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Anit [1.1K]
3 years ago
11

HELPPPPPPPPP PLEASE SHOW WORK DOWN BELOW

Mathematics
2 answers:
Yakvenalex [24]3 years ago
4 0

Answer:

12 blueberries, 4 strawberries

Step-by-step explanation:

For every 3 blueberries, there is 1 strawberry.  So 3 + 1 = 4

As there are a total of 16 berries in the bowl, 16/4 = 4 so there are 4 lots of "3 blueberries and 1 strawberry".

Therefore, there are 3 x 4 = 12 blueberries

1 x 4 = 4 strawberries

Zanzabum3 years ago
3 0
4 strawberries and 12 blueberries
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Answer:

Lol i think this is da answer man

4 0
3 years ago
What is the distance between the points (-3,-5) and (3, 3)?<br> A. 14 <br> B. 2<br> C. 10 <br> D. -2
solong [7]

Answer:

-2

Step-by-step explanation:

                             

3 0
3 years ago
Find the area of the following shape. Show work.
Serga [27]

Answer:

The total area:

12 units²

Step-by-step explanation:

There are three parts on this shape:

1 big right triangle

1 rectangle

1 small right triangle

  • Big right triangle:

area = (4*4)/2 = 16/2 = 8units²

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area = 2*1 = 2units²

  • Small right triangle:

area = (2*2)/2 = 2units²

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6 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
Recursive Sequences: If a1=8 and an=−4an−1 then find the value of a6.
Alona [7]
I believe it is a6=a36
8 0
3 years ago
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