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3 years ago

5. Janet has 5 marbles. She finds double that number of marbles in her art box. How many marbles does Janet have now?

Mathematics

2 answers:

JulijaS [17]3 years ago

8 0

She has 10 marbles, because 5*2 =10 marbles.

Alexeev081 [22]3 years ago

4 0

Janet has 10. How you solve this problem is 5 x 2 because it is doubled.

Your welcome :)

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3 years ago

Is 976 -522 more or less than 400? Explain how you can tell without actually subtracting.

liq [111]

You can round 976 to 1000 and 522 to 500. So it would be
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3 years ago

Read 2 more answers

App 19.study

lions [1.4K]

Answer:

$1 \to 22 \to 0.176$

$2 \to 13 \to 0.104$

$3 \to 18 \to 0.144$

$4 \to 29 \to 0.232$

$5 \to 37 \to 0.296$

$6 \to 6 \to 0.048$

Step-by-step explanation:

Given

$n = 125$

See attachment for proper table

Required

Complete the table

Experimental probability is calculated as:

$Pr = \frac{Frequency}{n}$

We use the above formula when the frequency is known.

For result of roll 2, 4 and 6

The frequencies are 13, 29 and 6, respectively

So, we have:

$Pr(2) = \frac{13}{125} = 0.104$

$Pr(4) = \frac{29}{125} = 0.232$

$Pr(6) = \frac{6}{125} = 0.048$

When the frequency is to be calculated, we use:

$Pr = \frac{Frequency}{n}$

$Frequency = n * Pr$

For result of roll 3 and 5

The probabilities are 0.144 and 0.296, respectively

So, we have:

$Frequency(3) = 125 * 0.144 = 18$

$Frequency(5) = 125 * 0.296 = 37$

For roll of 1 where the frequency and the probability are not known, we use:

$Total \ Frequency = 125$

So:

Frequency(1) added to others must equal 125

This gives:

$Frequency(1) + 13 + 18 + 29 + 37 + 6 = 125$

$Frequency(1) + 103 = 125$

Collect like terms

$Frequency(1) =- 103 + 125$

$Frequency(1) =22$

The probability is then calculated as:

$Pr(1) = \frac{22}{125}$

$Pr(1) = 0.176$

So, the complete table is:

$1 \to 22 \to 0.176$

$2 \to 13 \to 0.104$

$3 \to 18 \to 0.144$

$4 \to 29 \to 0.232$

$5 \to 37 \to 0.296$

$6 \to 6 \to 0.048$

5 0

3 years ago

A debate team of 4 members for a high school will be chosen randomly from a potential group of 15 students. Ten of the 15 studen

iragen [17]

Answer:

A) 0.1538

Step-by-step explanation:

4 members are going to be chosen:

M1 - M2 - M3 - M4

We have to find the probability that for each member, one without experience is chosen. There is no replacents.

Initially, we have 15 students, of which 10 have none experience. So the probability that M1 is chosen as a student with no experience is $\frac{10}{15}$ .

Then, there are 14 students, of which 9 have no experience. So the probability that M2 is chosen as a student with no experience is $\frac{9}{14}$ .

For the third member, there are 13 students, of which 8 have no experience. So the probability that M3 is chosen as a student with no experience is $\frac{8}{13}$ .

For the last member, there are 12 students, of which 7 have no experience. So the probability that M4 is chosen as a student with no experience is $\frac{7}{12}$ .

What is the probability that none of the members chosen for the team have any competition experience?

This is

$P = \frac{10}{15}*\frac{9}{14}*\frac{8}{13}*\frac{7}{12} = 0.1538$

The correct answer is:

A) 0.1538

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3 years ago

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MrRissso [65]

Answer:

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Step-by-step explanation:

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3 years ago