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Umnica [9.8K]
2 years ago
13

What is the slope of a line that is parallel to the line whose equation is y=4/5x-3

Mathematics
1 answer:
PolarNik [594]2 years ago
3 0

Answer: 4/5

Step-by-step explanation: For a line to be parallel to another line there must be two conditions true.

Condition 1: Both lines must have the same slope (i.e the same x in the form y = mx + b)

Condition 2: Both lines must have different y-intercepts (i.e different b in the form y = mx + b)

Thus, if you are looking for the slope of the line parallel to y = 4/5x - 3, we can look at condition 1, and see that the slope should also be 4/5 for the other line.

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A cooler has 12 apple juices and 15 grape juices. 9 of the apple juices are sugar free and 5 of the grape juices are sugar free.
Romashka-Z-Leto [24]

Answer:

<em>The probability that it is grape juice, given that the juice is not sugar free is </em><em>0.769 or 76.9%</em>

Step-by-step explanation:

P(\text{Grape }|\text{ Not sugar free})=\dfrac{P(\text{Grape }\cap \text{ Not sugar free})}{P(\text{Not sugar free})}

A cooler has 12 apple juices and 15 grape juices. So total 27 juices are there.

Out of 12 apple juices, 9 of the apple juices are sugar free, so 3 are not sugar free.

Out of 15 grape juices, 5 of the grape juices are sugar free, so 10 are not sugar free. So,

P(\text{Grape }\cap \text{ Not sugar free})=\dfrac{10}{27}

P(\text{Not sugar free})=\dfrac{10+3}{27}=\dfrac{13}{27}

Putting the values,

P(\text{Grape }|\text{ Not sugar free})=\dfrac{P(\text{Grape }\cap \text{ Not sugar free})}{P(\text{Not sugar free})}=\dfrac{\frac{10}{27}}{\frac{13}{27}}=\dfrac{10}{13}=0.769

5 0
4 years ago
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Maksim231197 [3]

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5 0
3 years ago
What is the equation for the line that passes through (3, 2) and has a slope<br> equal to 1/3
lana66690 [7]

Answer:

y=1/3x+1

Step-by-step explanation:

6 0
3 years ago
HELP ASAP!
zaharov [31]

Answer:

Question 1:   $115

Question 2:  40 + 25x

Step-by-step explanation:

Question 1: 40 + 25*3 = $115

6 0
3 years ago
A cylindrical beaker with diameter 10 in and height 12 in. is filled with water that is then poured into a rectangular
emmasim [6.3K]

Answer:

A. The volume of the cylindrical beaker = 942.5 in3.

B. Volume of the pan = 378 in3

C. The water in the cylindrical beaker will over flow the pan.

D. The height of the water in the beaker after the pan is filled is 7.2 in

Step-by-step explanation:

Data given from the question.

Diameter = 10 in

Height = 12 in

Dimension of the pan = 14 in x 9 in x 3 in

Determination of the volume of each solid.

A. Volume of the cylindrical beaker:

Volume = πr2h

Radius (r) = 10/2 = 5 in

Height (h) = 12 in

Volume (V) =?

V = πr2h

V = π x (5)^2 x 12

V = 942.5 in3.

The volume of the cylindrical beaker = 942.5 in3.

B. Volume of the rectangular pan

Volume = Length x Width x Height

The dimension for the rectangular pan = volume of the pan = 14 in x 9 in x 3 in

Volume of the pan = 378 in3

C. Since the volume of the cylindrical beaker is higher than that of the pan, the water in the cylindrical beaker will overflow the pan.

D. Determination of the height of the water in the cylindrical beaker after the pan is filled. This is illustrated below.

Let us calculate the volume of the water in the cylindrical beaker after the pan is filled

The volume of the cylindrical beaker = 942.5 in3.

Volume of the pan = 378 in3

Volume of water in the cylinder beaker after the pan is filled = 942.5 - 378 = 564.5 in3

With this new volume, we can calculate the new height of the water as follow:

Volume = 564.5 in3

Radius (r) = 5 in (the radius remains the same)

Height (h) =?

V = πr2h

h = V/πr2

h = 564.5/π(5)^2

h = 7.2 in

Therefore, the height of the water in the beaker after the pan is filled is 7.2 in

4 0
3 years ago
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