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2 years ago

two numbered 1 to 6 dice are thrown together and their scores are added. the probability that the sum will be 4 is?

Mathematics

1 answer:

natulia [17]2 years ago

5 0

Answer:

$Pr = \frac{3}{36}$

Step-by-step explanation:

Given

2 number die

Required

P(Sum = 4)

The sample space of 2 die is:

$S = \{(1, 1) (1, 2) (1, 3) (1, 4) (1, 5) (1, 6), (2, 1) (2, 2) (2, 3) (2, 4) (2, 5) (2, 6)$

$(3, 1) (3, 2) (3, 3) (3, 4) (3, 5) (3, 6), (4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (4, 6)$

$(5, 1) (5, 2) (5, 3) (5, 4) (5, 5) (5, 6), (6, 1) (6, 2) (6, 3) (6, 4) (6, 5) (6, 6)$

$n(S)= 36$

The pairs that adds up to 4 are:

$Sum(4) = \{(1,3), (2,2),(3,1)\}$

$n(Sum) =3$

So, the probability is:

$Pr = \frac{n(Sum)}{n(S)}$

$Pr = \frac{3}{36}$

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2 years ago

Is the given point interior , exterior, or on the circle k (x+2)2 + (y-3)2 =18 P (8,4)

Mnenie [13.5K]

One way would be to find the distance from the point to the center of the circle and compare it to the radius

for
$(x-h)^2+(y-k)^2=r^2$
the center is (h,k) and the radius is r

and the distance formula is
distance between $(x_1,y_1)$ and $(x_2,y_2)$ is
$D=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$

r=radius
D=distance form (8,4) to center

if r>D, then (8,4) is inside the circle
if r=D, then (8,4) is on the circle
if r<D, then (8,4) is outside the circle

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$(x-(-2))^2+(y-3)^2=(\sqrt{18})^2$
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find distance between (8,4) and (-2,3)

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$D=\sqrt{10^2+1}$
$D=\sqrt{100+1}$
$D=\sqrt{101}$

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$D=\sqrt{101}$ ≈10.04

do r<D

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two numbered 1 to 6 dice are thrown together and their scores are added. the probability that the sum will be 4 is?​

two numbered 1 to 6 dice are thrown together and their scores are added. the probability that the sum will be 4 is?