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myrzilka [38]
3 years ago
5

A car uses 7litres of pétrole to coker 28km . How many litres of pétrole does hé use to coker 64km

Mathematics
1 answer:
lions [1.4K]3 years ago
5 0

28/7=4km is car journey covered by car in 1l

64/4=16l

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Round 491,778 to the nearest hundred
hichkok12 [17]

Answer:

491,800!

Step-by-step explanation:

6 0
3 years ago
What is the period of the function f(x)=sin(x/3) ?
vitfil [10]
<h3>Answer: 6pi radians</h3>

(this is equivalent to 1080 degrees)

======================================

Explanation:

f(x) = sin(x/3)

is the same as

f(x) = 1*sin( (1/3)(x-0) )+0

and that is in the form

f(x) = A*sin( B(x-C) )+D

The letters A,B,C,D are explained below

A = helps find the amplitude

B = 2pi/T, where T is the period

C = determines phase shift (aka left/right shifting)

D = determines vertical shift = midline

All we care about is the value of B as that is the only thing that is connected to the period T

--------

Compare f(x) = 1*sin( (1/3)(x-0) )+0 with f(x) = A*sin( B(x-C) )+D and we see that B = 1/3, so,

B = 2pi/T

1/3 = 2pi/T

1*T = 3*2pi ... cross multiply

T = 6pi

The period is 6pi radians. This is equivalent to 1080 degrees. To convert from radians to degrees, you multiply by (180/pi).

6 0
3 years ago
A jar contains 21 yellow marbles and 22 orange marbles. A marble is drawn at random.<br> P(orange)
Lina20 [59]

Answer:

1/43 is the probability of drawing an orange marble.

7 0
3 years ago
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Given an=−6+(n−1)(23) a n = - 6 + ( n - 1 ) ( 2 3 ) what would the constant rate of change be?
KiRa [710]
23 is the constant rate of change..............
4 0
4 years ago
34​% of college students say they use credit cards because of the rewards program. You randomly select 10 college students and a
finlep [7]

Answer:

a) There is a 18.73% probability that exactly two students use credit cards because of the rewards program.

b) There is a 71.62% probability that more than two students use credit cards because of the rewards program.

c) There is a 82% probability that between two and five students, inclusive, use credit cards because of the rewards program.

Step-by-step explanation:

There are only two possible outcomes. Either the student use credit cards because of the rewards program, or they use for other reason. So, we can solve this problem by the binomial distribution.

Binomial probability

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}.\pi^{x}.(1-\pi)^{n-x}

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

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

And \pi is the probability of X happening.

In this problem, we have that:

10 student are sampled, so n = 10

34% of college students say they use credit cards because of the rewards program, so \pi = 0.34

(a) exactly​ two

This is P(X = 2).

P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}

P(X = 2) = C_{10,2}.(0.34)^{2}.(0.66)^{8} = 0.1873

There is a 18.73% probability that exactly two students use credit cards because of the rewards program.

(b) more than​ two

This is P(X > 2).

Either a value is larger than two, or it is smaller of equal. The sum of the decimal probabilities must be 1. So:

P(X \leq 2) + P(X > 2) = 1

P(X > 2) = 1 - P(X \leq 2)

In which

P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)

So

P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}

P(X = 0) = C_{10,0}.(0.34)^{0}.(0.66)^{10} = 0.0157

P(X = 1) = C_{10,1}.(0.34)^{1}.(0.66)^{9} = 0.0808

P(X = 2) = C_{10,2}.(0.34)^{2}.(0.66)^{8} = 0.1873

P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.0157 + 0.0808 + 0.1873 = 0.2838

P(X > 2) = 1 - P(X \leq 2) = 1 - 0.2838 = 0.7162

There is a 71.62% probability that more than two students use credit cards because of the rewards program.

(c) between two and five inclusive

This is:

P = P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)

P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}

P(X = 2) = C_{10,2}.(0.34)^{2}.(0.66)^{8} = 0.1873

P(X = 3) = C_{10,3}.(0.34)^{3}.(0.66)^{7} = 0.2573

P(X = 4) = C_{10,4}.(0.34)^{4}.(0.66)^{6} = 0.2320

P(X = 5) = C_{10,5}.(0.34)^{5}.(0.66)^{5} = 0.1434

P = P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) = 0.1873 + 0.2573 + 0.2320 + 0.1434 = 0.82

There is a 82% probability that between two and five students, inclusive, use credit cards because of the rewards program.

6 0
3 years ago
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