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3 years ago

Solve the following equation-8=(10x-22)/11

1 answer:

6 0

-8 = (10x - 22)/11

Multiply both sides by 11:
-88 = 10x - 22

Add 22 to both sides:
-66 = 10x

Divide both sides by 10
-6.6 = x

Hope it helped :)

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A loaded 8-sided die is loaded so that the number 4 occurs 3/10 of the time while the other numbers occur with equal frequency.

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Answer:

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Step-by-step explanation:

The sum of the probabilities of all possible outcomes is 1.

As the die is loaded so that the number 4 occurs 3/10 of the time, and the other numbers occur with equal frequency, then the probability of numbers 1, 2, 3, 5, 6, 7 and 8 being rolled is 1/10.

Create a probability distribution table for X, where X is the score on the loaded 8-sided die:

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Add a product row and a totals column:

$\begin{array}{ | c | c | c | c | c | c | c | c | c | c |}\cline{1-10}x & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 &\text{Total} \\ \cline{1-10} &&&&&&&&& \\ \text{P}(X=x) & \frac{1}{10} & \frac{1}{10} & \frac{1}{10} & \frac{3}{10} & \frac{1}{10} & \frac{1}{10} & \frac{1}{10} & \frac{1}{10} & 1\\ &&&&&&&&&\\ \cline{1-10} &&&&&&&&&\\\text{Product} & \frac{1}{10} & \frac{2}{10} & \frac{3}{10} & \frac{12}{10} & \frac{5}{10} & \frac{6}{10} & \frac{7}{10} & \frac{8}{10} & \frac{44}{10} \\ &&&&&&&&&\\\cline{1-10}\end{array}$

(The Product is row is the product of the <u>score on the die</u> and <u>its probability</u>).

The expected value (EV) is the sum of the product of each outcome and its probability.

Therefore, the expected value (EV) of this die is 44/10 = 4.4

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Given : A patient is given a solution containing 0.625 g of calcium carbonate.

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