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ryzh [129]
3 years ago
7

You are a running a concession stand selling hot dogs and sodas each hot dog costs .75 cents and each soda costs $1.25. You sold

27 more sodas than hot dogs. if you made a total of $257.75. How many sodas and hot dogs did you sell?
Mathematics
2 answers:
tatiyna3 years ago
6 0
<span>Each hot dog costs $ 1.50 and each soda costs $ .50. You made a total of $ 78.50. You sold a total of 87 hot dogs and sodas combined. You must report the number of hot dogs and soda's sold. How many hot dogs and soda's were sold? 

Quantity Equation: h + s = 87
Value Equation:: 1.5h+0.5s=78.50

Multiply thru the Quantity Equation by 15
Multiply thru the Value Equation by 10.

15h + 15s = 15*87
15h + 5s = 785</span>Subtract and solve for "s":<span>10s = 520
</span>s = 52 (# of soda's sold)
<span>
Solve for h:
h + s = 87
h + 52 = 87
h = 35 (# of hot dog's sold)</span>
anastassius [24]3 years ago
5 0
<span>Each hot dog costs $ 1.50 and each soda costs $ .50. You made a total of $ 78.50. You sold a total of 87 hot dogs and sodas combined. You must report the number of hot dogs and soda's sold. How many hot dogs and soda's were sold? 

Quantity Equation: h + s = 87
Value Equation:: 1.5h+0.5s=78.50

Multiply thru the Quantity Equation by 15
Multiply thru the Value Equation by 10.

15h + 15s = 15*87
15h + 5s = 785</span>Subtract and solve for "s":<span>10s = 520
</span>s = 52 (# of soda's sold)
<span>
Solve for h:
h + s = 87
h + 52 = 87
h = 35 (# of hot dog's sold)

hope i helped^-^</span>
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Answer:

\mu_p -\sigma_p = 0.74-0.0219=0.718

\mu_p +\sigma_p = 0.74+0.0219=0.762

68% of the rates are expected to be betwen 0.718 and 0.762

\mu_p -2*\sigma_p = 0.74-2*0.0219=0.696

\mu_p +2*\sigma_p = 0.74+2*0.0219=0.784

95% of the rates are expected to be betwen 0.696 and 0.784

\mu_p -3*\sigma_p = 0.74-3*0.0219=0.674

\mu_p +3*\sigma_p = 0.74+3*0.0219=0.806

99.7% of the rates are expected to be betwen 0.674 and 0.806

Step-by-step explanation:

Check for conditions

For this case in order to use the normal distribution for this case or the 68-95-99.7% rule we need to satisfy 3 conditions:

a) Independence : we assume that the random sample of 400 students each student is independent from the other.

b) 10% condition: We assume that the sample size on this case 400 is less than 10% of the real population size.

c) np= 400*0.74= 296>10

n(1-p) = 400*(1-0.74)=104>10

So then we have all the conditions satisfied.

Solution to the problem

For this case we know that the distribution for the population proportion is given by:

p \sim N(p, \sqrt{\frac{p(1-p)}{n}})

So then:

\mu_p = 0.74

\sigma_p =\sqrt{\frac{0.74(1-0.74)}{400}}=0.0219

The empirical rule, also referred to as the three-sigma rule or 68-95-99.7 rule, is a statistical rule which states that for a normal distribution, almost all data falls within three standard deviations (denoted by σ) of the mean (denoted by µ). Broken down, the empirical rule shows that 68% falls within the first standard deviation (µ ± σ), 95% within the first two standard deviations (µ ± 2σ), and 99.7% within the first three standard deviations (µ ± 3σ).

\mu_p -\sigma_p = 0.74-0.0219=0.718

\mu_p +\sigma_p = 0.74+0.0219=0.762

68% of the rates are expected to be betwen 0.718 and 0.762

\mu_p -2*\sigma_p = 0.74-2*0.0219=0.696

\mu_p +2*\sigma_p = 0.74+2*0.0219=0.784

95% of the rates are expected to be betwen 0.696 and 0.784

\mu_p -3*\sigma_p = 0.74-3*0.0219=0.674

\mu_p +3*\sigma_p = 0.74+3*0.0219=0.806

99.7% of the rates are expected to be betwen 0.674 and 0.806

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