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Degger [83]
3 years ago
8

As compared to its first year of operation, ABC Company grew 8% in the second year and an additional 2% per year for the next tw

o years. If total growth for years 2-4 was $45,000 over the first year’s sales of $150,000, how much did ABC Company grow in year 2?
Mathematics
1 answer:
QveST [7]3 years ago
4 0
Honestly, most of the information here is excess, and its only purpose is to confuse you. So, lets take a look at the important information.
What we're trying to find is the growth in the second year. We're told that the company grows 8% on the second year, and the starting sale on the first year was 150,000. Really, thats all you need to know!

Since you know the starting price (150,000) and the growth percentage (8%), all you really have to do is multiply the starting price with the percentage.

So that would be 150,000 *.08, which is 12000. Therefore, the growth in the second year was 12,000

Answer: $12,000

Hope this helped ^-^
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2 1/16, 10 11/16, 3 15/16

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Write an inequality for the following.<br> Jackie spent no more than $40 on a video
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x ≥ $40

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3 years ago
P(X&lt; ) 1-P(X&gt; ) A softball pitcher has a 0.626 probability of throwing a strike for each curve ball pitch. If the softball
Mashutka [201]

Answer:

19.49% probability that no more than 16 of them are strikes

Step-by-step explanation:

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

E(X) = np

The standard deviation of the binomial distribution is:

\sqrt{V(X)} = \sqrt{np(1-p)}

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that \mu = E(X), \sigma = \sqrt{V(X)}.

In this problem, we have that:

n = 30, p = 0.626

So

\mu = E(X) = np = 30*0.626 = 18.78

\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{30*0.626*0.374} = 2.65

What is the probability that no more than 16 of them are strikes?

Using continuity correction, this is P(X \leq 16 + 0.5) = P(X \leq 16.5), which is the pvalue of Z when X = 16.5. So

Z = \frac{X - \mu}{\sigma}

Z = \frac{16.5 - 18.78}{2.65}

Z = -0.86

Z = -0.86 has a pvalue of 0.1949

19.49% probability that no more than 16 of them are strikes

7 0
3 years ago
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