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Lesechka [4]
3 years ago
13

Brooke bought 5 plates for $44. If the price of the plates is always the same, which represents the total cost, y, of buying any

number of plates, x? 5 y = 44 y = 8.8 x On a coordinate plane, the x-axis is labeled number of plates and the y-axis is labeled total cost (in dollars). A line goes through points (2, 30) and (4, 40). On a coordinate plane, the x-axis is labeled number of plates and the y-axis is labeled total cost (in dollars). A line goes through points (10, 1) and (60, 7).

Mathematics
1 answer:
Illusion [34]3 years ago
6 0

Answer:

B

Step-by-step explanation:

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A circle's diameter is 4 feet.
Soloha48 [4]

Answer:

13

Step-by-step explanation:

The formula for cirmference is

C= π2 * r

we're using 3.14 as pi and r = radius.

the radius is always half of the diameter.

So in this case since the diamter is 4, the radius is 2. let's substitute these numbers.

C= 3.14 *2*2 and there your answer is 12.56 which rounded to the nearest tenth is 13.

Hope this helps :)

6 0
2 years ago
A class had a plate with 128 Cookies. The cookies where divided evenly among the students. Each student was given 4 cookies. How
Mariulka [41]

Answer: 32

Step-by-step explanation:

128 divided by 4 is 32 so u get 32 and 4 and multiply it and you get 32.

8 0
3 years ago
The probability that a professor arrives on time is 0.8 and the probability that a student arrives on time is 0.6. Assuming thes
saul85 [17]

Answer:

a)0.08  , b)0.4  , C) i)0.84  , ii)0.56

Step-by-step explanation:

Given data

P(A) =  professor arrives on time

P(A) = 0.8

P(B) =  Student aarive on time

P(B) = 0.6

According to the question A & B are Independent  

P(A∩B) = P(A) . P(B)

Therefore  

{A}' & {B}' is also independent

{A}' = 1-0.8 = 0.2

{B}' = 1-0.6 = 0.4

part a)

Probability of both student and the professor are late

P(A'∩B') = P(A') . P(B')  (only for independent cases)

= 0.2 x 0.4

= 0.08

Part b)

The probability that the student is late given that the professor is on time

P(\frac{B'}{A}) = \frac{P(B'\cap A)}{P(A)} = \frac{0.4\times 0.8}{0.8} = 0.4

Part c)

Assume the events are not independent

Given Data

P(\frac{{A}'}{{B}'}) = 0.4

=\frac{P({A}'\cap {B}')}{P({B}')} = 0.4

P({A}'\cap {B}') = 0.4 x P({B}')

= 0.4 x 0.4 = 0.16

P({A}'\cap {B}') = 0.16

i)

The probability that at least one of them is on time

P(A\cup B) = 1- P({A}'\cap {B}')  

=  1 - 0.16 = 0.84

ii)The probability that they are both on time

P(A\cap  B) = 1 - P({A}'\cup {B}') = 1 - [P({A}')+P({B}') - P({A}'\cap {B}')]

= 1 - [0.2+0.4-0.16] = 1-0.44 = 0.56

6 0
3 years ago
PLS HELP!! need the answer asap
julsineya [31]
The answer would be C) P,D and A
4 0
3 years ago
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(1, 1), (–1,5) <br> find slope
ziro4ka [17]

Answer:

<h2>slope = 0</h2>

<h2>MARK ME AS A BRAINLIST AND FOLLOW ME </h2>
7 0
2 years ago
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