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3 years ago

HELP!!!!!!!!!!

Mathematics

1 answer:

Vadim26 [7]3 years ago

7 0

Answer:

A)Points are A ,B,C,D,G,E

B) other names are AC, CA, ABC

C)the Rays are DE, DG

D) M refers to the plane in which these points and Rays,lines are mentioned

E) < GDE < EDG are other names for angle 1

Explanation:

A line is one which has arrow heads on both ends and extends infinitely .

A ray has arrow Head at one end.

An angle is formed by the joining of two rays or lines.

Points are specific locations on a line or ray .

A plane has two dimensions length and breadth which has a line etc drawn on it

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3 years ago

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valentina_108 [34]

Answer:

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3 years ago

(4,5) and (11,33) rate of change

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The computers of nine engineers at a certain company are to be replaced. Four of the engineers have selected laptops and the oth

Gala2k [10]

Answer:

(a) There are 70 different ways set up 4 computers out of 8.

(b) The probability that exactly three of the selected computers are desktops is 0.305.

Step-by-step explanation:

Of the 9 new computers 4 are laptops and 5 are desktop.

Let X = a laptop is selected and Y = a desktop is selected.

The probability of selecting a laptop is = $P(Laptop) = p_{X} = \frac{4}{9}$

The probability of selecting a desktop is = $P(Desktop) = p_{Y} = \frac{5}{9}$

Then both X and Y follows Binomial distribution.

$X\sim Bin(9, \frac{4}{9})\\ Y\sim Bin(9, \frac{5}{9})$

The probability function of a binomial distribution is:

$P(U=k)={n\choose k}\times(p)^{k}\times (1-p)^{n-k}$

(a)

Combination is used to determine the number of ways to select k objects from n distinct objects without replacement.

It is denotes as: ${n\choose k}=\frac{n!}{k!(n-k)!}$

In this case 4 computers are to selected of 8 to be set up. Since there cannot be replacement, i.e. we cannot set up one computer twice or thrice, use combinations to determine the number of ways to set up 4 computers of 8.

The number of ways to set up 4 computers of 8 is:

${8\choose 4}=\frac{8!}{4!(8-4)!}\\=\frac{8!}{4!\times 4!} \\=70$

Thus, there are 70 different ways set up 4 computers out of 8.

(b)

It is provided that 4 computers are randomly selected.

Compute the probability that exactly 3 of the 4 computers selected are desktops as follows:

$P(Y=3)={4\choose 3}\times(\frac{5}{9})^{3}\times (1-\frac{5}{9})^{4-3}\\=4\times\frac{125}{729}\times\frac{4}{9}\\ =0.304832\\\approx0.305$

Thus, the probability that exactly three of the selected computers are desktops is 0.305.

(c)

Compute the probability that of the 4 computers selected at least 3 are desktops as follows:

$P(Y\geq 3)=1-P(Y$

Thus, the probability that at least three of the selected computers are desktops is 0.401.

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3 years ago

The probability of successfully landing a plane using a flight simulator is p. Nine randomly

nekit [7.7K]

Success = p
failure = (1-p)
is the correct first step.

To find the probability of at least one success, P(X>0), it is much easier to first calculate the probability of P(X=0)=NO success. Then the complement of
P(X=0)'=1-P(X=0) = P(>0) = probability of at least one success.

By the multiplication rule,
P(X=0)=(1-p)^9
or the binomial distribution
P(X=0)=C(9,0)(p^0)(1-p)^9=1*1*(1-p)^9=(1-p)^9

So the probability of at least one success = 1-P(X=0)
=1-(1-p)^9

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3 years ago