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lesantik [10]
2 years ago
12

Lexie earned $64.38 for working from 8:00 AM to 4:15 PM. Within that time, she took a 1 hour lunch break for which she did not g

et paid. How much does Lexie earn per hour?
Mathematics
2 answers:
melamori03 [73]2 years ago
7 0
There are 8 hours and 15 minutes in her work day. If you take off one of those hours, it would be 7 hours and 15 minutes. If she is getting paid by the hour, you can take off the 15 minutes since it doesn't matter. Then, you divide her income ($64.38) by her hours (7). From this, you get about $9 (or if you wanna be specific, $9.20) per hour.
SCORPION-xisa [38]2 years ago
3 0
$8.88 per hour is the answer
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3 years ago
Why do you start with the greater number when adding on a number line
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3 years ago
This scatter plot shows the relationship between the number of sweatshirts sold and the temperature outside
Sphinxa [80]

Answer:

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6 0
2 years ago
Of the total population of the United States, 20% live in the northeast. If 200 residents of the United States are selected at r
Snezhnost [94]

Answer:

0.0465 = 4.65% probability that at least 50 live in the northeast.

Step-by-step explanation:

I am going to use the normal approximation to the binomial to solve this question.

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The expected value of the binomial distribution is:

E(X) = np

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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 = 200, p = 0.2

So

\mu = E(X) = np = 200*0.2 = 40

\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{200*0.2*0.8} = 5.65685

Approximate the probability that at least 50 live in the northeast.

Using continuity correction, this is P(X \geq 50 - 0.5) = P(X \geq 49.5, which is 1 subtracted by the pvalue of Z when X = 49.5. So

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

Z = \frac{49.5 - 40}{5.65685}

Z = 1.68

Z = 1.68 has a pvalue of 0.9535

1 - 0.9535 = 0.0465

0.0465 = 4.65% probability that at least 50 live in the northeast.

3 0
3 years ago
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