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Pavlova-9 [17]
3 years ago
11

a store is having a sale on chocolate chips and walnuts. for 3 pounds of chocolate chips and 5 pounds of walnuts the total cost

is$23. for 6 pounds of chocolate chips and 2 pounds of walnuts the total cost is $20. Find the cost for each pound of chocolate chips and each pound of walnuts
Mathematics
1 answer:
Delvig [45]3 years ago
6 0
If you set you the equations such as 3c+5w=23 and 6c+2w=20 and then solve the first equation for c, you get c= (23-5w)/3 Taking that equation and substituting it for 'c' in the second equation (so the second equation becomes: 6((23-5w)/3)+2w=20  ) you get w=3.25 If you plug that back into your first equation (so it now becomes: 3c+5(3.25)=23 ) you get c=2.25 So the answer would be walnuts cost $3.25 per lb and choc. chips cost $2.25 per lb. 
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0.0193 = 1.93% probability that there is equipment damage to the payload of at least one of five independently dropped parachutes.

Step-by-step explanation:

For each parachute, there are only two possible outcomes. Either there is damage, or there is not. The probability of there being damage on a parachute is independent of any other parachute, which means that the binomial probability distribution is used to solve this question.

To find the probability of damage on a parachute, the normal distribution is used.

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.

Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the z-score of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Probability of a parachute having damage.

The opening altitude actually has a normal distribution with mean value 185 and standard deviation 32 m, which means that \mu = 185, \sigma = 32

Equipment damage will occur if the parachute opens at an altitude of less than 100 m, which means that the probability of damage is the p-value of  Z when X = 100. So

Z = \frac{X - \mu}{\sigma}

Z = \frac{100 - 185}{32}

Z = -2.66

Z = -2.66 has a p-value of 0.0039.

What is the probability that there is equipment damage to the payload of at least one of five independently dropped parachutes?

0.0039 probability of a parachute having damage, which means that p = 0.0039

5 parachutes, which means that n = 5

This probability is:

P(X \geq 1) = 1 - P(X = 0)

In which

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

P(X = 0) = C_{5,0}.(0.0039)^{0}.(0.9961)^{5} = 0.9807

Then

P(X \geq 1) = 1 - P(X = 0) = 1 - 0.9807 = 0.0193

0.0193 = 1.93% probability that there is equipment damage to the payload of at least one of five independently dropped parachutes.

5 0
3 years ago
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