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Sunny_sXe [5.5K]
3 years ago
14

The U.S. produces about 5.5 million tons of apples each year. Of the total, $20\%$ is mixed with other products, with $50\%$ of

the remainder used for apple juice and the other $50\%$ sold fresh. How many million tons of apples are used for apple juice? Express your answer as a decimal to the nearest tenth.
Mathematics
2 answers:
blondinia [14]3 years ago
8 0

Answer:

2.2 Million

Step-by-step explanation:

First, we wish to determine what percentage of the tons of apples are used for apple juice. After 20% is mixed with other products, 80% remains. Half of this is used for apple juice; therefore, 40% of the tons of apples is used for apple juice. To calculate 40% of 5.5 million tons, we find 10% and then multiply by four. 10% = 0.55, and 0.55* 4 = 2.2. Thus, 2.2 million tons are used for apple juice.

mars1129 [50]3 years ago
7 0

Answer:

2.2 million tons

Step-by-step explanation:

20% of 5.5 million tons is 1.1 million tons. That is how many are mixed with other products. So that would leave you 4.4 million tons of apples.

50% of 4.4 is 2.2 so 2.2 million tons are used for apple juice.

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Sophia bought items from the bake sale at school. She bought a sea salt caramel brownie for $1.94 and a dozen chocolate chunk co
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Required information An article presents a new method for timing traffic signals in heavily traveled intersections. The effectiv
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Answer:

652.6-2.01\frac{311.7}{\sqrt{50}}=563.997    

652.6+2.01\frac{311.7}{\sqrt{50}}=741.203    

So on this case the 95% confidence interval would be given by (563.997;741.203)    

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

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".

\bar X=652.6 represent the sample mean for the sample  

\mu population mean (variable of interest)

s=311.7 represent the sample standard deviation

n=50 represent the sample size  

Soltuion to the problem

The confidence interval for the mean is given by the following formula:

\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}   (1)

In order to calculate the critical value t_{\alpha/2} we need to find first the degrees of freedom, given by:

df=n-1=50-1=49

Since the Confidence is 0.95 or 95%, the value of \alpha=0.05 and \alpha/2 =0.025, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,49)".And we see that t_{\alpha/2}=2.01

Now we have everything in order to replace into formula (1):

652.6-2.01\frac{311.7}{\sqrt{50}}=563.997    

652.6+2.01\frac{311.7}{\sqrt{50}}=741.203    

So on this case the 95% confidence interval would be given by (563.997;741.203)    

6 0
3 years ago
Concert tickets cost $15 for general admission, but only $9 with a student ID. Ticket sales total $450. Write and graph an equat
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The equation that models the situation is 15x + 9y = 450

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Given the following parameters:

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Given that the total sales is $450, the equation that models this situation will be expressed as:

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x is the those that admitted generally

y are those with student ID

<u />

<u>Get the x-intercept.</u>

This is a point where the y-axis is zero

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<u>Get the y-intercept.</u>

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