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3 years ago

Last One Please Help

Mathematics

1 answer:

tigry1 [53]3 years ago

7 0

Answer: A. 3/5

Step-by-step explanation:

Count all the add numbers in 1-10 then add the 4 and you get

6/10

Simply and you get 3/5

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2 years ago

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2 years ago

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If h is 36in, w is 28in, and A is 6112in to the second power, what is the length of the box

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The length of the box would be 12288.

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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%.

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2 years ago

Hamburgers come packed 12 in a package. Hamburger buns come packed 10 in a package. If we want one hamburger for each bun for a

BlackZzzverrR [31]

Answer:

5 hamburger packages and 6 bun packages

Step-by-step explanation:

It is asking for the highest common multiple of 12 and 10. the first time a multiple of 12 is divisible by 10 is at 12x5, which is 60. now divide 60 by 12 and 10, 5 and 6.

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1 year ago