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natali 33 [55]
3 years ago
15

Given a standard six-sided die, what is 1 point the probability of rolling a 2? Please use a fraction bar (/) OR a colon (:) to

represent your answer.
Mathematics
2 answers:
snow_tiger [21]3 years ago
6 0

Answer:

1/6

Step-by-step explanation:

We know that a die has 6 sides. What would be the probability to roll 1 of those sides? (to rephrase the question)

well. the probability to roll one of the sides on a 6 side dice is 1/6, 1 side on the 6 sided die.

Aleks [24]3 years ago
3 0

Answer:

1/6

Step-by-step explanation:

There are six possible outcomes when rolling a die

1,2,3,4,5,6

P(rolling a 2) = number of outcomes that is a 2/ total outcomes

                     = 1/6

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If 1-cosA = 1/2 , Then find the value of sinA.
zimovet [89]

Step-by-step explanation:

Given :

1- cosA = 1/2

or, CosA = 1 -1/2

Therefore ; CosA = 1/2 = b/h

According to the Pythagoras theorem,

P = root under h^2 - b^2

= root under (2)^2 - (1)^2

= root under 4 -1

= root 3

Again,

SinA = P/h

= root 3 / 2

8 0
2 years ago
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Zina [86]
LFE angles and GFE angles
4 0
3 years ago
3s – 12 = -21<br> Solve for s
NISA [10]

Answer:

S= -3

Step-by-step explanation:

3s-12=-21

3s= -21+12

3s= -9

s= -9/3

s= -3

7 0
3 years ago
Read 2 more answers
Points M, N, and P are respectively the midpoints of sides AC , BC , and AB of △ABC. Prove that the area of △MNP is on fourth of
Hunter-Best [27]

Answer:

The area of △MNP is one fourth of the area of △ABC.

Step-by-step explanation:

It is given that the points M, N, and P are the midpoints of sides AC, BC and AB respectively. It means AC, BC and AB are median of the triangle ABC.

Median divides the area of a triangle in two equal parts.

Since the points M, N, and P are the midpoints of sides AC, BC and AB respectively, therefore MN, NP and MP are midsegments of the triangle.

Midsegments are the line segment which are connecting the midpoints of tro sides and parallel to third side. According to midpoint theorem the length of midsegment is half of length of third side.

Since MN, NP and MP are midsegments of the triangle, therefore the length of these sides are half of AB, AC and BC respectively. In triangle ABC and MNP corresponding side are proportional.

\triangle ABC \sim \triangle NMP

MP\parallel BC

MP=\frac{BC}{2}

By the property of similar triangles,

\frac{\text{Area of }\triangle MNP}{\text{Area of }\triangle ABC}=\frac{PM^2}{BC^2}

\frac{\text{Area of }\triangle MNP}{\text{Area of }\triangle ABC}=\frac{(\frac{BC}{2})^2}{BC^2}

\frac{\text{Area of }\triangle MNP}{\text{Area of }\triangle ABC}=\frac{1}{4}

Hence proved.

5 0
3 years ago
Distribute 7 x (2+2x)
True [87]
28 x because parenthesis first and you can’t do 2 + 2x so 7x times 2 is 14x and 7 times 2x is 14x
4 0
3 years ago
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