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ehidna [41]
4 years ago
6

A candy company manufactures bags of colorful candies. The company reports that it makes 10% each of green and red candies, and

20% each of yellow, blue, and orange candies. The rest of the candies are brown If you pick a candy at random, what is the probability that it is either red or yellow?
Mathematics
1 answer:
Darina [25.2K]4 years ago
4 0

Answer:

30% probability that it is either red or yellow

Step-by-step explanation:

We have that:

10% of the candies are green

10% of the candies are red

20% of the candies are yellow

20% of the candies are blue

20% of the candies are orange.

The rest is brown. The percentage of brown candies is not important to this problem.

If you pick a candy at random, what is the probability that it is either red or yellow?

10% are red

20% are yellow

10 + 20 = 30%

30% probability that it is either red or yellow

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In a standard television set, the screen height is 0.75 times the screen width. If a television set measures 34 inches along the
S_A_V [24]
We can solve this by using the  P<span>ythagorean theorem which is below:

a^2 + b^2 = c^2

Or we can say
</span>w^2 + h^2 = d^2
<span>w = widht
h = height
d = diagonal measure


With that said, we know the height is .75 times the width so .75w. We also know d = 34, which is our diagonal measure.
w = don't know yet but need to find
h = .75w
d = 34
Now lets plugin the information we know into our equation

</span>w^2 + h^2 = d^2
w^2 + (.75w)^2 = 34^2
Now lets to the math
w^2 + (.75w)^2 = 34^2
w^2 + (.75w)^2 = 1156
w^2 + .5625w^2 = 1156
Combine like terms
w^2 + .5625w^2 = 1156
1.5625w^2 = 1156
Divide both sides of the equal sign by 1.5625
\frac{1.5625w^2}{1.5625} = \frac{1156}{1.5625}
w^2 = 739.84
Now take the square root on both sides of the equal sign
\sqrt{w^2} = \sqrt{739.84}
w = 27.2

So the width is 27.2
We can check this by putting 27.2 back into our original equation
w^2 + (.75w)^2 = 34^2
27.2^2 + (.75\times 27.2)^2 = 34^2



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I hope this helps, if it doesn't then just message me and ill be more than happy to help :)

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