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sladkih [1.3K]
3 years ago
8

Write two hundred thousand and fifty seven in figures

Mathematics
1 answer:
mihalych1998 [28]3 years ago
8 0
200,057
This is the how write
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Is this question correct?
Andrews [41]

Answer:

10^20

3^11

3^24

10^12

Step-by-step explanation:

(10^2)^10=10^2*10=10^20

3^3*3^8=3^3+8=3^11

(3^3)^8=3^3*8=3^24

10^2*10^10=10^2+10=10^12

6 0
3 years ago
12. What is the length of SQ? F.5 H.13 G.9 J.17​
netineya [11]

Answer:

Step-by-step explanation:

4 0
3 years ago
Which situation can be represented by the equation 4d + 5 = 29?
lys-0071 [83]

Answer:

d=6

Step-by-step explanation:

subtract 5 from both sides

then simplify what you get

then divide both sides by 4

then simplify to get answer

4 0
3 years ago
Patrick is designing a large stained glass window for the library. If the area of the window is 36 feet, how long is each side
vova2212 [387]
Divide 36 by 4 sides of the window. = 9 square feet.
4 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
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