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2 years ago

This graph shows how fast a race car can travel in a stock car race

Mathematics

1 answer:

Nookie1986 [14]2 years ago

7 0

Answer:

C. In 2 seconds, the race car travels 180 meters.

Step-by-step explanation:

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Please help me on #37. I know for A you would use the quadratic equation but I got the wrong answer.

ozzi

Δ = (8i)^2 - 4*(-25) => Δ = -36 +100 => Δ = 64 => x1 =(-8i + 8)/2 => x1 = -4i +4;
x2 = (-8i - 8)/2 => x2 = -4i-4; in this case, your solutions are complex conjugates.

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3 years ago

Please help me im stuck

Rudiy27

Answer:

41/100

Step-by-step explanation:

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2 years ago

Pls help i'll give brainlest

Darina [25.2K]

Answer: B. 183 cm^3

V = pi r^2 h/3

V = pi (5^2) (7/3)

V = pi (25) (2.33)

V = 183

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2 years ago

Read 2 more answers

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krek1111 [17]

Answer:

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Step-by-step explanation:

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4 0

3 years ago

Car colors preference change over the years and according to the particular model that the customer selects. In a recent year, 1

Margarita [4]

Answer:

99.05% probabilitiy that at most six cars are black.

Step-by-step explanation:

For each luxury car, there are only two possible outcomes. Either it is black, or it is not black. The probability of a luxury car being black is independent from other luxury cars. So we use the binomial probability distribution to solve this question.

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:

$p = 0.1, n = 25$

If 25 cars of that year and type are randomly selected, find the probabilitiy that at most six cars are black.

$P(X \leq 6) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6)$

In which

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

$P(X = 0) = C_{25,0}.(0.1)^{0}.(0.9)^{25} = 0.0718$

$P(X = 1) = C_{25,1}.(0.1)^{1}.(0.9)^{24} = 0.1994$

$P(X = 2) = C_{25,2}.(0.1)^{2}.(0.9)^{23} = 0.2659$

$P(X = 3) = C_{25,3}.(0.1)^{3}.(0.9)^{22} = 0.2265$

$P(X = 4) = C_{25,4}.(0.1)^{4}.(0.9)^{21} = 0.1384$

$P(X = 5) = C_{25,5}.(0.1)^{5}.(0.9)^{20} = 0.0646$

$P(X = 6) = C_{25,6}.(0.1)^{6}.(0.9)^{19} = 0.0239$

$P(X \leq 6) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) = 0.0718 + 0.1994 + 0.2659 + 0.2265 + 0.1384 + 0.0646 + 0.0239 = 0.9905$

99.05% probabilitiy that at most six cars are black.

7 0

3 years ago