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n200080 [17]
3 years ago
10

or the end-of-year party, Mt. Rose Middle School ordered 112 pizzas. There were eight fewer veggie pizzas than there were pepper

oni pizzas. There were three times as many combo pizzas as pepperoni pizzas. Use the 5-D Process to define a variable and write an equation for this situation. Then determine how many of each kind of pizza were ordered.

Mathematics
2 answers:
marshall27 [118]3 years ago
8 0

Let the number of pepperoni pizzas be x

pepperoni = x

veggie = x - 8    [There were eight fewer veggie than pepperoni]

combo = 3x       [There were three times as many combo as pepperoni]

Given that the total Pizza: 112

x + x - 8 + 3x = 112

5x - 8 = 112

5x = 112 + 8

5x = 120

x = 24

x= 24

x - 8 = 16

3x = 72

So there were 24 pepperoni pizza, 16 veggie pizza and 72 combo pizza

Tresset [83]3 years ago
3 0

Answer:

16 Veggie

24 Pepperoni

72 Combo

Step-by-step explanation:

Working shown in the pic.

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0.5 inches per a month

Step-by-step explanation:

you turn them to a decimal to make them easier and then divide them to get the answer.

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Sarah claims that the thickness of the spearmint gum she produces is 7.5 one-hundredths of an inch. A quality control specialist
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Answer:

t=\frac{7.55-7.5}{\frac{0.103}{\sqrt{10}}}=1.539    

p_v =2*P(t_{(9)}>1.539)=0.158  

If we compare the p value and the significance level assumed \alpha=0.05 we see that p_v>\alpha so we can conclude that we have enough evidence to fail reject the null hypothesis.  

Step-by-step explanation:

Data given and notation  

Data: 7.65, 7.60, 7.65, 7.7, 7.55, 7.55, 7.4, 7.4, 7.5, 7.5

We can begin calculating the sample mean given by:

\bar X = \frac{\sum_{i=1}^n X_i}{n}

And the sample deviation given by:

s=\sart{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}

And we got:

\bar X=7.55 represent the sample mean

s=0.103 represent the sample standard deviation

n=10 sample size  

\mu_o =7.5 represent the value that we want to test

\alpha=0.05 represent the significance level for the hypothesis test.  

t would represent the statistic (variable of interest)  

p_v represent the p value for the test (variable of interest)  

State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the mean is equal to 7.5 or no, the system of hypothesis would be:  

Null hypothesis:\mu = 7.5  

Alternative hypothesis:\mu \neq 7.5  

If we analyze the size for the sample is < 30 and we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:  

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}  (1)  

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

Calculate the statistic

We can replace in formula (1) the info given like this:  

t=\frac{7.55-7.5}{\frac{0.103}{\sqrt{10}}}=1.539    

P-value

The first step is calculate the degrees of freedom, on this case:  

df=n-1=10-1=9  

Since is a two sided test the p value would be:  

p_v =2*P(t_{(9)}>1.539)=0.158  

Conclusion  

If we compare the p value and the significance level assumed \alpha=0.05 we see that p_v>\alpha so we can conclude that we have enough evidence to fail reject the null hypothesis.  

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The diameter of a small gear is 16 cm. This is a 2.5 cm more than 1/4 of the diameter of a large gear. What is the diameter of t
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3 years ago
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Answer:

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Step-by-step explanation:

Let the amount he would have spent without the sales be x. Now, if there is a 65% discount, what this means is that he is exactly paying for 100 - 65% = 35%

Now, it is this 35% of X that is equal to the amount he paid

Thus, mathematically, we have the following;

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Answer:

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