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3 years ago

Use the numbers 12,12,12,12,23,23,31,31,49 to add up to 100 exactly(not necessary to use all the numbers)..

Mathematics

2 answers:

Radda [10]3 years ago

8 0

Answer:

are you missing any numbers ? or forgetting any of the rules ?

krek1111 [17]3 years ago

3 0

Answer:

49,12, 39

Step-by-step explanation:

49+12+39=100

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Given the parent function of y = | x |, list the values that would fill in the table of the transformed function y = | x - 4 |.

NeTakaya

I believe it is answer letter D.

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3 years ago

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A student ran out of time on a multiple choice exam and randomly guessed the answers for two problems. Each problem had 5 answer

choli [55]

Answer:

there is a 64% chance that the student got both problems wrong

a 32% chance that they got only 1 correct

and a 4% chance that they got both correct

Step-by-step explanation:

There are 25 total possible combinations of answers, with 8 possible combinations where the student would get 1 answer right, and 1 combination where the student would get both answers correct.

$25-9=16$

$\frac{16}{25} =\frac{x}{100}$

$\frac{64}{100}$

$64$ %

$\frac{8}{25} =\frac{y}{100}$

$\frac{32}{100}$

$32$ %

$\frac{1}{25} =\frac{z}{100}$

$\frac{4}{100}$

$4$ %

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3 years ago

What is the mean of the following data values?<br> 22, 37, 49, 15, 92

attashe74 [19]

Answer:

Step-by-step explanation:

First, you have to add 22+37+49+15+92 to get 215. To find the mean, you need to divide by how much numbers there are (5). So 215 divided by 5 is 43!

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3 years ago

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The prior probabilities for events A1 and A2 are P(A1) = 0.20 and P(A2) = 0.80. It is also known that P(A1 ∩ A2) = 0. Suppose P(

Umnica [9.8K]

Answer:

(a) $A_1$ and $A_2$ are indeed mutually-exclusive.

(b) $\displaystyle P(A_1\; \cap \; B) = \frac{1}{20}$ , whereas $\displaystyle P(A_2\; \cap \; B) = \frac{1}{25}$ .

(d) $\displaystyle P(A_1 \; |\; B) \approx \frac{5}{9}$ , whereas $P(A_1 \; |\; B) = \displaystyle \frac{4}{9}$

Step-by-step explanation:

$P(A_1 \; \cap \; A_2) = 0$ means that it is impossible for events $A_1$ and $A_2$ to happen at the same time. Therefore, event $A_1$ and $A_2$ are mutually-exclusive.

By the definition of conditional probability:

$\displaystyle P(B \; | \; A_1) = \frac{P(B \; \cap \; A_1)}{P(B)} = \frac{P(A_1 \; \cap \; B)}{P(B)}$ .

Rearrange to obtain:

$\displaystyle P(A_1 \; \cap \; B) = P(B \; |\; A_1) \cdot P(A_1) = 0.25 \times 0.20 = \frac{1}{20}$ .

Similarly:

$\displaystyle P(A_2 \; \cap \; B) = P(B \; |\; A_2) \cdot P(A_2) = 0.80 \times 0.05 = \frac{1}{25}$ .

Note that:

$\begin{aligned}P(A_1 \; \cup \; A_2) &= P(A_1) + P(A_2) - P(A_1 \; \cap \; A_2) = 0.20 + 0.80 = 1\end{aligned}$ .

In other words, $A_1$ and $A_2$ are collectively-exhaustive. Since $A_1$ and $A_2$ are collectively-exhaustive and mutually-exclusive at the same time:

$\displaystyle P(B) = P(B \; \cap \; A_1) + P(B \; \cap \; A_2) = \frac{1}{20} + \frac{1}{25} = \frac{9}{100}$ .

By Bayes' Theorem:

$\begin{aligned} P(A_1 \; |\; B) &= \frac{P(B \; | \; A_1) \cdot P(A_1)}{P(B)} \\ &= \frac{0.25 \times 0.20}{9/100} = \frac{0.05 \times 100}{9} = \frac{5}{9}\end{aligned}$ .

Similarly:

$\begin{aligned} P(A_2 \; |\; B) &= \frac{P(B \; | \; A_2) \cdot P(A_2)}{P(B)} \\ &= \frac{0.05 \times 0.80}{9/100} = \frac{0.04 \times 100}{9} = \frac{4}{9}\end{aligned}$ .

6 0

3 years ago

Please help me with this question!!

stepladder [879]

Answer:

rational
irrational
irrational
irrational
√7, it is irrational

Step-by-step explanation:

A <em>rational</em> number is one that can be expressed as the ratio of two integers. All fractions that have integer numerators and (non-zero) denominators are <em>rational</em> numbers. Any finite decimal number, or any repeating decimal number, is a rational number. These can always be expressed as the ratio of two integers. For example, 0.4040... = 40/99, and 0.286 = 286/1000.

To make an irrational sum, at least one of the contributors must be irrational. You want an irrational 2-number sum that has 7/8 as one of the contributors. Since 7/8 is rational, the other contributor must be irrational.

<u>Step 1</u>. The number 7/8 is <em>rational</em>.

<u>Step 2</u>. The desired sum is <em>irrational</em>.

<u>Step 3</u>. The rule <em>rational + </em><em>irrational</em><em> = irrational</em> applies.

<u>Step 4</u>. An <em>irrational</em> number must be chosen.

Step 5. √7 will produce an irrational sum, because <em>it is irrational</em>.

5 0

3 years ago