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andrey2020 [161]

2 years ago

12

For the school's sports day, a group of students prepared 12 1/2 litres of lemonade. At the end of the day they had 2 5/8 litres

left over. How many litres of lemonade were sold?

1 answer:

Hoochie [10]2 years ago

7 0

Given :

For the school's sports day, a group of students prepared 12 1/2 litres of lemonade. At the end of the day they had 2 5/8 litres left over.

To Find :

How many litres of lemonade were sold.

Solution :

Initial amount of lemonade, I = 12 1/2 = 25/2 litres.

Final amount of lemonade, F = 2 5/8 = 21/8 litres.

Amount of lemonade sold, A = I - F

A = 25/2 - 21/8 litres

A = 9.875 litres

Therefore, 9.875 litres of lemonade were sold.

Hence, this is the required solution.

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At Brighton Middle School, Mr. Yule asked 50 randomly selected students from each grade level about their favorite subject, and

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Answer:

Yes, he made a reasonable inference by saying about 25 percent of students choose the science.

Step-by-step explanation:

Given that

Number of students for survey = 50

Number of students who choose science = 12

Now

Percentage of student who choose science = $\frac{12}{50} x 100$

Percentage of student who choose science = $\frac{12*100}{50}$

Percentage of student who choose science = 24 %

Now as Mr. Yule said "about 25" percent of students choose the science it can be treated as reasonable approximation.

Now if the number of students in survey become 200, then we can expect 12 out of every 50 students would choose the science. So, in total 48 students would choose the science.

Now lets calculate the percentage of students who choose science.

Percentage of student who choose science = $\frac{48}{200} x 100$

Percentage of student who choose science = $\frac{48*100}{200}$

Percentage of student who choose science = 24 %

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2 years ago

The marching band is selling candy bars and fruit snacks for a fundraiser. Candy bars are sold for $2.50 and fruit snacks are $1

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Answer:

220 candy bars

Step-by-step explanation:

Let the number of candy bars sold=c

Let the number of fruit snacks sold=f

A total of 650 items were sold.

Therefore: c+f=650

Candy bars are sold for $2.50

Fruit snacks are sold for $1.50

Total Income =$1,195

Therefore: 2.5c+1.5f=1195

We solve the resulting system of equation for c, the number of candy bars sold.

c+f=650

2.5c+1.5f=1195

From the first equation, f=650-c

Substitute f=650-c into 2.5c+1.5f=1195

2.5c+1.5(650-c)=1195

2.5c+975-1.5c=1195

Collect like terms

2.5c-1.5c=1195-975

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Therefore, 220 candy bars were sold.

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2 years ago

During April of 2013, Gallup randomly surveyed 500 adults in the US, and 47% said that they were happy, and without a lot of str

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Answer:

number of successes

$k = 235$

number of failure

$y = 265$

The criteria are met

A

The sample proportion is $\r p = 0.47$

B

$E =4.4 \%$

C

What this mean is that for N number of times the survey is carried out that the which sample proportion obtain will differ from the true population proportion will not more than 4.4%

Ci

$r = 0.514 = 51.4 \%$

$v = 0.426 = 42.6 \%$

D

This 95% confidence interval mean that the the chance of the true population proportion of those that are happy to be exist within the upper and the lower limit is 95%

E

Given that 50% of the population proportion lie with the 95% confidence interval the it correct to say that it is reasonably likely that a majority of U.S. adults were happy at that time

F

Yes our result would support the claim because

$\frac{1}{3 } \ of N < \frac{1}{2} (50\%) \ of \ N , \ Where\ N \ is \ the \ population\ size$

Step-by-step explanation:

From the question we are told that

The sample size is $n = 500$

The sample proportion is $\r p = 0.47$

Generally the number of successes is mathematical represented as

$k = n * \r p$

substituting values

$k = 500 * 0.47$

$k = 235$

Generally the number of failure is mathematical represented as

$y = n * (1 -\r p )$

substituting values

$y = 500 * (1 - 0.47 )$

$y = 265$

for approximate normality for a confidence interval criteria to be satisfied

$np > 5 \ and \ n(1- p ) \ >5$

Given that the above is true for this survey then we can say that the criteria are met

Given that the confidence level is 95% then the level of confidence is mathematically evaluated as

$\alpha = 100 - 95$

$\alpha = 5 \%$

$\alpha =0.05$

Next we obtain the critical value of $\frac{\alpha }{2}$ from the normal distribution table, the value is

$Z_{\frac{ \alpha }{2} } = 1.96$

Generally the margin of error is mathematically represented as

$E = Z_{\frac{\alpha }{2} } * \sqrt{ \frac{\r p (1- \r p}{n} }$

substituting values

$E = 1.96 * \sqrt{ \frac{0.47 (1- 0.47}{500} }$

$E = 0.044$

=> $E =4.4 \%$

What this mean is that for N number of times the survey is carried out that the proportion obtain will differ from the true population proportion of those that are happy by more than 4.4%

The 95% confidence interval is mathematically represented as

$\r p - E < p < \r p + E$

substituting values

$0.47 - 0.044 < p < 0.47 + 0.044$

$0.426 < p < 0.514$

The upper limit of the 95% confidence interval is $r = 0.514 = 51.4 \%$

The lower limit of the 95% confidence interval is $v = 0.426 = 42.6 \%$

This 95% confidence interval mean that the the chance of the true population proportion of those that are happy to be exist within the upper and the lower limit is 95%

Given that 50% of the population proportion lie with the 95% confidence interval the it correct to say that it is reasonably likely that a majority of U.S. adults were happy at that time

Yes our result would support the claim because

$\frac{1}{3 } < \frac{1}{2} (50\%)$

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