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nydimaria [60]
3 years ago
7

Two kinds of tickets to an outdoor concert were sold: lawn tickets and seat tickets. Less than 400 tickets in total were sold. L

awn tickets cost $30 each and seat tickets cost $50 each. The organizers want to make at least $14,000 from ticket sales.
Write a system of inequalities to represent the situation. Be sure to define your variables.
Mathematics
1 answer:
miskamm [114]3 years ago
7 0

Answer:

Tickets to sit on the bench (b) = 150

Tickets to sit on the lawn (l) = 200

Step-by-step explanation:

Tickets to sit on a bench (b)cost $75 each.

Tickets to sit on the lawn (l) cost $40 each

350 tickets had been sold

$19,250 had been raised through tickets sales

This forms a simultaneous equation:

b + l = 350 ... (i)

75b + 40l = 19,250 ... (ii)

Multiplying (i) by 40 and (ii) by 1 we get;

40b + 40l = 14,000 ... (i)

75b + 40l = 19,250 ... (ii)

Subtracting (ii) - (i) we get;

35b = 5250

b = 5250 ÷ 35 = 150

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Step-by-step explanation: If the scale factor of a dilation is between 0 and 1, the image will be smaller than the object, a reduction. It would only be an enlargement if the scale factor is greater than 1.

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2 years ago
Consider the population of all 1-gallon cans of dusty rose paint manufactured by a particular paint company. Suppose that a norm
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Answer:

a) 0.5.

b) 0.8413

c) 0.8413

d) 0.6826

e) 0.9332

f) 1

Step-by-step explanation:

Problems of normally distributed samples are 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 = 6, \sigma = 0.2

(a) P(x > 6) =

This is 1 subtracted by the pvalue of Z when X = 6. So

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

Z = \frac{6-6}{0.2}

Z = 0

Z = 0 has a pvalue of 0.5.

1 - 0.5 = 0.5.

(b) P(x < 6.2)=

This is the pvalue of Z when X = 6.2. So

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

Z = \frac{6.2-6}{0.2}

Z = 1

Z = 1 has a pvalue of 0.8413

(c) P(x ≤ 6.2) =

In the normal distribution, the probability of an exact value, for example, P(X = 6.2), is always zero, which means that P(x ≤ 6.2) = P(x < 6.2) = 0.8413.

(d) P(5.8 < x < 6.2) =

This is the pvalue of Z when X = 6.2 subtracted by the pvalue of Z when X  5.8.

X = 6.2

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

Z = \frac{6.2-6}{0.2}

Z = 1

Z = 1 has a pvalue of 0.8413

X = 5.8

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

Z = \frac{5.8-6}{0.2}

Z = -1

Z = -1 has a pvalue of 0.1587

0.8413 - 0.1587 = 0.6826

(e) P(x > 5.7) =

This is 1 subtracted by the pvalue of Z when X = 5.7.

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

Z = \frac{5.8-6}{0.2}

Z = -1.5

Z = -1.5 has a pvalue of 0.0668

1 - 0.0668 = 0.9332

(f) P(x > 5) =

This is 1 subtracted by the pvalue of Z when X = 5.

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

Z = \frac{5-6}{0.2}

Z = -5

Z = -5 has a pvalue of 0.

1 - 0 = 1

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