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tankabanditka [31]
3 years ago
6

What is the value of lim 2x+3/4x-5 A. 3/5 B. 1 C. 1/2 D. 0

Mathematics
1 answer:
Reptile [31]3 years ago
5 0

First of all, thank you for sharing the illustration of this problem. Without the illustration it would be hard to come up with an answer, since there are various kinds of limit and your own, typewritten instructions did not specify which.

Here you're finding the limit as x approaches infinity. Both 2x and 4x (as shown in the illustration grow larger continually and without bound, as x increases. As this happens, the other terms (+3 and -5) are simply overpowered by the x terms. If you choose to ignore these other terms for this reason, then your expression will be

2x

----- and this approaches the value 1/2 as x grows increasingly large in the

4x original expression.

So the limit of that expression, as x grows large without bound, is 1/2.

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Help for both a and b
kipiarov [429]

Hi!

Remember that an x-intercept is a point in which the line touches the x-axis (the horizontal line). And, the y-intercept is a point in which the line touches the y-axis (the vertical/up and down line)

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For A)

The coordinate of the y-intercept is (0,1)

The coordinate of the x-intercept is (3,0)

For B)

The coordinate of the y-intercept is (0,0)

The coordinate of the x-intercept is (0,0) !

*both the x and y axis meet at the origin. So, a line that goes through the origin (0,0) is intersecting with BOTH the x and y-axis.

Hope I helped! Comment if you have any questions or concerns.

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Choice the correct answer
PSYCHO15rus [73]

Answer:

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Step-by-step explanation:

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A machine that produces a major part for an airplane engine is monitored closely. In the past, 10% of the parts produced would b
nata0808 [166]

Answer:

With a .95 probability, the sample size that needs to be taken if the desired margin of error is .04 or less is of at least 216.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of \pi, and a confidence level of 1-\alpha, we have the following confidence interval of proportions.

\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}

In which

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The margin of error:

M = z\sqrt{\frac{\pi(1-\pi)}{n}}

For this problem, we have that:

p = 0.1

95% confidence level

So \alpha = 0.05, z is the value of Z that has a pvalue of 1 - \frac{0.05}{2} = 0.975, so Z = 1.96.

With a .95 probability, the sample size that needs to be taken if the desired margin of error is .04 or less is

We need a sample size of at least n, in which n is found M = 0.04.

M = z\sqrt{\frac{\pi(1-\pi)}{n}}

0.04 = 1.96\sqrt{\frac{0.1*0.9}{n}}

0.04\sqrt{n} = 0.588

\sqrt{n} = \frac{0.588}{0.04}

\sqrt{n} = 14.7

(\sqrt{n})^{2} = (14.7)^{2}

n = 216

With a .95 probability, the sample size that needs to be taken if the desired margin of error is .04 or less is of at least 216.

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Step-by-step explanation:

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