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Verizon [17]
2 years ago
12

Suri made 6 gallons of lemonade to sell at her lemonade stand. in one day she sold 2/3 of the lemonade. how much lemonade did sh

e sell?

Mathematics
2 answers:
Zina [86]2 years ago
7 0
I hope this helps you

nikdorinn [45]2 years ago
6 0

Answer: There are 4 lemonades that she sold.

Step-by-step explanation:

Since we have given that

Number of gallons of lemonade to sell = 6

Fraction of lemonade she sold = \dfrac{2}{3}

So, the number of lemonades she sold is given by

\dfrac{2}{3}\times 6\\\\=\dfrac{2\times 6}{3}\\\\=\dfrac{12}{3}\\\\=4

Hence, there are 4 lemonades that she sold.

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A hospital bill is estimated to be $462.00. It ends up actually costing the patient $525.00. What is the percent error in the bi
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In order to find the percent error, we need to first find the difference between what was expected and what is actually costed. We do this by subtracting:

525-462=63

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1 year ago
Dalgliesh the detective fancies himself a shrewd judge of human nature. In careful tests, it has been discovered that he is righ
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Answer:

The probability that Jones is lying is 6/7

Step-by-step explanation:

First we will list out 2 different cases when the outcome is a lie

1.probability that Jones tells lies is = 0.6 and probability that dalgiliesh analyses it correctly is 0.8

So the probability that dagliesh correctly analyses that he is telling lies is 0.8*0.6=0.48

2.Probability that Jones tells truth is 0.4 and if dagliesh analyses it incorrectly (which has a probability of 0.2) the outcome(as analysed by dagiliesh) is a lie

So probability that dagliesh analyses Jones truth as a lie is 0.2*.0.4=0.08

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"Majesty Video Production Inc. wants the mean length of its advertisements to be 30 seconds. Assume the distribution of ad lengt
Westkost [7]

Answer:

a) \bar X \sim N(\mu=30, \frac{2}{\sqrt{16}})

b) Se=\frac{\sigma}{\sqrt{n}}=\frac{2}{\sqrt{16}}=0.5

c) P(\bar X >31.25)=0.006=0.6\%

d) P(\bar X >28.25)=0.9997=99.97\%

e) P(28.25

Step-by-step explanation:

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

Let X the random variabl length of advertisements produced by Majesty Video Production Inc. We know from the problem that the distribution for the random variable X is given by:

X\sim N(\mu =30,\sigma =2)

We take a sample of n=16 . That represent the sample size.

a. What can we say about the shape of the distribution of the sample mean time?

From the central limit theorem we know that the distribution for the sample mean \bar X is also normal and is given by:

\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})

\bar X \sim N(\mu=30, \frac{2}{\sqrt{16}})

b. What is the standard error of the mean time?

The standard error is given by this formula:

Se=\frac{\sigma}{\sqrt{n}}=\frac{2}{\sqrt{16}}=0.5

c. What percent of the sample means will be greater than 31.25 seconds?

In order to answer this question we can use the z score in order to find the probabilities, the formula given by:

z=\frac{\bar X- \mu}{\frac{\sigma}{\sqrt{n}}}

And we want to find this probability:

P(\bar X >31.25)=1-P(\bar X

d. What percent of the sample means will be greater than 28.25 seconds?

In order to answer this question we can use the z score in order to find the probabilities, the formula is given by:

z=\frac{\bar X- \mu}{\frac{\sigma}{\sqrt{n}}}

And we want to find this probability:

P(\bar X >28.25)=1-P(\bar X

e. What percent of the sample means will be greater than 28.25 but less than 31.25 seconds?"

We want this probability:

P(28.25

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