

$=\sqrt{(9\times 6)(a^{6+1})(b^{2+1})$

$=\sqrt{(3^2\times 6)(a^6.a^1)(b^2.b^1)$ (by using the identity $a^{m+n}=a^m.a^n$

$=\sqrt{3^2\times 6\times ((a^3)^2.a)(b^2.b)$ (by using the identity $a^{mn}=(a^m)^n$ )

$=3ab\sqrt{6ab}$

Therefore $\sqrt{54a^7b^3}=3ab\sqrt{6ab}$ where $a\geqslant 0$

<h3>The correct answer is $3ab\sqrt{6ab}$ where $a\geqslant 0$ </h3>

Sadie's error is " she made error in step 2 $=\sqrt{3^2\times 6\times a^2\times a^5\times b^2\times b}$ " where $a\geqslant 0$

Because she made error in splitting the powers to simplify the square root

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Answer:

The farmer needs 24 crates to hold the carrots.

Step-by-step explanation:

To determine how many crates the farmer needs to hold the carrots, knowing that his farm has 4.8 acres of land, where he can harvest 120 pounds per acre, and that each crates can hold up to 24 pounds of carrots, the following calculation must be performed:

(4.8 x 120) / 24 = X

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Thus, the farmer needs 24 crates to hold the carrots.

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Answer:

The expected payoff for this game is -$1.22.

Step-by-step explanation:

It is given that a pair of honest dice is rolled.

Possible outcomes for a dice = 1,2,3,4,5,6

Two dices are rolled then the total number of outcomes = 6 × 6 = 36.

$\{(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(2,1),(2,2),(2,3),(2,4),(2,5),(2,6),\\(3,1),(3,2),(3,3),(3,4),(3,5),(3,6),(4,1),(4,2),(4,3),(4,4),(4,5),(4,6),\\(5,1),(5,2),(5,3),(5,4),(5,5),(5,6),(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)\}$

The possible ways of getting a total of 7,

{ (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) }

Number of favorable outcomes = 7

Formula for probability:

$Probability=\frac{\text{Favorable outcomes}}{\text{Total outcomes}}$

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So, the probability of getting the sum of numbers other than 7 or 11 = $\frac{28}{36}$ = $\frac{7}{9}$

Since, for the sum of 7, $ 22 will earn, for the sum of 11, $ 66 will earn while for any other total loss is $11,

Hence, the expected value for this game is

$\frac{1}{6}\times 22+\frac{1}{18}\times 66-\frac{7}{9}\times 11$

$\frac{11}{3}+\frac{11}{3}-\frac{77}{9}$

$\frac{22}{3}-\frac{77}{9}$

$\frac{66-77}{9}$

$-\frac{11}{9}$

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Therefore the expected payoff for this game is -$1.22.

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Answer:

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Step-by-step explanation:

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