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Kamila [148]
3 years ago
11

Atile installer charges 515 per soruare foot to install floor tile What is the cost to install tile in a room of the size shown?

Mathematics
1 answer:
Temka [501]3 years ago
4 0

Answer:

it gayy

Step-by-step explanation:

that the answer

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What’s the horizontal line and vertical line?-use X=blank and Y=blank.
vivado [14]

Answer:

Part 1) Vertical line : x=1

Part 2) Horizontal line :  y=-4

Step-by-step explanation:

Part 1) Write the equation for the vertical line passing through the point (1,-4)

we know that

The equation of a vertical line (parallel to the y-axis) is equal to the x-coordinate of the point that passes through it

so

The x-coordinate is 1

therefore

The equation of the line is

x=1

Part 2) Write the equation for the horizontal line passing through the point (1,-4)

we know that

The equation of a horizontal line (parallel to the x-axis) is equal to the y-coordinate of the point that passes through it

so

The y-coordinate is -4

therefore

The equation of the line is

y=-4

8 0
4 years ago
4(-8x + 5) - (-33x - 26) =
rosijanka [135]
<h2>4(-8x + 5) - (-33x - 26) =</h2><h2>-32x + 20 + 33x + 26=                      (distributive)</h2><h2>-32x  + 33x + 26 + 20=                     (combine like terms)</h2><h2>  x + 46</h2>

Hope it helps.

8 0
3 years ago
Here are the scores that Jenna has received on her seven quizzes what is the upper quartile
Ivenika [448]
The upper quartile is 18
3 0
4 years ago
A survey of athletes at a high school is conducted, and the following facts are discovered: 13% of the athletes are football pla
nekit [7.7K]

Answer:

The probability that an athlete chosen is either a football player or a basketball player is 56%.

Step-by-step explanation:

Let the athletes which are Football player be 'A'

Let the athletes which are Basket ball player be 'B'

Given:

Football players (A) = 13%

Basketball players (B) = 52%

Both football and basket ball players = 9%

We need to find probability that an athlete chosen is either a football player or a basketball player.

Solution:

The probability that athlete is a football player = P(A)= \frac{13}{100}=0.13

The probability that athlete is a basketball player = P(B)= \frac{52}{100}=0.52

The probability that athlete is both basket ball player and  football player = P(A\cap B) = \frac{9}{100}=0.09

We have to find the probability that an athlete chosen is either a football player or a basketball player P(A\cup B).

Now we know that;

P(A\cup B)= P(A) + P(B) - P(A \cap B)\\\\P(A\cup B) = 0.13+0.52-0.09=0.56\\\\P(A\cup B) = \frac{0.56}{100}=56\%

Hence The probability that an athlete chosen is either a football player or a basketball player is 56%.

5 0
4 years ago
A classic counting problem is to determine the number of different ways that the letters of "generally""generally" can be arrang
soldier1979 [14.2K]
The word generally can be arranged more than 5 times.
6 0
3 years ago
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