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Goryan [66]
3 years ago
8

The amount of daily time that teenagers spend on a brand A cell phone is normally distributed with a given mean mc027-1.jpg = 2.

5 hr and standard deviation mc027-2.jpg = 0.6 hr. What percentage of the teenagers spend more than 3.1 hr?
Mathematics
2 answers:
Anarel [89]3 years ago
7 0
<span><span>If I read this properly, mean is 2.5 hr and sd is 0.6 hr.
z=(x-mean)/sd

=0.6/0.6, or 1
What is probability z>1?
It is 0.1587 or 16%. </span><span>
</span></span>
ivolga24 [154]3 years ago
4 0
<span>16% is the answer Fam.....................................</span>
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The problem is asking how much each person will need to pay. Simplifying the problem into an equation with variables (an algorithm) will greatly help you solve it:

S = Sales Tax = $ 7.18 per any purchase
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F = Food = $ 35.50 purchases for two people

We know the cost for one person was: (22.50) + [(35.50/2) + 7.18] =
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Method A)
[2A + (F + 2S)] / 2 = [ (2)(22.50) + [35.50 + (2)(7.18)] ]/ 2 = $47.43
Method A is the correct answer

Method B)
[(2A + (1/2)F + 2S) /2 = [(2)(22.50) + 35.50(1/2) + (2)7.18] / 2 = $38.55
Wrong answer. This method is incorrect because the tax for both tickets bought are not being used in the equation.

Method C)
[(A + F) / 2 ]+ S = [(22.50 + 35.50) / 2 ] + 7.18 = $35.93
Wrong answer. Incorrect Method. The food cost is being reduced to the cost of one person but admission price is set for two people.
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\text{Let}\ k:y=m_1x+b_1\ \text{and}\ l:y=m_2x+b_2.\\\\l\ \parallel\ k\iff m_1=m_2\\\\\text{We have}\\\\L_1:y=5x+1\to m_1=5\\\\L_2:2y-10x-3=0\qquad\text{add 10x and 3 to both sides}\\\\L_2:2y=10x+3\qquad\text{divide both sides by 2}\\\\L_2:y=5x+1.5\to m_2=5\\\\m_1=m_2\ \text{therefore}\ l_1\ \text{and}\ L_2\ \text{are parallel.}

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