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Mrac [35]
4 years ago
15

Find the unit rate. A 6oz jar of jalapeños cost $2.34

Mathematics
2 answers:
saveliy_v [14]4 years ago
8 0

Answer:

$0.39 per oz

Step-by-step explanation:

\frac{6}{2.34} =\frac{1}{x} \\\\6x=2.34\\\\x=\frac{2.34}{6} \\\\x=0.39

Levart [38]4 years ago
5 0

Answer:

1oz cost $0.39 which will be the unite rate

Step-by-step explanation:

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Due to the increase of posts on social media, the number of flip cell phones is in exponential decline. Data collected from 2009
gizmo_the_mogwai [7]

Answer:

y=256e^{-0.288x}

where y is in thousands

Step-by-step explanation:

The equation in two variables that can be used to represent the situation given can be determined using computer software such Stat-Crunch or Ms. Excel or advanced calculators.

I shall use Ms. Excel for this question.

In Ms. Excel, enter the data into any two adjacent columns. Next, highlight the data, then click the insert ribbon and select the scatter-plot option.

Excel returns a scatter-plot chart as shown in the attachment below.

After obtaining the scatter-plot, we shall need to add a trend line in order to determine the mathematical equation that best fits the scenario given.

Click anywhere inside the chart, then select the design tab under chart tools. Click on the Add Chart element in the upper left corner of the excel workbook and select more trend-line options. This feature will enable us to fit any trend-line to our data.

Select the exponential option, since we were informed that he number of flip cell phones is in exponential decline,ensuring you check the box; Display Equation on chart.

Find the attached.

The equation in two variables is thus;

y=256e^{-0.288x}

5 0
4 years ago
Historically, the proportion of people who trade in their old car to a car dealer when purchasing a new car is 48%. Over the pre
choli [55]

Answer:

z=\frac{0.4 -0.48}{\sqrt{\frac{0.48(1-0.48)}{115}}}=-1.717  

p_v =P(z  

So the p value obtained was a very low value and using the significance level given \alpha=0.1 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 the proportion of people that have traded in their old car is lower than 0.48 or 48%.  

Step-by-step explanation:

Data given and notation

n=115 represent the random sample taken

X=46 represent the number of people that have traded in their old car.

\hat p=\frac{46}{115}=0.4 estimated proportion of people that have traded in their old car

p_o=0.48 is the value that we want to test

\alpha=0.1 represent the significance level

Confidence=90% or 0.9

z would represent the statistic (variable of interest)

p_v represent the p value (variable of interest)  

Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that the proportion is less than 0.48.:  

Null hypothesis:p\geq 0.48  

Alternative hypothesis:p < 0.48  

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.

Calculate the statistic  

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

z=\frac{0.4 -0.48}{\sqrt{\frac{0.48(1-0.48)}{115}}}=-1.717  

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 provided \alpha=0.1. The next step would be calculate the p value for this test.  

Since is a left tailed test the p value would be:  

p_v =P(z  

So the p value obtained was a very low value and using the significance level given \alpha=0.1 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 the proportion of people that have traded in their old car is lower than 0.48 or 48%.  

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