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Stolb23 [73]
3 years ago
8

During a catered lunch, an average of 4 cups of tea are poured per minute. The lunch will last 2 hours. How many gallons of tea

should the caterer bring if there are 16 cups in one gallon?
Mathematics
2 answers:
mina [271]3 years ago
8 0
1 min = 4cups
60 min = 1hr
4×60 =240
1hr=240 cups
2hrs = 480 cups
1 gallon = 16 cups
480/16 = 30 gallons

The caterer should bring 30 gallons of tea
hram777 [196]3 years ago
8 0
In 2 hours you would need 30 gallons of tea.
Explanation:
4cups/per minute x 1 gallon/16cups x 60minutes/1 hour = 240gallons/16hours
240gallons/16hours = 15gallons/1 hour
15gallons/1 hour x 2 = 30 gallons/2 hours
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Suppose that X is a random variable with mean and variance both equal to 20. What can be said about P{0 ≤ X ≤ 40}.
Dennis_Churaev [7]

Answer:

P(0 \leq X \40) = \frac{19}{20}

Step-by-step explanation:

We know that:

P(|X - 20| < 20) + P(|X - 20| \geq 20) = 1

So

P(0 \leq X \leq 40) = P(0-20 \leq X-20 \leq 40-20) = P(-20 \leq X \leq 20) = P(|X - 20| \leq 20)

P(|X - 20| \leq 20) = 1 - P(|X - 20| > 20) = 1 - \frac{20}{20^{2}} = \frac{19}{20}

So

P(0 \leq X \40) = \frac{19}{20}

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3 years ago
Find the next term in the following number pattern: 1, 16, 81, 256, 625, ____
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1^4 = 1
2^4 = 16
3^4 = 81
4^4 = 256
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For the answer to the question above,

From the data given, generate first the quadratic equation which best describe the height of the ball. The quadratic function has a general form of 
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A coin is tossed 20 times. A person, who claims to have extrasensory perception, is asked to predict the outcome of each flip in
Korolek [52]

Answer:

5.77% probability of being correct 14 or more times by guessing

Step-by-step explanation:

For each time the coin is tossed, there are only two possible outcomes. Either the person predicts the correct outcome, or she does not. The probability of predicting the correct outcome in a toss is independent of other tosses. 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.

A coin is tossed 20 times.

This means that n = 20

Fair coin:

Equally as likely to be heads or tails, so p = \frac{1}{2} = 0.5

What is the probability of being correct 14 or more times by guessing

P(X \geq 14) = P(X = 14) + P(X = 15) + P(X = 16) + P(X = 17) + P(X = 18) + P(X = 19) + P(X = 20)

In which

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

P(X = 14) = C_{20,14}.(0.5)^{14}.(0.5)^{6} = 0.0370

P(X = 15) = C_{20,15}.(0.5)^{15}.(0.5)^{5} = 0.0148

P(X = 16) = C_{20,16}.(0.5)^{16}.(0.5)^{4} = 0.0046

P(X = 17) = C_{20,17}.(0.5)^{17}.(0.5)^{3} = 0.0011

P(X = 18) = C_{20,18}.(0.5)^{18}.(0.5)^{2} = 0.0002

P(X = 19) = C_{20,19}.(0.5)^{19}.(0.5)^{1} \approx 0

P(X = 20) = C_{20,20}.(0.5)^{20}.(0.5)^{0} \approx 0

Then

P(X \geq 14) = P(X = 14) + P(X = 15) + P(X = 16) + P(X = 17) + P(X = 18) + P(X = 19) + P(X = 20) = 0.0370 + 0.0148 + 0.0046 + 0.0011 + 0.0002 + 0 + 0 = 0.0577

5.77% probability of being correct 14 or more times by guessing

5 0
3 years ago
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