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3 years ago

PART B: Uber was also available for Mike to use to get home from the airport. Uber

Mathematics

1 answer:

Ber [7]3 years ago

3 0

Answer:

taxi

Step-by-step explanation:

you would take the taxi because it would cost less than Uber

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What is the volume of this rectangular prism? e) 4 in. 4 in. 5 in.

Alika [10]

Answer:

5*4*4=80

Step-by-step explanation:

brainliest?

7 0

3 years ago

Read 2 more answers

PLEASE ANSWER ASAP!! What is the equation of the line that best fits the given data?

stepladder [879]

Answer:

it should be A if not I'm so so sry

3 0

2 years ago

Naval intelligence reports that 99 enemy vessels in a fleet of 1818 are carrying nuclear weapons. If 99 vessels are randomly tar

oee [108]

Answer:

0.001687 = 0.1687% probability that no more than 1 vessel transporting nuclear weapons was destroyed.

Step-by-step explanation:

The vessels are destroyed and then not replaced, which means that the hypergeometric distribution is used to solve this question.

Hypergeometric distribution:

The probability of x successes is given by the following formula:

$P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}$

In which:

x is the number of successes.

N is the size of the population.

n is the size of the sample.

k is the total number of desired outcomes.

Combinations formula:

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

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

In this question:

Fleet of 18 means that $N = 18$

9 are carrying nuclear weapons, which means that $k = 9$

9 are destroyed, which means that $n = 9$

What is the probability that no more than 1 vessel transporting nuclear weapons was destroyed?

This is:

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

In which

$P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}$

$P(X = 0) = h(0,18,9,9) = \frac{C_{9,0}*C_{9,9}}{C_{18,9}} = 0.000021$

$P(X = 1) = h(1,18,9,9) = \frac{C_{9,1}*C_{9,8}}{C_{18,9}} = 0.001666$

Then

$P(X \leq 1) = P(X = 0) + P(X = 1) = 0.000021 + 0.001666 = 0.001687$

0.001687 = 0.1687% probability that no more than 1 vessel transporting nuclear weapons was destroyed.

4 0

3 years ago

The length of a parking Lot is 5 yards more than the width. If the area of the parking lot is 66 square yards, find the dimensio

Brums [2.3K]

Answer:

6 yards by 11 yards

Step-by-step explanation:

Factors of 66 are ...

66 = 1·66 = 2·33 = 3·22 = 6·11

The last pair of factors differ by 5, so represent the solution to the problem.

The width of the parking lot is 6 yards; the length is 11 yards.

7 0

3 years ago

Before every flight, the pilot must verify that the total weight of the load is less than the maximum allowable load for the ai

ss7ja [257]

Answer:

The probability that the plane is oveloaded is P=0.9983.

The pilot should take out the baggage and send it in another plain or have less passengers in the plain to not overload.

Step-by-step explanation:

The aircraft will be overloaded if the mean weight of the passengers is greater than 163 lb.

If the plane is full, we have 41 men in the plane. This is our sample size.

The weights of men are normally distributed with a mean of 180.5 lb and a standard deviation of 38.2.

So the mean of the sample is 180.5 lb (equal to the population mean).

The standard deviation is:

$\sigma=\frac{\sigma}{\sqrt{N}} =\frac{38.2}{\sqrt{41}}=\frac{38.2}{6.4} =5.97$

Then, we can calculate the z value for x=163 lb.

$z=\frac{x-\mu}{\sigma}=\frac{163-180.5}{5.97}=\frac{-17.5}{5.97}= -2.93$

The probability that the mean weight of the men in the airplane is below 163 lb is P=0.0017

$P(\bar X$

Then the probability that the plane is oveloaded is P=0.9983:

$P(overloaded)=1-P(X$

The pilot should take out the baggage or have less passengers in the plain to not overload.

7 0

3 years ago