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anyanavicka [17]
2 years ago
7

I need help ring the center, radius, and diameter.

Mathematics
1 answer:
faust18 [17]2 years ago
4 0

Answer:

(-2, 5), 6, 12

Step-by-step explanation:

Center: (-2, 5)

Radius: 6

Diameter: 12

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Find the odds for and the odds against the event rolling a fair die and getting a 4, a 2, or a 1.
HACTEHA [7]

Answer: Both in favor and against: 3/6 OR 1/2 OR 50%.

Step-by-step explanation: Both in favor and against: 3/6 OR 1/2 OR 50%.

The probability of rolling one die and getting any digit is \frac{1}{6}. That is because the desired outcome is only 1 digit out of the 6 total digits on each of the six sides.

That means the probability of rolling either a 1, a 2, or a 4 <em>if you are looking for any of those numbers specifically, </em>is \frac{1}{6}. But! You are looking to roll a 1, a 2, or a 4, without any preference among those three. Therefore, you are going to add the desired outcomes.

\frac{1}{6} + \frac{1}{6}  + \frac{1}{6} = \frac{3}{6} = \frac{1}{2} = 0.5 = 50%

Above, are added the probability of rolling a 1 PLUS the probability of rolling a 2 PLUS the probability of rolling a 4, yielding a 50% probability.

Think about it. There are six total digits on a die. You are okay with rolling three of those. Three is half of six. Therefore, the probability of rolling the number you'd like is one-half, fifty-percent.

Likewise, since the probability of all complement events happening is 1 (or 100%), The odds against rolling either 1, 2, or 4 is also fifty-percent. 100% (which is rolling <em>any </em>number on a die) MINUS 50% (which is rolling a 1, a 2, or a 4 with no preference) EQUALS TO 50% (which is rolling a 3, a 5, or a 6).

8 0
3 years ago
Which point is a solution to the inequality shown in this graph?
kykrilka [37]

Answer:

The only solution can be (0,-3) point.

Step-by-step explanation:

We have to judge whether the points in options are the solution to the graphed inequality or not.

The first point is (5,-5) which not included in the shaded region of the graph. Hence, it can not be a solution.

The second point is (6,0) which not included in the shaded region of the graph. Hence, it can not be a solution.

The third point is (0,-5) which not included in the shaded region of the graph. Hence, it can not be a solution.

The fourth point is (0,-3). It is on the firm red line which is included in the shaded region of the graph. Hence, it is a solution.

Therefore, the only solution can be (0,-3) point. (Answer)

5 0
2 years ago
Which of the following would best represent a cosine function with an amplitude of 3, a period of pi/2 , and a midline at y = –4
Grace [21]
To transform the function f(x)= cos(x) to have the amplitude of 3, we need to multiply the constant 3 to the function f(x), so we have y=3f(x)

To transform the function f(x)=cos(x) to have the midline y=-4 we need to subtract f(x) by 4, so we have y=f(x)-4, 

To transform the function f(x)=cos(x) to have period of \frac{ \pi }{2}, we need to divide the original period 2 \pi by 4, so we have y=f(4x). Note that it is the 4x gives the effect of dividing the points on x-axes by 4 and the period is read on x-axes

Hence, the full transformation is given y=3f(4x)-4 which is the last option
8 0
3 years ago
Read 2 more answers
Lucy places five cards that are labeled 1 to 6 face down on the table and mixes them up. What is the likelihood that her friend
melisa1 [442]

Answer:

1/2 or 0.5

Step-by-step explanation:

Within 1 to 6,

Even numbers are 2, 4, 6

While odd numbers are 1, 3, 5

Assuming cards are mixed up properly and the probabilities of drawing a card with any number from 1 to 6 is constant,

Probability of drawing even number cards = number of units of even number cards ÷ total number of units of cards from 1 to 6 aka total number of cards

= 3 / 6

= 1 / 2

4 0
3 years ago
Name the intersection of plane QMW and plane RMW
Norma-Jean [14]
The answer is "C", "MW".

In the given problem, the place QMW and plane RMW. These planes intersect at MW, in which intersection is either a point, line or curve that an entity or entities both possess or is in contact with  but if we see in Euclidean<span> geometry, the intersection of two planes is called a “line”. </span>In the plane we can understand that the common line for both plane QMW and plane RMW is MW.
6 0
3 years ago
Read 2 more answers
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