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3 years ago

PLSSS HELP WILL GIVE BRAINLIEST FOR FIRST CORRECT ANSWER!!!!!!!!!

Mathematics

1 answer:

Kryger [21]3 years ago

6 0

Answer:

A fair die has 6 sides.

There are 3 odd numbers on a fair die so the probability would be 3/6

There are 4 numbers greater than 2 on a fair die o the probability of that would be 4/6

There is only one number that will make it square and that is 4. So the probability would be 1/6

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$$\frac{\left(\frac{(4 r)^{3}}{15 t^{4}}\right)}{\left(\frac{16 r}{(3 t)^{2}}\right)}$

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Step 2: Simplify the denominator

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Step 3: Using step 1 and step 2

$$\frac{\left(\frac{(4 r)^{3}}{15 t^{4}}\right)}{\left(\frac{16 r}{(3 t)^{2}}\right)}=\frac{\left(\frac{64 r^{3}}{15 t^{4}}\right)}{\left(\frac{16 r}{9 t^{2}} \right)}$

Step 4: Using fraction rule:

$$\frac{\frac{a}{b}}{\frac{c}{d}}=\frac{a \cdot d}{b \cdot c}$

$$\frac{\left(\frac{64 r^{3}}{15 t^{4}}\right)}{\left(\frac{16 r}{9 t^{2}}\right)}=\frac{64r^3 \cdot 9t^2}{16 r \cdot 15 t^4}$

Cancel the common factor r and t², we get

$$=\frac{64 r^{2} \cdot 9 }{16 \cdot 15 t^2 }$

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$$=\frac{4 r^{2} \cdot 3 }{ 5 t^2 }$

$$=\frac{12 r^{2} }{ 5 t^2 }$

$$\frac{\left(\frac{(4 r)^{3}}{15 t^{4}}\right)}{\left(\frac{16 r}{(3 t)^{2}}\right)}=\frac{12 r^{2} }{ 5 t^2 }$

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