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Gala2k [10]
3 years ago
10

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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B $4.00
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valentina_108 [34]

Answer:

y= -6

Step-by-step explanation:

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(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.

(c) The probability that at least three of the selected computers are desktops is 0.401.

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 <em>k</em> objects from <em>n</em> 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.

6 0
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

5 0
3 years ago
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