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Lunna [17]
3 years ago
15

Anyone please help!!!

Mathematics
2 answers:
kvasek [131]3 years ago
8 0

The guy above me is correct but for the margin of error, I got 34%, 58% for the 90% interval and 32%, 60% for 95% interval. He just didnt take his margin of errors numbers and times them by 100 to get a percentage, and then round it so you get 12% for 90% and 14% for 95%. Then add and minus 12 percent to the 46% and thats how you get it. Then do the same with 95%

Tanya [424]3 years ago
7 0

Problem 1

Answers:

Percentage of patients that were dogs = 46%

Standard Error = 0.07048404074682

Margin of error for 90% confidence interval = 0.11594624702851

Margin of error for 95% confidence interval = 0.13814871986377

Round the decimal values however you need them

------------

To get the first answer, you add up the numbers given (7,4,5,5,2) and divide that over 50. So 7+4+5+5+2 = 23 which leads to 23/50 = 0.46 = 46%; therefore phat = 0.46 is the sample proportion of dogs.

Use the SE (standard error) formula given to you with phat = 0.46 and n = 50 to get SE = 0.07048404074682

The critical z value at 90% confidence is 1.645; this value is found in your Z table (back of your stats textbook). Multiply the SE value by 1.645 to get 0.07048404074682*1.645 = 0.11594624702851

Also found in your textbook is 1.960 which is the z critical value at 95% confidence. Multiply this with the SE value to get 0.07048404074682*1.960 = 0.13814871986377

===============================================

Problem 2

Answer: Choice B) picking balls from a bin; the 60 randomly selected get chosen for the first bus, while the remaining 60 go to the second bus

-----------

Choice A is fairly vague on what the lower and upper boundaries are. What is the smallest number allowed? What about the largest? This isn't clear so it's possible that we could end up with more positive numbers than negative (eg: if we had an interval -10 < x < 110). So choice A is false. A similar issue shows up with choice D.

Choice B is true. Assuming the selection process is random and not biased, then each ball is equally likely for each selection. The fact that the balls are colored seems to be extra info which I'm not sure why your teacher threw that in there.

Choice C is false because choice B is true

Choice D is false for similar reasons as choice A. It's not clear where we start and where we end. If we had the interval 2 < x < 6 then x could take on the values {3, 4, 5} and we see that picking an odd number is twice as likely than picking an even. In this example, there is bias.

===============================================

Problem 3

Answer: Choice B) Roll a die; each number corresponds to a different class

------------

Choice A is false because choice B being the answer contradicts it

Choice B is true: there are 6 sides on the die, and each side is equally likely to be landed on, so each class is equally likely

Choice C is false because 2*2*2 = 8 represents the number of combos you can have when you flip three coins (one combo being HTH for heads tail heads) but there are 6 classes, not 8

Choice D is false because while we want 6 regions on the spinner. Each region must have the same area; otherwise, one class is weighted heavier than the others making it more likely you select that particular class.


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"More" means addition.

2n + 3

Hope this helps you!

7 0
3 years ago
Ten pounds of mixed birdseed sells for 8.24 per pound. The mixture is obtained from two kinds of birdseed, with one variety pric
Kipish [7]

Answer:

2 pounds of 1st kind and 8 pounds of 2nd kind are mixed.

Explanation:

Given:

Ten pounds of mixed birdseed sells for 8.24 per pound.  

Price of the 1st variety of birdseed=5.68 per pound

Price of the 2nd variety of the birdseed=8.88 per pound.  

To find :

Pounds of 1st variety of birdseed are used in the mixture=?

Pounds of 2ndvariety of birdseed are used in the mixture=?

Solution:

Let the weight of 1st kind added in mixture be n pounds.

Then weight of 2nd kind will be 10 – n pounds.

Now, total cost of mixture = cost of 1st kind birdseed + cost of 2nd kind birdseed

Weight of mixture x cost per pound = weight of 1st kind x cost per pound + weight of 2nd kind x cost per pound.

Substituting the values,

10\times8.24 = n\times5.68 + (10-n)\times8.88

82.4 = 5.68n + 88.8 – 8.88n  

8.88n – 5.68n = 88.8 – 82.4  

3.2n = 6.4  

n = \frac{3.2}{6.4}

n = 2

Then, weight of 2nd kind is 10 – 2 = 8 pounds.

Hence, 2 pounds of 1st kind and 8 pounds of 2nd kind are mixed.

5 0
3 years ago
What is the percent of decrease from 25 to 15? Round your answer to the nearest tenth.
Dahasolnce [82]

Answer: 60%

Step-by-step explanation:

25=100%

50%=12.5

10%=2.5

12.5(50%) +2.5(10%)= 50%(15)

so the answer is 60%

7 0
2 years ago
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Norma-Jean [14]

Answer:

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Step-by-step explanation:

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4 years ago
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sasho [114]

Answer:

Firstly, rewrite the equation:

⅓ (18 + 27) = 81

Substitute x for the given number of it's supposed equivalent.

In this case x = 12.

⅓ (18(12) + 27) = 81

Solve using PEMDAS and simplify what is in the parenthesis first. Then, multiply.

(18 x 12) + 27 = 243

Now, solving using PEMDAS, multiply the total of what you got that was originally in the parenthesis by ⅓ .

⅓ (243) = 81

When you multiple these number they are equivalent to 81.

81 = 81

Since the equation given, when substituted x for 12, is equivalent to 81, this proves that substituting x for 12 makes this equation true.

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3 years ago
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