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dusya [7]
3 years ago
15

Students in 6 classes, displayed below, ate the same ratio of cheese pizza slices to pepperoni pizza slices. Complete the follow

ing table, which represents the number of slices of pizza students in each class ate.
Mathematics
1 answer:
Vikki [24]3 years ago
5 0
<span>Slices of Cheese Pizza 2 , 6, 8, 5 ½, 3 1/3, 3/5 Slices of Pepperoni Pizza 5, 15, 20, 13 ¾, 8 1/3, 1 ½ Total Slices of Pizza 7, 21, 28, 19 ¼, 11 2/3, 2 1/10</span>
Dionne Woods Freeman
2 years ago
Its acually Total Slices of Pizza 7, 21, 28, 19 ¼, 11 2/3, 2 2/10
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Can anybody give real answers please? Thanks.
Kazeer [188]
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3 years ago
Which function grows the fastest for large values of x? f(x)=8x f(x)=3x f(x)=4x2+3 f(x)=1.5x 20 points
Aleonysh [2.5K]
The 4 functions are:
f_1 (x) = 8x
f_2(x)=3x
f_3(x)=4x^2+3
f_4(x)=1.5 x

Let's keep in mind that for large values of x, a quadratic function grows faster than a linear function:
ax^2 \ \textgreater \  kx for large values of x

In this problem, we can see that the only quadratic function is f_3(x), while all the others are linear functions, so the function that grows faster for large values of x is
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7 0
3 years ago
A 90 percent confidence interval for the slope of a regression line is determined to be (-0.181, 1.529). Which of the following
GenaCL600 [577]

Answer: a. The correlation coefficient of the data is positive.

Step-by-step explanation:

Estimated slope of sample regression line  = \dfrac{Upper \ limit + lower\ limit}{2}

Here , confidence interval :  (-0.181, 1.529)

Estimated slope of sample regression line  = \dfrac{-0.181+1.529}{2}

=\dfrac{1.348}{2}\\\\=0.674\ \ \ \ [\text{ positive}]

⇒Correlation coefficient(r) must be positive, So a. is true.

But, d. and e. are wrong(0.674 ≠ 0 or 1.348).

We cannot check residuals or its sum from confidence interval of slope of a regression line, so b is wrong.

We cannot say that scatterplot is linear as we cannot determine it from interval, so c. is wrong

So, the correct option : a. The correlation coefficient of the data is positive.

7 0
3 years ago
Coffee: A popular chain of cafes has been receiving online complaints about one store location. Regular customers complained tha
spin [16.1K]

Answer:

z=\frac{0.18 -0.1}{\sqrt{\frac{0.1(1-0.1)}{100}}}=2.67  

Step-by-step explanation:

1) Data given and notation

n=100 represent the random sample taken

X=18 represent the ounce cups of coffee that were underfilled

\hat p=\frac{18}{100}=0.18 estimated proportion of ounce cups of coffee that were underfilled

p_o=0.1 is the value that we want to test

\alpha represent the significance level  

z would represent the statistic (variable of interest)

p_v represent the p value (variable of interest)  

2) Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that the true proportion it's higher than 0.1 or 10%:  

Null hypothesis:p\leq 0.1  

Alternative hypothesis:p > 0.1  

When we conduct a proportion test we need to use the z statistic, and the is given by:  

z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}} (1)  

The One-Sample Proportion Test is used to assess whether a population proportion \hat p is significantly different from a hypothesized value p_o.

3) Calculate the statistic  

Since we have all the info requires we can replace in formula (1) like this:  

z=\frac{0.18 -0.1}{\sqrt{\frac{0.1(1-0.1)}{100}}}=2.67  

4) Statistical decision  

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.  

The significance level assumed for this case is \alpha=0.05. The next step would be calculate the p value for this test.  

Since is a one right tailed test the p value would be:  

p_v =P(Z>2.67)=0.0037  

So the p value obtained was a very low value and using the significance level given \alpha=0.05 we have p_v so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance thetrue proportion is not significanlty higher than 0.1 or 10% .  

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