On a baseball field, the pitcher's mound is 60.5 feet from home plate. During practice, a batter hits a ball 214 feet at an
1 answer:
9514 1404 393
Answer:
169 ft
Step-by-step explanation:
The law of cosines can be used to find the distance from the outfielder to the pitcher. It tells you for triangle ABC, the length of side c can be found from ...
c² = a² +b² -2ab·cos(C)
Here, we hve a=60.5, b=214, and C=36°. Then the desired distance is ...
c = √(60.5² +214² -2·60.5·214·cos(36°)) ≈ √28507.56 ≈ 168.84
The outfielder throws the ball about 169 feet.
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Answer:
a) There is 17.64% probability that exactly two of the selected bulbs are rated 23-watt.
b) There is a 8.65% probability that all three of the bulbs have the same rating.
c) There is a 12.45% probability that one bulb of each type is selected.
Step-by-step explanation:
There are 24 compact fluorescent lightbulbs in the box, of which:
8 are rated 13-watt
9 are rated 18-watt
7 are rated 23-watt
(a) What is the probability that exactly two of the selected bulbs are rated 23-watt?
There are 7 rated 23-watt among 23. There are no replacements(so the denominators in the multiplication decrease). Then can be chosen in different orders, so we have to permutate.
It is a permutation of 3(bulbs selected) with 2(23-watt) and 1(13 or 18 watt) repetitions. So
There is 17.64% probability that exactly two of the selected bulbs are rated 23-watt.
(b) What is the probability that all three of the bulbs have the same rating?
is the probability that all three of them are 13-watt. So:
is the probability that all three of them are 18-watt. So:
is the probability that all three of them are 23-watt. So:
There is a 8.65% probability that all three of the bulbs have the same rating.
(c) What is the probability that one bulb of each type is selected?
We have to permutate, permutation of 3(bulbs), with (1,1,1) repetitions(one for each type). So
There is a 12.45% probability that one bulb of each type is selected.