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2 years ago

If there are opportunities for 120 girls to participate in athletics, how many 5 points

Mathematics

1 answer:

svetoff [14.1K]2 years ago

8 0

Answer:

120..?

Step-by-step explanation:

I'd presume that if there's 120 for girls there must also be 120 for boys too? Is there any previous like examples/numbers regarding this question? Odd question..

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Julia can finish a 20-mile bike ride in 1.2 hours. Katie can finish the same bike ride in 1.6 hours. How much faster does Julia

Nimfa-mama [501]

We know that, $speed = \frac{distance}{time}$

Julia can finish a 20-mile bike ride in 1.2 hours.

The distance Julia travels is 20 miles and the time she takes is 1.2 hours.

So, Julia's speed = $\frac{20}{1.2}$ = 16.67 mph

Katie can finish the same bike ride in 1.6 hours.

The distance Katie travels is 20 miles and the time she takes is 1.6 hours.

So, Katie's speed = $\frac{20}{1.6}$ = 12.5 mph

Now, to find how much faster Julia rides than Katie we subtract Katie's speed from Julia's speed.

So, 16.67 mph - 12.5 mph = 4.17 mph = 4.2 mph (approximately)

Thus, Julia rides 4.2 mph faster than Katie.

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2 years ago

In a bag of m&m's there are 5 brown 6 yellow 4 blue 3 green and 2 orange. What's the probability of getting 3 yellow m&m

olasank [31]

There are 5+6+4+3+2=20 m&m's in the bag.
Calculate in how many ways you can choose 3 m&m's from 20:
$_{20} C _3=\frac{20!}{3!(20-3)!}=\frac{20!}{3! \times 17!}=\frac{17! \times 18 \times 19 \times 20}{6 \times 17!}=\frac{18 \times 19 \times 20}{6}=3 \times 19 \times 20= \\
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There are 6 yellow m&m's.
Calculate in how many ways you can choose 3 m&m's from 6:
$_6 C _3 = \frac{6!}{3!(6-3)!}=\frac{6!}{3! \times 3!}=\frac{3! \times 4 \times 5 \times 6}{3! \times 6}=\frac{4 \times 5 \times 6}{6}=4 \times 5=20$

The probability is the number of ways of choosing 3 m&m's from 6 m&m's divided by the number of ways of choosing 3 m&m's from 20 m&m's.
$P=
\frac{20}{1140}=\frac{20 \div 20}{1140 \div 20}=\frac{1}{57}$

The probability is 1/57.

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2 years ago

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