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Anon25 [30]
2 years ago
15

A car completes a journey of 570 km at an average speed of 75 km/h calculate the time taken for the journey in hours and minutes

Mathematics
1 answer:
Evgesh-ka [11]2 years ago
4 0
570/75=7 hours and 36 minutes
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What is the product?
Kazeer [188]

Answer:

2) -81t² +16

Step-by-step explanation:

(9t -4)(-9t -4) = (9t x -9t) + (9t x -4) + (-4 x -9t) + (-4 x -4)

                     = -81t² - 36t + 36t + 16

                     = -81t² + 16

4 0
2 years ago
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We have just moved to geometry. <br> Any help? <br><br> This is just practice as well
Vera_Pavlovna [14]

Answer:

rotation of 270 counterclock wise

Step-by-step explanation:

Since none of the other ones are correct and man i miss geometry i'm in Ap calc and i don't recommend haha.

8 0
2 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
3 years ago
In Ryan Corporation, the first shift produced 5 1/2 times as many lightbulbs as the second shift. If the total lightbulbs produc
Andru [333]

Answer:

13750

Step-by-step explanation:

Let 'x' be the number of bulbs produced in second shift

Number of bulbs produced in 1st shift= 5.5x

Total number of bulbs= 5.5x+x

16250=5.5x+x

x=2500

Number of bulbs produced in shift 1= 5.5×2500

                                                            = 13750

7 0
2 years ago
:
Degger [83]

Let x = pounds of $12 per pound coffee

then

20-x = pounds of $9 per pound coffee

.

12x + 9(20-x) = 10(20)

12x + 180 - 9x = 200

3x + 180 = 200

3x = 20

x = 20/3

x = 6 and 2/3 pounds of $12 per pound coffee

.

Amount of $9 per pound coffee:

20-x = 20-20/3 = 60/3 - 20/3 = 40/3

or

13 and 1/3 pounds of $9 per pound coffee

3 0
2 years ago
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