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3 years ago

13

Ann has 12 game cards. She wants to deal each of 5 players the same number of cards. How many cards will each player get? How ma

ny cards will be remaining?

2 answers:

Nataly_w [17]3 years ago

7 0

The five will each get two and there will be two left for Ann. Not a good card strategy.

katrin2010 [14]3 years ago

5 0

Each player will have 2 cards and she will have 2 left over. This is obviously not Middle School Mathamatics ;)

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Put these numbers in order from greatest to least.

Lesechka [4]

Answer: $3\frac{21}{30} ,3\frac{14}{28},\frac{15}{25},0.5,-0.7$

Step-by-step explanation:

$-0.7=-0.7\\\\3\frac{21}{30}=3.7\\\\3\frac{14}{28}=3.5\\\\0.5=0.5\\ \\\frac{15}{25}=0.6$

6 0

3 years ago

a farmer has a rectangular field that measures 125 feet by 200 feet. He wants to enclose the field with a fence. What is the tot

grandymaker [24]

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4 0

3 years ago

Tanya [424]

Answer:

b140

Step-by-step explanation:

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4 0

3 years ago

Read 2 more answers

The diagram shows a right-angled triangle.

Alexus [3.1K]

Step-by-step explanation:

there is no diagram so I'm just going to guess this however

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8 0

3 years ago

student randomly receive 1 of 4 versions(A, B, C, D) of a math test. What is the probability that at least 3 of the 5 student te

alexdok [17]

Answer:

1.2%

Step-by-step explanation:

We are given that the students receive different versions of the math namely A, B, C and D.

So, the probability that a student receives version A = $\frac{1}{4}$ .

Thus, the probability that the student does not receive version A = $1-\frac{1}{4}$ = $\frac{3}{4}$ .

So, the possibilities that at-least 3 out of 5 students receive version A are,

1) 3 receives version A and 2 does not receive version A

2) 4 receives version A and 1 does not receive version A

3) All 5 students receive version A

Then the probability that at-least 3 out of 5 students receive version A is given by,

$\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{3}{4}\times \frac{3}{4}$ + $\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{3}{4}$ + $\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}$

= $(\frac{1}{4})^3\times (\frac{3}{4})^2+(\frac{1}{4})^4\times (\frac{3}{4})+(\frac{1}{4})^5$

= $(\frac{1}{4})^3\times (\frac{3}{4})[\frac{3}{4}+\frac{1}{4}+(\frac{1}{4})^2]$

= $(\frac{3}{4^4})[1+\frac{1}{16}]$

= $(\frac{3}{256})[\frac{17}{16}]$

= 0.01171875 × 1.0625

= 0.01245

Thus, the probability that at least 3 out of 5 students receive version A is 0.0124

So, in percent the probability is 0.0124 × 100 = 1.24%

To the nearest tenth, the required probability is 1.2%.

4 0

3 years ago