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Murljashka [212]
3 years ago
12

Well buying organic is expensive because the product is much much harder to grow. No pesticides or fertilizer means there's a ve

ry good chance of an unsuccessful harvenst
Mathematics
1 answer:
kolezko [41]3 years ago
6 0
That is true. It is much cheaper and economically safer for farms to not grow organic foods. I’m fact, sometimes it is even better for the environment for food to not be grown organically as it puts less stress on the farmland.
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22 x 20 is 440 ,

8x16 is 128 ,

440 - 128 is 312

Step-by-step explanation:

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|x/4|=12<br><br> solve absolute value equation
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Question on image down below
denpristay [2]

Answer:

The temperature at sunrise was lower on the day in 2012 than on the day in 2013.

The difference between the 2012 and 2013 rates of temperature increase was 0.1 degrees per hour.

Both days had a constant rate of change in temperature per hour over the time periods shown.

Step-by-step explanation:

I don't know of that's right but . . .

5 0
3 years ago
A nationwide study of American homeowners revealed that 65% have one or more lawn mowers. A lawnequipment manufacturer, located
GenaCL600 [577]

Answer:

z=\frac{0.684 -0.65}{\sqrt{\frac{0.65(1-0.65)}{497}}}=1.589  

p_v =P(z>1.589)=0.056  

If we compare the p value obtained with the significance level assumed \alpha=0.05 we see that p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the proportion of homes in Omaha with one or more lawn mowers is not ignificantly higher than 0.65

Step-by-step explanation:

Data given and notation

n=497 represent the random sample taken

X=340 represent the homes in Omaha with one or more lawn mowers

\hat p=\frac{340}{497}=0.684 estimated proportion of homes in Omaha with one or more lawn mowers

p_o=0.65 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)  

Concepts and formulas to use  

We need to conduct a hypothesis in order to test the true proportion of homes in Omaha with one or more lawn mowers is higher than 0.65.:  

Null hypothesis:p\leq 0.65  

Alternative hypothesis:p > 0.65  

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{0.684 -0.65}{\sqrt{\frac{0.65(1-0.65)}{497}}}=1.589  

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 next step would be calculate the p value for this test.  

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

p_v =P(z>1.589)=0.056  

If we compare the p value obtained with the significance level assumed \alpha=0.05 we see that p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the proportion of homes in Omaha with one or more lawn mowers is not ignificantly higher than 0.65

7 0
3 years ago
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