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laila [671]
4 years ago
8

On Friday,2,364 cars parked in the garage at the mall. On Sunday 2,455 more cars parked In the garage than on Friday. How many c

ars parked in the garage during the two days?
Mathematics
1 answer:
Lorico [155]4 years ago
5 0

Answer:

7183 cars

Step-by-step explanation:

On Friday, 2364 cars parked in the garage.

On Sunday, 2,455 more cars parked In the garage than on Friday.

This means that the number of cars parked on Sunday is 2455 plus the number of cars parked on Friday, that is:

2455 + 2364 = 4819 cars

Therefore, the number of cars parked in the garage in the two days is:

4819 + 2364 = 7183 cars

7183 cars were parked during the two days.

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Calculate the length of the diagonal of a rectangle whose sides measure : (a) 12 meters and 48 meters (b) 11 centimetres and 60
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A sequence of Bernoulli trials consists of choosing components at random from a batch of components. A selected component is eit
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Answer:

a) 0.0287 = 2.87% probability that three non-defective components in a batch of seven components.

b) 0.0336 = 3.36% probability that 8 non-defective components are drawn before the first defective component is chosen.

Step-by-step explanation:

A sequence of Bernoulli trials composes the binomial distribution.

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.

The probability that a selected component is non-defective is 0.8

This means that p = 0.8

a) Three non-defective components in a batch of seven components.

This is P(X = 3) when n = 7. So

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

P(X = 3) = C_{7,3}.(0.8)^{3}.(0.2)^{4} = 0.0287

0.0287 = 2.87% probability that three non-defective components in a batch of seven components.

b) 8 non-defective components are drawn before the first defective component is chosen.

Now the order is important, so the we just multiply the probabilities.

8 non-defective, each with probability 0.8, and then a defective, with probability 0.2. So

p = (0.8)^8*0.2 = 0.0336

0.0336 = 3.36% probability that 8 non-defective components are drawn before the first defective component is chosen.

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