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Ber [7]
3 years ago
5

The five members of the Traynor family each buy train tickets. During the train ride, each family member buys a boxed lunch for

$6.50. If the total cost of the trip is $248.50, what is the price of each train ticket?
Mathematics
2 answers:
larisa [96]3 years ago
8 0

Let's assume

the price of each train ticket is x

we are given

total number of family members =5

so, total cost of ticket = total number of family members*the price of each ticket

so, total cost of ticket =5*x

so, total cost of ticket =5x

now, we have

each family member buys a boxed lunch for $6.50

so, total cost of lunch box = total number of family members*price of each lunch box

so, total cost of lunch box = 5*6.50

total cost of trip = total cost of ticket+total cost of lunch box

we are given

total cost of trip is $248.50

now, we can plug values

and we get

248.50=5x+5*6.50

now, we can solve for x

248.50=5x+32.5

248.50-32.5=5x+32.5-32.5

216=5x

\frac{5x}{5} =\frac{216}{5}

x=43.2

so,

the price of each train ticket is $43.2..........Answer

anyanavicka [17]3 years ago
5 0
So 5(6.50) = 32.5
248.5 - 32.5 = 216
216/5 = 43.2
each ticket is $43.20
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Answer: The value of g(x) is x+3.

Step-by-step explanation:

Since we have given that

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We need to find the value of g(x):

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I need help?! Please
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Step-by-step explanation:

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A college admits a freshman class of 330 students at the beginning of each academic year and graduates 145 students twice a year
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Answer:

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Step-by-step explanation:

<u>Change in one year:</u>

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<u>Change after 5 years:</u>

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.Mariam went to the store to buy some apples. The price per pound of the apples is $6 per pound and she has a coupon for $2.25 o
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A manufacturer produces bearings, but because of variability in the production process, not all of the bearings have the same di
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Answer:

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Step-by-step explanation:

We are given that the diameters have a normal distribution with a mean of 1.3 centimeters (cm) and a standard deviation of 0.01 cm i.e.;

Mean, \mu = 1.3 cm            and           Standard deviation, \sigma = 0.01 cm

Also, since distribution is normal;

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Let X = range of diameters

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  P(X <= 1.27) = P( \frac{X -\mu}{\sigma} < \frac{1.27 -1.3}{0.01} ) = P(Z < -3) = 1 - P(Z < 3) = 1 - 0.99865

                                                                                            = 0.00135

 P(1.27 < X < 1.33) = 0.99865 - 0.00135 = 0.9973 .

Therefore, proportion of all bearings that falls in this acceptable range is 99.73% .

7 0
4 years ago
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