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Naily [24]
3 years ago
8

NEED YOUR HELP ASAP!!!!

Mathematics
2 answers:
Degger [83]3 years ago
8 0

Answer:

One blue square equals 4.2 grams

Step-by-step explanation:

If the triangle is equal to three of the blue squares then you would have to divide the triangles value by three to get the squares value.

12.6/3 = 4.2

MAVERICK [17]3 years ago
3 0

Answer:

dont braly me do it for him

Step-by-step explanation:

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How to determine if 3 points are collinear?
Verdich [7]
If the points are on the same line they are co linear. when working with coordinates if the all have the same y as in (x,y) they are on the same line for example, (4,5) (2,5) (9,6) are collinear
5 0
3 years ago
(a) Consider a class with 30 students. Compute the probability that at least two of them have their birthdays on the same day. (
Galina-37 [17]

Answer:

a.) 0.7063

b.) 23

Step-by-step explanation:

a.)

Let X be an event in which at least 2 students have same birthday

     Y be an event in which no student have same birthday.

Now,

P(X) + P(Y) = 1

⇒P(X) = 1 - P(Y)

as we know that,

Probability of no one has birthday on same day = P(Y)

⇒P(Y) = \frac{365!}{(365)^{n} (365-n)! }      where there are n people in a group

As given,

n = 30

⇒P(Y) = \frac{365!}{(365)^{30} (365-30)! } = \frac{365!}{(365)^{30} (335)! } = 0.2937

∴ we get

P(X) = 1 - 0.2937 = 0.7063

So,

The probability that at least two of them have their birthdays on the same day  =  0.7063

b.)

Given, P(X) > 0.5

As

P(X) + P(Y) = 1

⇒P(Y) ≤ 0.5

As

P(Y) = \frac{365!}{(365)^{n} (365-n)! }

We use hit and trial method

If n = 1 , then

P(Y) = \frac{365!}{(365)^{1} (365-1)! } = \frac{365!}{(365)^{1} (364)! }  = 1 \nleq 0.5

If n = 5 , then

P(Y) = \frac{365!}{(365)^{5} (365-5)! } = \frac{365!}{(365)^{5} (360)! }  = 0.97 \nleq 0.5

If n = 10 , then

P(Y) = \frac{365!}{(365)^{10} (365-10)! } = \frac{365!}{(365)^{10} (354)! }  = 0.88 \nleq 0.5

If n = 15 , then

P(Y) = \frac{365!}{(365)^{15} (365-15)! } = \frac{365!}{(365)^{15} (350)! }  = 0.75 \nleq 0.5

If n = 20 , then

P(Y) = \frac{365!}{(365)^{20} (365-20)! } = \frac{365!}{(365)^{20} (345)! }  = 0.588 \nleq 0.5

If n = 22 , then

P(Y) = \frac{365!}{(365)^{22} (365-22)! } = \frac{365!}{(365)^{22} (343)! }  = 0.52 \nleq 0.5

If n = 23 , then

P(Y) = \frac{365!}{(365)^{23} (365-23)! } = \frac{365!}{(365)^{23} (342)! }  = 0.49 \nleq 0.5

∴ we get

Number of students should be in class in order to have this probability above 0.5 = 23

5 0
3 years ago
Which equation is an identity? (Can someone explain to me how they got their answer, I don't get this.)
Greeley [361]
5w + 8 - w = 6w - 2(w -4).

Let's reduce the equation on the left:

4w + 8; (1)

and now let's reduce the equation on the right

6w - 2w +8 = 4w +8 (2).

We notice that (1) ≈ (2), and this is the only IDENTITY

7 0
3 years ago
Which quadrilateral MUST be a parallelogram?
Anuta_ua [19.1K]
A quadrilateral MUST be a parallelogram if it has one pair of opposite sides both parallel and congruent. A quadrilateral MUST be a parallelogram if it has both pairs of its opposite angles congruent (or equal in measure). A quadrilateral MUST be a parallelogram if it has both diagonals bisecting each other.
3 0
3 years ago
Room 1
nexus9112 [7]

Answer:

Hi... Your answer is... B

7 0
3 years ago
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