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3 years ago

Whats a multiplication expression with a product of five to the 13th power

1 answer:

8 0

A multiplication expression with a product of 5 to the 13th power is...
- 5 • 5 • 5 • 5 • 5 • 5 • 5 • 5 • 5 • 5 • 5 • 5 • 5 = 1,220,703, 125 (answer)
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I am 99.9% sure that this is the correct answer, but if it happens not to be, I am truly sorry, and I hope you can forgive me. Thank You!

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The vehicles petrol tank has a capacity of 40 litres. Will the vehicle be able to cover the full distance of 340km with on petro

mixer [17]

The vehicle be able to cover the full distance of 340km with on petrol tank if the petrol usage is 14 km/1.

Since the vehicles petrol tank has a capacity of 40 litres and the bvehicles wants to cover a distance of 340 with a pertol usage of 14 km/l.

We need to determine if the petrol in the tank will be enough to cover that distance.

<h3>Distance</h3>

So, distance = petrol usage × volume of tank

Since

petrol usage = 14 km/l and
volume of tank = 40 l,

Substituting the values of the variables into the equation, we have

distance = petrol usage × volume of tank

distance = 14 km/l × 40 l

distance = 560 km

Since the distance covered = 560 km which is greater than 340 km, the vehicle will be able to cover the full distance.

So, the vehicle be able to cover the full distance of 340km with on petrol tank if the petrol usage is 14 km/1.

Learn more about distance here:

brainly.com/question/10428039

5 0

2 years ago

Compare 5/6 and 1/3 using the benchmark fraction 1/2

fenix001 [56]

Answer:

Step-by-step explanation:

Denominators: 6 is 2 times more than 3

Make them common fractions

5/6 and 2/6

They are 3/6 apart, or 1/2 apart

8 0

4 years ago

Please help asap!!!!!!!!!!!!!!!!

vovangra [49]

Im not sure but i think its A because you have no value to multiply 3 so you just multiply it by itself.It would be a negative because you cant divide 3 by 2.so it would be A.-x9y2.
i hope i was correct and helped if not you can go ahead and delete my answer:)

6 0

3 years ago

2)<br> 3<br> 3 cm<br> 9 cm<br> Find area

sergey [27]

Answer:
27cm^2

Explanation:
3 x 9 = 27

8 0

3 years ago

Every day your friend commutes to school on the subway at 9 AM. If the subway is on time, she will stop for a $3 coffee on the w

Shtirlitz [24]

Answer:

1.02% probability of spending 0 dollars on coffee over the course of a five day week

7.68% probability of spending 3 dollars on coffee over the course of a five day week

23.04% probability of spending 6 dollars on coffee over the course of a five day week

34.56% probability of spending 9 dollars on coffee over the course of a five day week

25.92% probability of spending 12 dollars on coffee over the course of a five day week

7.78% probability of spending 12 dollars on coffee over the course of a five day week

Step-by-step explanation:

For each day, there are only two possible outcomes. Either the subway is on time, or it is not. Each day, the probability of the train being on time is independent from other days. So we use the binomial probability distribution to solve this problem.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

In this problem we have that:

The probability that the subway is delayed is 40%. 100-40 = 60% of the train being on time, so $p = 0.6$

The week has 5 days, so $n = 5$

She spends 3 dollars on coffee each day the train is on time.

Probabability that she spends 0 dollars on coffee:

This is the probability of the train being late all 5 days, so it is P(X = 0).

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{5,0}.(0.6)^{0}.(0.4)^{5} = 0.0102$

1.02% probability of spending 0 dollars on coffee over the course of a five day week

Probabability that she spends 3 dollars on coffee:

This is the probability of the train being late for 4 days and on time for 1, so it is P(X = 1).

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 1) = C_{5,1}.(0.6)^{1}.(0.4)^{4} = 0.0768$

7.68% probability of spending 3 dollars on coffee over the course of a five day week

Probabability that she spends 6 dollars on coffee:

This is the probability of the train being late for 3 days and on time for 2, so it is P(X = 2).

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 2) = C_{5,2}.(0.6)^{2}.(0.4)^{3} = 0.2304$

23.04% probability of spending 6 dollars on coffee over the course of a five day week

Probabability that she spends 9 dollars on coffee:

This is the probability of the train being late for 2 days and on time for 3, so it is P(X = 3).

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 3) = C_{5,3}.(0.6)^{3}.(0.4)^{2} = 0.3456$

34.56% probability of spending 9 dollars on coffee over the course of a five day week

Probabability that she spends 12 dollars on coffee:

This is the probability of the train being late for 1 day and on time for 4, so it is P(X = 4).

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 4) = C_{5,4}.(0.6)^{4}.(0.4)^{1} = 0.2592$

25.92% probability of spending 12 dollars on coffee over the course of a five day week

Probabability that she spends 15 dollars on coffee:

Probability that the subway is on time all days of the week, so $P(X = 5)$ .

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 5) = C_{5,5}.(0.6)^{5}.(0.4)^{0} = 0.0778$

7.78% probability of spending 12 dollars on coffee over the course of a five day week

8 0

4 years ago