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NeX [460]
4 years ago
12

How long is a diagonal in the figure

Mathematics
1 answer:
Harman [31]4 years ago
8 0
The diagonal of the rectangle is equivalent to finding the length of the hypotenuse of a right triangle with sides 3 and 4. Using the Pythagorean Theorem:
32+42=hypotenuse2
25=hypotenuse2
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3 years ago
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A box contains 12 chocolates, 3 of which are white chocolate, 4 milk chocolate, and 5 dark chocolate. I am sharing the box with
Kay [80]

Answer:

Step-by-step explanation:

Given that a box contains 12 chocolates, 3 of which are white chocolate, 4 milk chocolate, and 5 dark chocolate.

a)  the chance that I draw a dark chocolate

b) Given that none of my friends draws a dark chocolate,  the chance that I draw a dark chocolate is

\frac{5}{9} since now all 5 dark chocolates would be there and total changed to 9 because 3 friends took 1 each.

c) All four drawn milk chocolates would be

\frac{4C4}{12C4} \\=\frac{1}{495}

d) the chance that the friend who gets to select first draws a dark chocolate and I draw a dark chocolate too

=P(first friend draws a dark chocolate) * P(I draw a dark chocolate)

When it comes to me chances are either there are 4 dark, or 3 dark, or 1 dark depending upon the other two friends drew black or not

= \frac{5}{12} *\frac{7}{11} *\frac{6}{11} +2(\frac{5}{12} *\frac{4}{11} *\frac{7}{11} ) +\frac{5}{12} *\frac{4}{11} *\frac{4}{11} \\= \frac{140+280+80}{1452} \\=\frac{500}{1452} \\=\frac{125}{363}

4 0
3 years ago
Compare the value of the 3 in 6300 and 530
mixer [17]
The value of 3 in 6300<span> is </span>10<span> times the value of 3 in </span>530.
<span>It's because 3 in 530 is tens and 3 in 6300 is hundreds.</span>
8 0
4 years ago
Using the digits 1 to 9, at most one time each, fill in the blanks to make two different pairs of two-digit numbers that have a
agasfer [191]
Do you mean no repeating numbers within the two sets? Because if so, I don't think it's possible.

I started by trying to figure out what the numbers in the tenths place should be. I used subtraction: 71 - ___ = ____. If you try it out, you can't subtract anything with a 6, 7, 8, or 9 in the tenths place because it will leave you with 11 (a repeating digit number), 10 (has a 0), or less (1-digit numbers). Also, a 3 cannot go into the tenths place because when you do 71 - 3_ , your answer will always begin with a 3 (problem because the 3 repeats), or it will contain a 0. 

Therefore, the numbers left for the tenths place are: 1, 2, 4, and 5. 1 and 5 pair up, leaving 4 and 2.
71 - 5_ = 1_   and 71 - 4_ = 2_.

Then, I tried to figure out what numbers go in the ones place. That lead me to realize they act in pairs. The pairs possible in the ones place are 2 and 9, 3 and 8, 4 and 7, 5 and 6. These numbers always go together to result with the final "1" in the "71". Using this information, I looked at the numbers I already used: 1, 2, 4, and 5. Now, looking at the pairs, I eliminated the pairs containing a number already used. This leaves me with only one pair: 3 and 8. Obviously, you need two more pairs to solve the problem, which leads me to my point of saying: This problem is impossible to solve.

I really hope someone can prove me wrong! But this is the solution I have reached for now. :)
4 0
3 years ago
I don’t understand 23-26can someone help?
svetoff [14.1K]

Answer:

If 23-26, that is -3

Well, because 23 can't minus 26, that's why you have to rearrange so that you can find the correct answer by letting -26 + 23.

3 0
3 years ago
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