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Novay_Z [31]
3 years ago
11

Tickets to a school play are $4.50 for an adult and $3.00 for a student. If 300 tickets are sold and $1087.50 collected, how man

y tickets of each were sold
Mathematics
1 answer:
Kryger [21]3 years ago
3 0

Answer:

<h2>125 tickets for an adult</h2><h2>and 175 tickets for a student</h2>

Step-by-step explanation:

a-\text{number of the tickets for an adult}\\s-\text{number of the tickets for a student}\\\\(1)\qquad a+s=300\\(2)\qquad4.5a+3s=1087.5\\------------\\(1)\\a+s=300\qquad\text{subtstitute}\ s\ \text{from both sides}\\a=300-s\\\\\text{Put it to (2):}\\\\4.5(300-s)+3s=1087.5\qquad\text{use distributive property}\\\\(4.5)(300)+(4.5)(-s)+3s=1087.5\\\\1350-4.5s+3s=1087.5\qquad\text{subtract 1350 from both sides}\\\\-1.5s=-262.5\qquad\text{divide both sides by (-1.5)}\\\\\boxed{s=175}

\text{Put the value of}\ s\ \text{to (1):}\\\\a=300-175\\\\\boxed{a=125}

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Previously, an organization reported that teenagers spent 24.5 hours per week, on average, on the phone. The organization thinks
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Answer:

We need to develop a one-tail t-student test ( test to the right )

We reject H₀  we find evidence that student spent more than 24,5 hours on the phone

Step-by-step explanation:

Sample size  n = 15     n < 30

And we were asked if the mean is higher than, therefore is a one-tail t-student test ( test to the right )

Population mean   μ₀  = 24,5

Sample mean   μ  =  25,7

Sample standard deviation s = 2

Hypothesis Test:

Null Hypothesis      H₀                             μ  =  μ₀

Alternative Hypothesis     Hₐ                  μ  >  μ₀

t (c) =  ?

We will define CI = 95 %  then   α = 5 %   α = 0,05    α/2 =  0,025

n = 15     then degree of freedom    df = 14

From t-student table  we get:  t(c) = 2,1448

And  t(s)

t(s) = ( μ  -  μ₀  ) / s/√n

t(s) = (25,7 - 24,5) /2/√15

t(s) = 2,3237

Now we compare   t(c)   and  t(s)

t(c)  =  2,1448         t(s)  = 2,3237

t(s) > t(c)

Then we are in the rejection region we reject H₀   we have evidence at 95% of CI that students spend more than 24,5 hours per week on the phone

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3 years ago
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