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2 years ago

8

tickets to a production of a midsummer night's dream at broward college cost $5 for general admission or $4 with a student ID. I

f 184 people paid to see a performance, & $812 was collected, how many of wach type of ticket was sold?

1 answer:

Elena-2011 [213]2 years ago

7 0

Answer:

76 *5 = 380 $

184 - 76 = 108

108 * 4 = 432

432 + 380 = 812

Step-by-step explanation:

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Isosceles △ABC (AC=BC) is inscribed in the circle k(O). Prove that the tangent to the circle at point C is parallel to AB .

timofeeve [1]

Explanation:

Let M be the midpoint of AB. Then CM is the perpendicular bisector of AB. As such, center O is on CM, and OC is a radius (and CM). The tangent is perpendicular to that radius (and CM), so is parallel to AB, which is also perpendicular to CM.

If you need to go any further, you can show that triangles CMA and CMB are congruent, so (linear) angles CMA and CMB are congruent, hence both 90°.

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2 years ago

Korey’s comic book store has been up and running for 4 years. Korey feels that his store has been successful and is considering

Inessa05 [86]

Year Net Profit
1 <span>$14,250.00
2 $15,390.00
3 $16,621.20
4 $17,950.90</span>2

We need to get the increase of the net profit of the current year from the previous year.

Percentage increase = (Current year - Previous Year)/ Previous Year * 100%

Year 2: (15,390 - 14, 250) / 14,250 * 100% = 0.08 * 100% = 8%
Year 3: (16,621.20 - 15,390) / 15,390 * 100% = 0.08 * 100% = 8%
Year 4: (17,950.90 - 16,621.20) / 16,621.20 * 100% = 0.08 * 100% = 8%

Every year the net income increases by 8%. So, the net income in Year 5 will be:

17,950.90 x 1.08 = 19,386.97 Choice D.

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2 years ago

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The ratio of length to width in a rectangle is 3 to 1. If the perimeter of the rectangle is 128 feet, what is the length of the

Serga [27]

Answer: B

Step-by-step explanation:

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2 years ago

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Jeanne wants to start collecting coins and orders a coin collection starter kit. The kit comes with three coins chosen at random

lesya692 [45]

Conditional probability is a measure of the probability of an event given that another event has occurred. If the event of interest is A and the event B is known or assumed to have occurred, "the conditional probability of A given B", or "the probability of A under the condition B", is usually written as P(A|B), or sometimes $P_B(A)$ .

The conditional probability of event A happening, given that event B has happened, written as P(A|B) is given by
$P(A|B)= \frac{P(A \cap B)}{P(B)}$

In the question, we were told that there are three randomly selected coins which can be a nickel, a dime or a quarter.

The probability of selecting one coin is $\frac{1}{3}$

Part A:
To find <span>the probability that all three coins are quarters if the first two envelopes Jeanne opens each contain a quarter, let the event that all three coins are quarters be A and the event that the first two envelopes Jeanne opens each contain a quarter be B.

P(A) means that the first envelope contains a quarter AND the second envelope contains a quarter AND the third envelope contains a quarter.

Thus $P(A)= \frac{1}{3} \times \frac{1}{3} \times \frac{1}{3} = \frac{1}{27}$

</span><span>P(B) means that the first envelope contains a quarter AND the second envelope contains a quarter

</span><span>Thus $P(B)= \frac{1}{3} \times \frac{1}{3} = \frac{1}{9}$

Therefore, $P(A|B)=\left( \frac{ \frac{1}{27} }{ \frac{1}{9} } \right)= \frac{1}{3}$

Part B:
</span>To find the probability that all three coins are different if the first envelope Jeanne opens contains a dime<span>, let the event that all three coins are different be C and the event that the first envelope Jeanne opens contains a dime be D.
</span><span>
$P(C)= \frac{3}{3} \times \frac{2}{3} \times \frac{1}{3} = \frac{6}{27} = \frac{2}{9}$

</span><span> $P(D)= \frac{1}{3}$ </span><span>

Therefore, $P(C|D)=\left( \frac{ \frac{2}{9} }{ \frac{1}{3} } \right)= \frac{2}{3}$ </span>

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2 years ago

Jordan had the following scores on her math tests last quarter: 96, 89, 79, 85, 87, 94,

Lisa [10]

Answer:

a) 5.5

b) None

Step-by-step explanation:

The given data set is {96,89,79,85,87,94,96,98}

First we must find the mean.

$\bar X=\frac{96+89+79+85+87+94+96+98}{8}=\frac{724}{8}=90.5$

We now find the absolute value of the distance of each value from the mean.

This is called the absolute deviation

{ $|96-90.5|,|89-90.5|,|79-90.5|,|85-90.5|,|87-90.5|,|94-90.5|,|96-90.5|,|98-90.5|$ }

{ $5.5,1.5,11.5,5.5,3.5,3.5,5.5,7.5$ }

We now find the mean of the absolute deviations

$MAD=\frac{5.5+1.5+11.5+5.5+3.5+3.5+5.5+7.5}{8} =\frac{44}{8} =5.5$

The least absolute deviation is 1.5. This is not within one absolute deviation.

Therefore none of the data set is closer than one mean absolute deviation away from the mean.

5 0

2 years ago