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rodikova [14]
4 years ago
15

Flies cannot see much farther than 24 to 36 inches away for their eyes, but that is not due to a lack of lenses-dragonflies have

thousands of lenses! In one eye, dragonflies have 2000 more lenses than seven times the number of lenses that houseflies have in one eye if one housefly and one dragonfly eye together have 34000 lenses, how many lenses do houseflies have in one eye? How many do dragonflies have in one eye
Mathematics
1 answer:
olga55 [171]4 years ago
6 0

Answer:

The number of lenses in one eye of the dragonfly is 30,000

Step-by-step explanation:

Before we calculate the quantity of lenses in one eye of the dragonfly, let us denote the number of lenses in one eye of the dragonfly by "x" and denote the quantity of lenses in one eye of the housefly by "y".

Now, since dragonflies in one eye have 2,000 more lenses than seven times the number that houseflies have, we can represent this mathematically:-

x = 2000 + [7×(y)]

x = 2000 + 7y

Again we learnt that the number of lenses in one eye of the dragonfly plus the number of lenses in the housefly are equal to thirty - four thousand lenses. mathematically:-

x + y = 34000

We now have a set of two equations which are:

2000 + 7y = x ------------ 1

x + y = 34000 ----------- 2

The resulting simultaneous equation can then be solved using elimination or substitution method. Let us use substitution method to solve solve for "y".

In equation 1, x = 2000 + 7y.

We can substitute "x" for 2000 + 7y in equation 2

2000 + 7y + y = 34,000

2000 + 8y = 34,000

8y = 34,000 - 2,000

8y = 32,000

y = 32000/8

y = 4,000 lenses. This is the number of lenses in one eye of the housefly.

Now, substitute y for 4,000 in equation 1 in order to find "x".

x = 2000 + 7(4000)

x = 2000 + 28000

x = 30000 lenses.

Therefore the number of lenses in one eye of the dragonfly is 30,000 lenses.

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Step-by-step explanation:

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Answer:

a) 0.68 - 2.58 \sqrt{\frac{0.68(1-0.68)}{575}}=0.630  

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And the 99% confidence interval would be given (0.630;0.730).  

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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".  

p represent the real population proportion of interest  

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n=575 is the sample size required (variable of interest)  

z represent the critical value for the margin of error  

The population proportion have the following distribution  

p \sim N(p,\sqrt{\frac{\hat p(1-\hat p)}{n}})  

Part a

The confidence interval would be given by this formula  

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

For the 99% confidence interval the value of \alpha=1-0.99=0.01 and \alpha/2=0.005, with that value we can find the quantile required for the interval in the normal standard distribution.  

z_{\alpha/2}=2.58  

And replacing into the confidence interval formula we got:  

0.68 - 2.58 \sqrt{\frac{0.68(1-0.68)}{575}}=0.630  

0.68 + 2.58 \sqrt{\frac{0.68(1-0.68)}{575}}=0.730  

And the 99% confidence interval would be given (0.630;0.730).  

Part b

We are 99% confident that the true proportion of infected chickens are between 0.63 (63%) and 0.73 (73%)

Part c

For this case the answer would be No. Since all the required assumptions in order to construct the confidence interval  are satisfied, so then the results can be assumed for all the population

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