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konstantin123 [22]
3 years ago
7

FAST ASAP!! WILL GIVE BRAINLIST ALOT OF POINTS

Mathematics
1 answer:
Alik [6]3 years ago
3 0

Answer:

= 3 3/4 m^2

or 15/4 m^2

Step-by-step explanation:

The base is the triangle.  

A =1/2 bh

A = 1/2  (3 1/3 * 9/4)

Change 3 1/3 to an improper fraction (3*3+1)/3 = 10/3

A = 1/2 * 10/3 *9/4

Re arranging

   =  1/2* 9/3 * 10/4

 = 1/2 (3) 5/2

 =15/4

Now writing as a mixed number

= 3 3/4 m^2

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Pepsi [2]
Given that in 2008 the Better Business Bureau settled 75% of complaints they received (USA Today, March 2, 2009). Suppose you have been hired by the Better Business Bureau to investigate the complaints they received this year involving new car dealers. You plan to select a sample of new car dealer complaints to estimate the proportion of complaints the Better Business Bureau is able to settle. Assume the population proportion of complaints settled for new car dealers is .75, the same as the overall proportion of complaints settled in 2008.


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Suppose you select a sample of 450 complaints involving new car dealers. Show the sampling distribution of (to 4 decimals).

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Part b) :

Based upon a sample of 450 complaints, what is the probability that the sample <span>proportion will be within .04 of the population proportion (to 4 decimals)?

The probability </span><span>that the sample proportion will be within 0.04 of the population proportion</span> is given by:

P(p\ \textless \ 0.75\pm0.04)=2P\left(z\ \textless \  \frac{0.04}{0.0204} \right)-1 \\  \\ 2P(z\ \textless \ 1.961)-1=2(0.97505)-1=1.9501-1 \\  \\ =\bold{0.9501}

Therefore, the probability that the sample <span>proportion will be within .04 of the population proportion is 0.9501.



Part C:

</span><span>Suppose you select a sample of 200 complaints involving new car dealers. Show the sampling distribution of (to 4 decimals).

</span><span>Population proportion = 0.75
sample size. n = 200
Standard\ error=\sqrt{\frac{p(1-p)}{n}}  \\  \\ =\sqrt{\frac{(0.75)(0.25)}{200}} = 0.0306

Therefore, the sampling distribution has a proportion of 0.75 and a standard error of 0.0306.



Part D):

Based upon the smaller sample of only 200 complaints, what is the probability that the sample proportion will be within .04 of the population proportion (to 4 decimals)?

</span><span><span>
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P(p\&#10; \textless \ 0.75\pm0.04)=2P\left(z\ \textless \  \frac{0.04}{0.0306} &#10;\right)-1 \\  \\ 2P(z\ \textless \ 1.306)-1=2(0.90429)-1=1.80858-1 \\  \\&#10; =\bold{0.8086}

Therefore, the probability that the sample <span>proportion will be within .04 of the population proportion is 0.8086.</span></span>
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Answer:

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