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3 years ago

Guesstimate how much total time do Americans spend driving in one year? Express the answer in hours, years and lifetimes.

Mathematics

1 answer:

Drupady [299]3 years ago

7 0

Answer:

1,460 hours per year.

113,880 hours in a lifetime.

Step-by-step explanation:

Great question, it is always good to ask away and get rid of any doubts that you may be having.

Let's assume that on an average day Americans drive 4 hours a day. This includes going and coming from work as well as any groceries they might need to do. Since there are 365 days in a year we can multiply this by the hours a day driven.

$365*4 = 1,460$

We can "guesstimate" that on average an American drives 1,460 hours per year.

Based on my research the average lifetime of an American is 78 years old. We can now multiply this by the number of hours driven in a year to have an idea of the number of hours driven in a lifetime.

$1,460*78 = 113,880$

On average we "guesstimate" that an American drives 113,880 hours in a lifetime.

I hope this answered your question. If you have any more questions feel free to ask away at Brainly.

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Suppose that we have a given point (x, y)

If we translate this point N units to the right, then the new coordinates of the point are:

(x + N, y)

Ok, now if we know that the vertices of the trapezoid are:

(1, 2), (3, 1), (3, 5), (1, 4)

And we move the whole figure 4 units to the right, then all the vertices are moved 4 units to the right.

Then the new vertices of the figure will be:

(1 + 4, 2) = (5, 2)

(3 + 4, 1) = (7, 1)

(3 + 4, 5) = (7, 5)

(1 + 4, 4) = (5, 4)

Then the coordinates of the image of the trapezoid (of the new vertices) are:

(5, 2), (7, 1), (7, 5), (5, 4)

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A fair die is rolled 12 times. the number of times an even number occurs on the 12 rolls has

bonufazy [111]

Answer:

Step-by-step explanation:

For a fair die, there are six likely options; 1, 2, 3, 4, 5, and 6

the probability of a even number is 3/6 = 0.5

Since the results of the die roll is independent and each trial is mutually exclusive, the distribution to explain the probability of occurrence will follow a binomial distribution such that n is the number of trials

x = number of successful throws

therefore for a Binomial distribution where

P(X =x) = nCx . P^x . (1-P)^ (n-x)

since p = 0.5, and n = 12, the distribution follows

P(X = x) = 12Cx . 0.5^x . (1 - 0.5)^(12- x)

= 12Cx . 0.5^x . 0.5)^(12- x)

where x = (0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12)

since we are interested in the probability of the number of times an even number occurs

it can occur either as P(X = 0), P(X =1), P(X =2), P(X =3), P(X =4), P(X =5), P(X =6), P(X =7), P(X =8), P(X =9), P(X =10), P(X =11), and P(X =12)

For no even number in 12 rolls,

P(X = 0) = 12C0 . 0.5^0 . 0.5^(12- 0) = 0.000244

For one even number in 12 rolls,

P(X = 1) = 12C1 . 0.5^1 . 0.5^(12- 1) = 0.002930

For two even number in 12 rolls,

P(X = 2) = 12C2 . 0.5^2 . 0.5^(12- 2) = 0.016113

For three even number in 12 rolls,

P(X = 3) = 12C3 . 0.5^3 . 0.5^(12- 3) = 0.053711

For four even number in 12 rolls,

P(X = 4) = 12C4 . 0.5^4 . 0.5^(12- 4) = 0.120850

For five even number in 12 rolls,

P(X = 5) = 12C5 . 0.5^5 . 0.5^(12- 5) = 0.193359

For six even number in 12 rolls,

P(X = 6) = 12C6 . 0.5^6 . 0.5^(12- 6) = 0.225586

For seven even number in 12 rolls,

P(X = 7) = 12C7 . 0.5^7 . 0.5^(12- 7) = 0.193359

For eight even number in 12 rolls,

P(X = 8) = 12C8 . 0.5^8 . 0.5^(12- 8) = 0.120850

For nine even number in 12 rolls,

P(X = 9) = 12C9 . 0.5^9 . 0.5^(12- 9) = 0.053711

For ten even number in 12 rolls,

P(X = 10) = 12C10 . 0.5^10 . 0.5^(12- 10) = 0.016113

For eleven even number in 12 rolls,

P(X = 11) = 12C11 . 0.5^11 . 0.5^(12- 11) = 0.002930

For twelve even number in 12 rolls,

P(X = 12) = 12C12 . 0.5^12 . 0.5^(12- 12) = 0.000244

Final test summation[P(X)] = 1

i.e.

P(X = 0) + P(X =1) + P(X =2) + P(X =3) + P(X =4) + P(X =5) + P(X =6) + P(X =7) + P(X =8) + P(X =9) + P(X =10) + P(X =11) + P(X =12) = 1

Hence since 0.000244 + 0.002930 + 0.016113 + 0.053711 + 0.120850 + 0.193359 + 0.225586 + 0.193359 + 0.120850 + 0.053711 + 0.016113 + 0.002930 + 0.000244 = 1.000000,

the probability value stands

7 0

3 years ago