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nevsk [136]
3 years ago
14

A zookeeper predicted that the weight of a newborn lion would be 2.8 pounds. when the zoo's lion gave birth, the newborn weighed

3.5 pounds. what is the zookeepers's percent error?
Mathematics
2 answers:
bekas [8.4K]3 years ago
7 0

Answer: 20%

Step-by-step explanation:

According to the  zookeeper ,

The predicted weight of the newborn lion = 2.8 pounds

The actual weight of the newborn lion = 3.5 pounds

Now , the zookeepers's percent error = \dfrac{|Predicted-actual|}{actual}\times100

=\dfrac{|2.8-3.5|}{3.5}\times100\%\\\\=\dfrac{|-0.7|}{3.5}\times100\%\\\\=\dfrac{0.7}{3.5}\times100\%\\\\=\dfrac{100}{5}\%=20\%

Hence, the zookeepers's percent error = 20%.

Sonja [21]3 years ago
5 0

Answer:

20%

Step-by-step explanation:2.8-3.5 then divide that by 3.5 and multiple the answer by 100 to get the percent


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3 years ago
If the population density for the city is 10,000 people per square mile, what is the population of the city?
ololo11 [35]

Answer:

10,000 * (square miles in the city)

Step-by-step explanation:

You've been given a population of people for every square mile (10,000) and you're trying to figure out how many people live in the city.

To convert people per a square mile into people, you'll need to know how many square miles you have. I assume you've been given this as the size of the city or how many square miles the city takes up. Just multiply the amount of people per a square mile by the amount of square miles in the city, and it'll give you the population of people in every square mile of the city (the population).

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6 0
3 years ago
According to a study conducted in one​ city, 3838​% of adults in the city have credit card debts of more than​ $2000. A simple r
Maksim231197 [3]

Answer:

The sampling distribution of the sample proportion of adults who have credit card debts of more than​ $2000 will be normally distributed with mean = 0.38 and standard deviation of 0.034.

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For a proportion p, the sampling distribution will be normally distributed with mean \mu = p and standard deviation s = \sqrt{\frac{p(1-p)}{n}}

In this problem, we have that:

n = 200, p = 0.38

So

s = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.38*0.62}{200}} = 0.034

The sampling distribution of the sample proportion of adults who have credit card debts of more than​ $2000 will be normally distributed with mean = 0.38 and standard deviation of 0.034.

5 0
3 years ago
What is 2 divided by 346
swat32
As this is a relatively complex division, it is worth doing this on the calculator.
2/346= 0.00578034682
Rounded to 4 decimal places, this is 0.0058
Hope this helps :) 
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3 years ago
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There are 2 14 year olds. The answer is (A).
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The average should be 12. 11 is one less than 12. You have four 11 year olds. You get a deficit of 1 with each 11 year old, so with four of them the deficit is 4. To compensate for this and being the average to 12, you need to make a surplus of 4. If you add one 14 year old, you make a surplus of 2 (because 14 is 2 more than 12). If you add another 14 year old, you now have a surplus of 4 and that balances it out.
***
if it's hard to understand tell me and I'll explain it in the simpler way
4 0
3 years ago
Read 2 more answers
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