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Marianna [84]
3 years ago
14

Students who buy tickets at a local movie theater receive a 10% discount on the price of a ticket. what is the total price for t

he 4 students tickets if the regular cost per ticket is $6.00.
options- $21.60, $26.40, $6.60, or $5.40
Mathematics
2 answers:
Andreyy893 years ago
4 0

Answer:

$21.60

Step-by-step explanation:

marshall27 [118]3 years ago
4 0

Answer:

Answer is $ 21. 60

Step-by-step explanation:

$22 or 21.60 in decimal form

Step-by-step explanation: $6.00 - 10% is 5.4

5.4 + 5.4 +5.4+5.4 is 21.60

you can also round this off and get the answer as $ 22 .

Hope this answer helps you :)

Have a great day

Mark brainliest

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Ty's social security tax is
  6.2% * $3000 = $186.00

The best choice is ...
  A. $186.00
6 0
3 years ago
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Points (−3, 5) and (3, 5) on the coordinate grid below show the positions of two players on a tennis court:
Anna35 [415]

Answer:

4.  Player 2's position is Player 1's position reflected across the y-axis; only the signs of the x-coordinates of Player 1 and Player 2 are different.

Step-by-step explanation:

Player 1's position is (-3, 5).

It means that it is 3 units left from the origin and 5 units above the origin.

Player 2's position is (3, 5).

It means that it is 3 units right from the origin and 5 units above the origin.

Hence, the two points are on the same horizontal line bisected by the y-axis.

So, Player 2's position is Player 1's position reflected across the y-axis; only the signs of the x-coordinates of Player 1 and Player 2 are different.

3 0
3 years ago
A person's blood glucose level and diabetes are closely related. Let x be a random variable measured in milligrams of glucose pe
soldier1979 [14.2K]

Using the normal distribution, it is found that:

a) 0.8599 = 85.99% probability that x is more than 60.

b) 0.1788 = 17.88% probability that x is less than 110.

c) 0.6811 = 68.11% probability that x is between 60 and 110.

d) 0.0643 = 6.43% probability that x is greater than 125.

In a normal distribution with mean \mu and standard deviation \sigma, the z-score of a measure X is given by:

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

  • It 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, which is the percentile of X.

In this problem:

  • The mean is of 87, thus \mu = 87.
  • The standard deviation is of 25, thus \sigma = 25.

Item a:

This probability is <u>1 subtracted by the p-value of Z when X = 60</u>, thus:

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

Z = \frac{60 - 87}{25}

Z = -1.08

Z = -1.08 has a p-value of 0.1401.

1 - 0.1401 = 0.8599

0.8599 = 85.99% probability that x is more than 60.

Item b:

This probability is the <u>p-value of Z when X = 110</u>, thus:

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

Z = \frac{110 - 87}{25}

Z = 0.92

Z = 0.92 has a p-value of 0.8212.

1 - 0.8212 = 0.1788.

0.1788 = 17.88% probability that x is less than 110.

Item c:

This probability is the <u>p-value of Z when X = 110 subtracted by the p-value of Z when X = 60</u>.

From the previous two items, 0.8212 - 0.1401 = 0.6811.

0.6811 = 68.11% probability that x is between 60 and 110.

Item d:

This probability is <u>1 subtracted by the p-value of Z when X = 125</u>, thus:

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

Z = \frac{125 - 87}{25}

Z = 1.52

Z = 1.52 has a p-value of 0.9357.

1 - 0.9357 = 0.0643.

0.0643 = 6.43% probability that x is greater than 125.

A similar problem is given at brainly.com/question/24863330

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Answer is b hope this helps
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The length of a rectangular poster is 9 more inches than two times its width. The area of the poster is 161 square inches. Solve
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Answer:

<u>Width = 7 inches</u>

<u>Length = 23 inches</u>

Step-by-step explanation:

Let :

  • length = 2x + 9
  • width = x

Equating to area :

  • (2x + 9)(x) = 161
  • 2x² + 9x = 161
  • 2x² + 9x - 161 = 0
  • 2x² - 14x + 23x - 161 = 0
  • 2x(x - 7) + 23(x - 7) = 0
  • x - 7 = 0
  • x = 7

Dimensions are :

  • <u>Width = 7 inches</u>
  • Length = 2(7) + 9
  • <u>Length = 23 inches</u>
8 0
2 years ago
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