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amid [387]
3 years ago
8

Dan is a software salesman Y represent his total pay in dollars let X represent the number of copies English is fun he sells sup

pose that X and y are related by the equation 80x+2400=y
Mathematics
1 answer:
notka56 [123]3 years ago
8 0

Answer:

1. 80

2. 2400

Step-by-step explanation:

1. what is the change in Dan's total pay for each copy of math is fun he sells?

2. what is Dan's total pay if he doesn't sell any copies of math is fun?

Given

y=80x + 2400

When x=0

y=80(0)+2400

=0+2400

=2400

When x=1

y=80x + 2400

=80(1) + 2400

=80+2400

=2480

change in Dan's total pay for each copy of math is fun he sells is

= 2480 - 2400

=80

2. Dan's total pay if he doesn't sell any copies of math is fun?

y=80x+2400

When x=0

y=80(0)+2400

=0+2400

=2400

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exis [7]

Answer:

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Step-by-step explanation:Parameters given:

Area of the rectangular garden = 16.02m^2

Length of the rectangular garden = 4.5m

The Area of a rectangle can be calculated for using the formula below

Area (square meters) = length (meters) × width (meters)

Therefore, 16.02m^2 = 4.5m × width

Width = 16.02m^2 / 4.5m

Width = 3.56m

Therefore the width in meters of the garden is 3.56m

See the attachment below for an illustrative diagram

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3 years ago
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. Exam scores for a large introductory statistics class follow an approximate normal distribution with an average score of 56 an
Andru [333]

Answer:

0.1% probability that the average score of a random sample of 20 students exceeds 59.5.

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a random variable X, with mean \mu and standard deviation \sigma, a large sample size can be approximated to a normal distribution with mean \mu and standard deviation \frac{\sigma}{\sqrt{n}}

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.

In this problem, we have that:

\mu = 56, \sigma = 5, n = 20, s = \frac{5}{\sqrt{20}} = 1.12

What is the probability that the average score of a random sample of 20 students exceeds 59.5?

This is 1 subtracted by the pvalue of Z when X = 59.5. So

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

Z = \frac{59.5 - 56}{1.12}

Z = 3.1

Z = 3.1 has a pvalue of 0.9990.

So there is a 1-0.9990 = 0.001 = 0.1% probability that the average score of a random sample of 20 students exceeds 59.5.

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Answer:

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