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Amiraneli [1.4K]

3 years ago

8

If 1st base is 90degree angle throw from the Pitcher's mound can you tell me how far it is from the Home Plate to the Pitcher's

Mound?

1 answer:

SIZIF [17.4K]3 years ago

6 0

Answer:

100

Step-by-step explanation:

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Help me determine if relations 1, 2, 3, and 4 are functions or not please!

Reil [10]

They are all not functions

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2 years ago

I need help ! What's 2+2-8^2+208

Nata [24]

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148

Step-by-step explanation:

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3 years ago

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What is the probability that 6 rolls of a fair die will show four exactly 5 times?

storchak [24]

Answer:

very less

Step-by-step explanation:

When rolling a fair 6 sided die 5 times there are 6^5 possible combinations.

Of those, there are 5 ways to roll exactly four sixes, and each of those ways have 5 possible results on the die which has not rolled a six, plus there is 1 way to roll five sixes. 5*5+1=26 combinations with at least four sixes.

Now, take the number of results which fit your conditions, divide by the number of possible results and you get your answer: 26/7776 which can be reduced to 13/3338, a bit less than 0.4%

6 0

3 years ago

Evaluate the expression for the given value of the variable.

Fudgin [204]

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Step-by-step explanation:

4 0

2 years ago

Neptune is an average distance of 4.5 × 10^9 km from the Sun. Estimate the length of the Neptunian year using the fact that the

zlopas [31]

Answer:

Answer:

164.32 earth year

Step-by-step explanation:

distance of Neptune, Rn = 4.5 x 10^9 km

distance of earth, Re = 1.5 x 10^8 km

time period of earth, Te = 1 year

let the time period of Neptune is Tn.

According to the Kepler's third law

T² ∝ R³

$\left ( \frac{T_{n}}{T_{e}} \right )^{2}=\left ( \frac{R_{n}}{R_{e}} \right )^{3}$

$\left ( \frac{T_{n}}{1} \right )^{2}=\left ( \frac{4.5\times10^{9}}}{1.5\times10^{8}}} \right )^{3}$

Tn = 164.32 earth years

Thus, the neptune year is equal to 164.32 earth year.

Step-by-step explanation:

3 0

3 years ago