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densk [106]
3 years ago
5

A bag of potatoes weighs 7 and 1/2 pounds. Of the potatoes in the bag, 1/6 are rotten. What is the weight of the good potatoes?

Mathematics
1 answer:
Finger [1]3 years ago
3 0

Answer:


one sixth of 7.5 pounds is 1.25, so 7.5 minus the rotten 1.25 is 6.25.

Step-by-step explanation:


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Giovanna, Frances, and their dad each carried a 10 pound bag of soil. After putting soil into the first flower bed, Giovanna's b
KIM [24]
(5/8)*10lbs = 6.25 lbs
(2/5)*10lbs = 4 lbs
(3/4)*10lbs = 7.5 lbs
then add and convert to ounces.

ignore my reply it is wrong.
5 0
3 years ago
HELP ME PLEASE!!!!!!!
kotegsom [21]

Answer:

1. (0,0) (3,1) (6,2)

2. proportinal

3. im so sorry but i cant see that one i really am sorry!

Step-by-step explanation:

i hope i helped brainliest?

3 0
3 years ago
Angles U and V are supplementary angles. The ratio of their measures is 7:13. Find the measure of each angle
Advocard [28]
Supplementary=add to 180. U/V=7/13, U+V=180. Solving, U=63 and V=117
6 0
3 years ago
Community college students conduct a survey at their college. They ask "Do you plan to transfer to a university to pursue a bacc
Tamiku [17]

Answer:

0.8 - 2.58\sqrt{\frac{0.8(1-0.8)}{100}}=0.697

0.8 + 2.58\sqrt{\frac{0.8(1-0.8)}{100}}=0.903

The 99% confidence interval would be given by (0.697;0.903)

And the best option is : 0.697 to 0.903

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

Solution to the problem

For this case the conditions we have the conditions to assume a normal distribution for the population proportion since:

np=100*0.8=80 >10 , n(1-p) = 10*(1-0.8) =20 >10

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 99% of confidence, our significance level would be given by \alpha=1-0.99=0.01 and \alpha/2 =0.005. And the critical value would be given by:

z_{\alpha/2}=-2.58, z_{1-\alpha/2}=2.58

The confidence interval for the mean is given by the following formula:  

\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}

The estimated proportion is given by \hat p =\frac{80}{100} =0.8. If we replace the values obtained we got:

0.8 - 2.58\sqrt{\frac{0.8(1-0.8)}{100}}=0.697

0.8 + 2.58\sqrt{\frac{0.8(1-0.8)}{100}}=0.903

The 99% confidence interval would be given by (0.697;0.903)

And the best option is : 0.697 to 0.903

6 0
3 years ago
Jessica has one third of a bag of dog food to divide evenly between her 2 dogs what faction of one whole bag doesJessica give ea
Anna35 [415]
1/6
1/3 divided by 2 is 1/6

hope this helps :)

6 0
3 years ago
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