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kirill115 [55]
3 years ago
5

A container holds 4 gallons of lemonade. A large lemonade contains 16 ounces. How many large lemonades could the restaurant cell

before they ran out of lemonade?
Mathematics
1 answer:
Shalnov [3]3 years ago
3 0

Given:

A container holds 4 gallons of lemonade.

A large lemonade contains 16 ounces.

To find:

The number of large lemonades that the restaurant can cell before they ran out of lemonade.

Solution:

We know that,

1 gallon = 128 ounces

4 gallon = 512 ounces

So, a container holds 512 ounces of lemonade.

A large lemonade contains 16 ounces. So,

\text{Number of large lemonades}=\dfrac{\text{Total lemonade in container}}{\text{Lemonade in a large container}}

\text{Number of large lemonades}=\dfrac{512}{16}

\text{Number of large lemonades}=32

Therefore, the required number of lemonades is 32.

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Answer:

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Frequency(f):      1          5        11       15      10     6        2

           xf:           62       315    704    975  660  402    136

b) mean(X) = ∑xf/N = 3254/50 = 65.08

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   (x - X)²:        9.49     4.33      1.17       0.0064    0.85   3.69    8.53

ii) Standard deviation(S.D.):  √(∑(x - X)²/N = √(28.0664)/50 = √0.5613

 ∴ S.D.= 0.75

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3 years ago
Choose all of the expressions that are equal to 1,024.
Arte-miy333 [17]
The only one that gives 1024 is A.
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Answer:

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DE ≅ QP

∠E ≅ ∠Q

etc.

Not B), because, while BC ≅ GH, the angles are not all the same, which makes it non-congruent.

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blondinia [14]
The given number which is 2A193A411 is written in Base 11. Base 11 number is also called undecimal system. The digits of Base 11 number are 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, and 10. Since 10 is not accepted as a digit, then we have to substitute a variable which is A = 10. Hence, the digits of a Base 11 number are 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, and A.
3 0
3 years ago
Huck and Jim are waiting for a raft. The number of rafts floating by over intervals of time is a Poisson process with a rate of
Harrizon [31]

Answer: 0.0081

Step-by-step explanation:

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Given : The mean number of rafts floating : \lambda=0.4 rafts per day .

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The formula to calculate the Poisson distribution is given by :_

P(X=x)=\dfrac{e^{-\lambda_1}\lambda_1^x}{x!}

Now, the  probability that they will have to wait more than a week is given by :-

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Hence, the required probability : 0.0081

4 0
3 years ago
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