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3 years ago

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The spinner has eight spaces of equal size. What is the mathematical probability of the pointer's stopping in the space marked 1

? If you spin the pointer 40 times, how many times would you expect it to stop on 1?

2 answers:

Advocard [28]3 years ago

6 0

Answer:

10/40

Step-by-step explanation:

serg [7]3 years ago

6 0

Answer:

The probability that the pointer stops on 1 is 1/8, since there are 8 sections.

If you spun it 40 times, you would expect it to land on 1 5 times.

Step-by-step explanation:

There are equal pieces, so the probability of landing on each is the same : 1/8

To find the amount of times you would expect it to land on 1 if you spun it 40 times, simply multiply the probability by 40. 40 * 1/8 = 5.

Hope this helps :)

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Given the center of the circle (-3,4) and a point on the circle (-6,2), (10,4) is on the circle

Anastasy [175]

Answer:

Part 1) False

Part 2) False

Step-by-step explanation:

we know that

The equation of the circle in standard form is equal to

$(x-h)^{2} +(y-k)^{2}=r^{2}$

where

(h,k) is the center and r is the radius

In this problem the distance between the center and a point on the circle is equal to the radius

The formula to calculate the distance between two points is equal to

$d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}$

Part 1) given the center of the circle (-3,4) and a point on the circle (-6,2), (10,4) is on the circle.

true or false

substitute the center of the circle in the equation in standard form

$(x+3)^{2} +(y-4)^{2}=r^{2}$

Find the distance (radius) between the center (-3,4) and (-6,2)

substitute in the formula of distance

$r=\sqrt{(2-4)^{2}+(-6+3)^{2}}$

$r=\sqrt{(-2)^{2}+(-3)^{2}}$

$r=\sqrt{13}\ units$

The equation of the circle is equal to

$(x+3)^{2} +(y-4)^{2}=(\sqrt{13}){2}$

$(x+3)^{2} +(y-4)^{2}=13$

Verify if the point (10,4) is on the circle

we know that

If a ordered pair is on the circle, then the ordered pair must satisfy the equation of the circle

For x=10,y=4

substitute

$(10+3)^{2} +(4-4)^{2}=13$

$(13)^{2} +(0)^{2}=13$

$169=13$ -----> is not true

therefore

The point is not on the circle

The statement is false

Part 2) given the center of the circle (1,3) and a point on the circle (2,6), (11,5) is on the circle.

true or false

substitute the center of the circle in the equation in standard form

$(x-1)^{2} +(y-3)^{2}=r^{2}$

Find the distance (radius) between the center (1,3) and (2,6)

substitute in the formula of distance

$r=\sqrt{(6-3)^{2}+(2-1)^{2}}$

$r=\sqrt{(3)^{2}+(1)^{2}}$

$r=\sqrt{10}\ units$

The equation of the circle is equal to

$(x-1)^{2} +(y-3)^{2}=(\sqrt{10}){2}$

$(x-1)^{2} +(y-3)^{2}=10$

Verify if the point (11,5) is on the circle

we know that

If a ordered pair is on the circle, then the ordered pair must satisfy the equation of the circle

For x=11,y=5

substitute

$(11-1)^{2} +(5-3)^{2}=10$

$(10)^{2} +(2)^{2}=10$

$104=10$ -----> is not true

therefore

The point is not on the circle

The statement is false

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4 years ago

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11111nata11111 [884]

Answer:

Step-by-step explanation:

first the parenthesis

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anything raised to the power of 0 is 1

(3^4)^2=3^8 you multiply the powers

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Marysya12 [62]

Answer:

D. 3.5

Step-by-step explanation:

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2 years ago