Anyone can help me with this ?

Hey! I don’t have the energy to do this right now and I’m really hoping one of you could help, please I’m begging. I’ll steal yo

Zolol [24]

D
3 apples and 1 banana cause each apple is 1 dollar so just do 1x3=3 and add one banana and it would be 3.50

8 0

3 years ago

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For every 25 litres of petrol a car can run for 60 km.

Dovator [93]

43.75 liters of petrol

6 0

3 years ago

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L,m,n share money in the ratio 2:3:5. What fraction of money does m receive?

vaieri [72.5K]

Marcus has created a budget for his upcoming trip to the theme park. Admission is 40% of the budget. He plans to spend 32% of his money on food, 23% on souvenirs, and save 5% for emergencies. He knows the admission will be $6 less than he will spend on food and souvenirs. How much money will Marcus need to take to the park?

solve my question giving 18 points

5 0

3 years ago

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Suppose weights of the checked baggage of airline passengers follow a nearly normal distribution with mean 46 pounds and standar

Alik [6]

Answer:

85.99% of airline passengers incur this fee.

Step-by-step explanation:

When the distribution is normal, we use the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore 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 pvalue, we get the probability that the value of the measure is greater than X.

Mean 46 pounds and standard deviation 3.7 pounds.

This means that $\mu = 46, \sigma = 3.7$

Most airlines charge a fee for baggage that weigh in excess of 50 pounds. Determine what percent of airline passengers incur this fee.

As a proportion, this is 1 subtracted by the pvalue of Z when X = 50. So

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

$Z = \frac{50 - 46}{3.7}$

$Z = 1.08$

$Z = 1.08$ has a pvalue of 0.8599

0.8599*100% = 85.99%

85.99% of airline passengers incur this fee.

3 0

3 years ago

6. A submarine has three navigational devices but can remain at sea if at least two are working. Suppose that the failure times

horsena [70]

Answer:

The three survival functions are e-t, e-2/3t and e-1/3t

Then, the rate of the second death equals (multiplying all combinations of 1 survival + 1 death * the rate of death of the third device)

(e-t*(1-e-2/3t)+(1-e-t)e-2/3t)1/3e-1/3t +

(e-t*(1-e-1/3t)+(1-e-t)e-1/3t)2/3e-2/3t +

(e-2/3t*(1-e-1/3t)+(1-e-2/3t)e-1/3t)e-t =

1/3e-4/3t + 1/3e-t -2/3e -2t +

2/3e -5/3t + 2/3e-t -4/3e -2t +

e -4/3 + e -5/3 -2e -2t =

e-t + 4/3e-4/3t + 5/3e-5/3t - 4e-2t

Note that this integrates to 1, thus satisfying one of the rules of a density. We also can see that this function is greater than 0, as e-t + 4/3e-4/3t + 5/3e-5/3t - 4e-2t =

e-t + 4/3e-4/3t + 5/3e-5/3t - e-2t - 4/3e-2t - 5/3e-2t =

e-t - e-2t + 4/3e-4/3t - 4/3e-2t + 5/3e-5/3t- 5/3e-2t =

e-t (1 - e-t) + 4/3e-4/3t(1 - e-2/3t) + 5/3e-5/3t(1 - e-1/3t) > 0 for t > 0

Then, to find the average length that the ship is at sea, we must integrate

t(e-t + 4/3e-4/3t + 5/3e-5/3t - 4e-2t) from 0 to infinity.

However, as a shortcut, note that we can easily identify the integral of the first three functions,

te-t, 4/3te-4/3t, and 5/3te-5/3t, as calculating the expected value of the exponential distribution with parameters 1, 4/3, and 5/3, and the expected value is 1, 3/4, and 3/5

For the final function, 4te-2t = 2*2e-2t , this is twice the expected value of the exponential distribution with parameter 2 = 2 * 1/2 = 1

1 + 3/4 + 3/5 - 1 = 27/20

Step-by-step explanation:

5 0

4 years ago