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Alinara [238K]
2 years ago
13

What is the answer? please tell me i am desperate!?

Mathematics
1 answer:
solniwko [45]2 years ago
6 0
Okay so this is easy conversion.
40 quarters=10$
60 dimes =24 quarters
30 dimes= 12 quarters=3$
200 dimes=80 quarters
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Mary Sue and Betty have 603 necklaces altogether. Mary Sue has 115 more necklaces than Betty. How many necklaces does Betty have
Blizzard [7]

Let Mary Sue have m necklaces and Betty have b necklaces.

And they have 603 necklaces altogether, that is

m + b =603

And Mary Sue has 115 more necklaces than Betty, that is

m=b +115

Substituting this value in first equation we will get

b+115+b=603

Subtracting 115

2b= 488

b= 244

So Betty have 244 necklaces .

6 0
3 years ago
The perimeter of the entire garden.
Leno4ka [110]
Add up all the lengths of the garden and you will be able to get the perimeter.
8 0
1 year ago
5q + 5/14 + 8z + 2/14 + 4q - 4z
SVEN [57.7K]

Answer:

9q+4z+12

Step-by-step explanation:

7 0
2 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
Write the equation of a line that passes through the points ( 2 , - 5 ) and ( 8 , - 2 ).
Zielflug [23.3K]

Answer:

y = 1/2x - 6

Step-by-step explanation:

y2 - y1 / x2 - x1

-2 - (-5) / 8 - 2

3 / 6

= 1/2

y = 1/2x + b

-5 = 1/2(2) + b

-5 = 1 + b

-6 = b

3 0
3 years ago
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