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schepotkina [342]
3 years ago
7

Each of the faces of a fair six-sided number cube is numbered with one of the numbers 1 through 6, with a different number appea

ring on each face. Two such number cubes will be tossed, and the sum of the numbers appearing on the faces that land up will be recorded. What is the probability that the sum will be 4, given that the sum is less than or equal to 6 ?
Mathematics
1 answer:
makkiz [27]3 years ago
3 0

Answer:

0.2

Step-by-step explanation:

If two number cubes as tossed then the total possible out comes are

Total = {(1,1),(2,1),(3,1),(4,1),(5,1),(6,1),(1,2),(2,2),(3,2),(4,2),(5,2),(6,2),(1,3),(2,3),(3,3),(4,3),(5,3),(6,3),(1,4),(2,4),(3,4),(4,4),(5,4),(6,4),(1,5),(2,5),(3,5),(4,5),(5,5),(6,5),(1,6),(2,6),(3,6),(4,6),(5,6),(6,6)}

Number of elements in sample space are

n(S)=36

Let A and B represent the following events.

A = The sum will be 4.

A = {(1,3),(2,2),(3,1)}

n(A) = 3

B = The sum is less than or equal to 6.

B = {(1,1),(1,2),(1,3),(1,4),(1,5),(2,1),(2,2),(2,3),(2,4),(3,1),(3,2),(3,3),(4,1),(4,2),(5,1)}

n(B)=15

Intersection of both events is

A∩B = {(1,3),(2,2),(3,1)}

n(A∩B) = 3

P(A)=\dfrac{n(A)}{n(S)}=\dfrac{3}{36}

P(B)=\dfrac{n(B)}{n(S)}=\dfrac{15}{36}

P(A\cap B)=\dfrac{n(A\cap B)}{n(S)}=\dfrac{3}{36}

We need to find the probability that the sum will be 4, given that the sum is less than or equal to 6.

P(\frac{A}{B})=\dfrac{P(A\cap B)}{P(B)}

P(\frac{A}{B})=\dfrac{\frac{3}{36}}{\frac{15}{36}}

P(\frac{A}{B})=\dfrac{3}{15}

P(\frac{A}{B})=\dfrac{1}{5}

P(\frac{A}{B})=0.2

Therefore, the required probability is 0.2.

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(08.04 MC)
skelet666 [1.2K]

Answer:

292,968

Step-by-step explanation:

As we know,

Sum of a geometric sequence (S) = \frac{a(1-r^{n}) }{(1-r)}

where,

a = first term of sequence,

r = the constant ratio,

n = number of terms in sequence.

So, according to the question,

a = 3,

r = 5,

n = 8.

by substituting the values in the above formula, we get;

⇒ S=\frac{3(1-5^{8}) }{(1-5)}

⇒ 292,968

7 0
2 years ago
Explain how to find the surface area of a can of fruit with a height of 80 mm and a diameter of 60 mm. Give the area in your exp
sdas [7]

Answer:

S=6600\pi \text{ mm}^2

S\approx 20724 \text{ mm}^2

Step-by-step explanation:

A can of fruit is geometrically a cylinder.

We are talking about the <u>surface area of a cylinder</u>:

S=2\pi rh+2\pi r^2

\pi \approx 3.14

r: \text{radius}

h: \text{height}

Factoring, we have:

S=2\pi r (h+r)

Calculating the surface area of the can of fruit:

S=2\pi (30) (80+30)

S=2\pi (30) (110)

S=6600\pi

S\approx 20724 \text{ mm}^2

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If both of those times were on the same date and in
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The length of the measure of YZ is 80cm

<h3>Perpendicular lines</h3>

Perpendicular lines are lines that are at angle 90 degrees to each other. From the given diagram;

WX = 82cm

Required

Length. of YZ

Using the pythagoras theorem

YV = VZ = √41²-9²

YV = VZ = √1600

YV = VZ = 40

Since YZ = YV + VZ

YZ = 40cm + 40cm

YZ= 80cm

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5 0
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