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andrew11 [14]
3 years ago
10

A baseball player hit 60 home runs in a season. Of the 60 home runs, 19 went to right field, 18 went to right center field, 12 w

ent to center field, 10 went to left center field, and 1 went left field.
(a) What is the probability that a randomly selected home run was hit to leftleft ​field?
(b) Was it unusual for this player to hit a home run to leftleft ​field?
Physics
1 answer:
vlada-n [284]3 years ago
5 0

Answer:

(a) 1/60

(b) Yes, it was unusual.

Explanation:

Number that went to the right field = 19

Number that went to the right center field = 18

Number that went to the center field = 12

Number that went to the left center field = 10

Number that went to the left field = 1

Total number of home runs = 60

(a) The probability that a randomly selected home run was hit to left ​field is

P(left field) = 1/60

(b) It was unusual because whenever a player hit a home run, it most times went to the right field,which is why the right field had the highest probability, 19/60.

On the other hand, it went to the left field the least number of times, which is why the left field had the least probability, 1/60.

So, it was unusual.

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Answer:

See below

Explanation:

Net force acting on the plane after overcoming frictional forces is

2.7 x 10^6   -  6.8 x 10^4   -  8.5 x 10^5 = 1 782 000 N

Work = F x d

        = 1 782 000 x 1400 m = 2.5 x 10^9   J       or  2.5 x 10^6  kJ

4 0
2 years ago
If an object in space is giving off a frequency of 10^13 wavelength of 10^-6 what will scientist be looking for?
aev [14]

Answer:

The scientist will be looking for the velocity of the wave in air which is equivalent to 10^7m/s

Explanation:

If an object in space is giving off a frequency of 10^13Hz and wavelength of 10^-6m then the scientist will be looking for the velocity of the object in air.

The relationship between the frequency (f) of a wave, the wavelength (¶) and the velocity of the wave in air(v) is expressed as;

v = f¶

Given f = 10^13Hz and ¶ = 10^-6m,

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4 years ago
What is the acceleration of an object with a mass of
aev [14]

Heya!!

For calculate aceleration, lets applicate second law of Newton:

                                                    \boxed{F=ma}

                                                 <u>Δ   Being   Δ</u>

                                             F = Force = 183 N

                                            m = Mass = 367 kg

                                             a = Aceleration = ?

⇒ Let's replace according the formula and clear "a":

\boxed{a=183 \ N / 367\ kg}

⇒ Resolving

\boxed{ a = 0.49 \ m/s^{2}}

Result:

The aceleration is <u>0,49 meters per second squared (m/s²)</u>

Good Luck!!

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