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Luda [366]
2 years ago
9

The recommended daily calcium intake for a 20-year-old is 1,000 milligrams (mg). One cup of milk contains 299 mg of calcium and

one cup of juice contains 261 mg of calcium. Which of the following inequalities represents the possible number of cups of milk m and cups of juice j a 20-year-old could drink in a day to meet or exceed the recommended daily calcium intake from these drinks alone?
Mathematics
1 answer:
Alex787 [66]2 years ago
5 0
One <u>possible inequality</u> is:

299m+261j ≥ 1000.

Explanation:

m is the number of cups of milk and j is the number of cups of juice.

Each cup of milk contains 299 mg of calcium.  This gives us 299m mg of calcium for m cups of milk drank.

Each cup of juice contains 261 mg of calcium.  This gives us 261j mg of calcium for j cups of juice drank.

Together, this would be a total of 299m+261j mg of calcium.

We want to meet or exceed 1000 mg of calcium.  This means we could have more than 1000 or equal to 1000; this gives us the inequality symbol "greater than or equal to":

299m+261j ≥ 1000
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answer A radio station located 120 miles due east of Collinsville has a listening radius of 100 miles. A straight road joins Col
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Answer:

168.7602 miles

Step-by-step explanation:

One way to solve this problem is by using an equation that describes the listening radius of the station, and another for the road, then the points where this two-equation intersect each other will represent when the driver starts and stops listening to the station, and the distance between the points is the miles that the driver will receive the signal.

The equation for the listening radius (the radio station is at (0,0)):

x^2+y^2=100^2

The equation for the road that past through the points (-120,0) and (80,100) (Collinsville and Harmony respectively):

m=\frac{y_2-y_1}{x_2-x_1} =\frac{100-0}{80-(-120)}=\frac{100}{200}=\frac{1}{2}

y-y_1=m(x-x_1)\\y-0=\frac{1}{2}(x-(-120))\\ y=\frac{1}{2}x+60

Substitutes the value of y in the equation of the circle:

x^2+(\frac{1}{2}x+60)^2=100^2\\x^2+\frac{1}{4} x^2+60x+3600=10000\\\frac{5}{4} x^2+60x+3600=10000\\\frac{5}{4} x^2+60x+3600-10000=0\\\frac{5}{4} x^2+60x-6400=0\\5 x^2+240x-25600=0\\x^2+48x-5120=0\\

The formula to solve second-degree equations:

x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac} }{2a} \\x_{1,2}=\frac{-48\pm\sqrt{48^2-4(1)(-5120)} }{2(1)}\\x_{1,2}=\frac{-48\pm\sqrt{2304+20480} }{2}\\x_{1,2}=\frac{-48\pm\sqrt{22784} }{2}\\x_{1,2}=\frac{-48\pm16\sqrt{89} }{2}\\x_{1,2}=-24\pm8\sqrt{89} \\x_1=-24+8\sqrt{89}\approx51.4718\\x_2=-24-8\sqrt{89}\approx-99.4718\\

Using the values in x to find the values in y:

y_1=\frac{1}{2}x_1+60\\y_1=\frac{1}{2}(-24+8\sqrt{89} )+60\\y_1=-12+4\sqrt{89}+60\\ y_1=48+4\sqrt{89}\approx85.7359

y_2=\frac{1}{2}x_2+60\\y_2=\frac{1}{2}(-24-8\sqrt{89} )+60\\y_1=-12-4\sqrt{89}+60\\ y_1=48-4\sqrt{89}\approx10.2641

The distance between the points (51.4718,85.7359) and (-99.4718,10.2641) :

d=\sqrt{(x_1 -x_2 )^2+(y_1 -y_2)^2} \\d=\sqrt{(-24+8\sqrt{89} -(-24-8\sqrt{89}) )^2+(48+4\sqrt{89} -(48-4\sqrt{89}) )^2}\\d=\sqrt{(-24+8\sqrt{89} +24+8\sqrt{89} )^2+(48+4\sqrt{89} -48+4\sqrt{89} )^2}\\d=\sqrt{(16\sqrt{89} )^2+(8\sqrt{89} )^2}\\d=\sqrt{22784+5696}\\d=\sqrt{28480}\\d=8\sqrt{445}\approx168.7602miles

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Step-by-step explanation:


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Step-by-step explanation:

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The best conclusion that can be made is - The difference of the means is not significant because the re-randomizations show that it is within the range of what could happen by chance.

As we know, in statistics, nothing happens by chance. So, this option is correct.

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