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valentina_108 [34]
3 years ago
8

What are the first 4 prime numbers

Mathematics
2 answers:
ddd [48]3 years ago
7 0
2, 3, 5 and 7 the the first 4 prime numbers
GarryVolchara [31]3 years ago
6 0
2 3 5 7 that is the list of prime numbers

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Help Pre-Calc
Alex73 [517]

Answer:

B) vertical stretch

Step-by-step explanation:

Given that the cost of manufacturing x refrigerators is given by the function f(x)= 2x^2 + 6000.

Now it says that the cost doubles for the year 2011 owing to higher raw material prices. So that means cost function will change from f(x) to 2f(x).

Now we need to decide the type of transformation from f(x) to 2f(x).

Remember that when f(x) changes to a*f(x) where value of "a" is greater than 1 then it stretches the function vertically by factor of "a".

in 2f(x), we have a=2

So that means "B) vertical stretch" is the correct choice.

6 0
3 years ago
HELP NEEDED PLEASE!!
kykrilka [37]
<YZW = <X = 63

hope it helps
3 0
3 years ago
Read 2 more answers
Please Help!!!!
mixas84 [53]

Step-by-step explanation:

kak kak jm ok bro open your eyes men open your eyes men open your eyes men open your eyes men open your eyes men open your eyes men open your eyes men

4 0
2 years ago
Read 2 more answers
Among 8 PS4s, four are good and four have defects. Unaware of this, a customer buys 5 PS4s.
astraxan [27]

Answer:

(a) The probability of exactly 2 defective PS4s among them is 0.3125.

(b) The probability that exactly ​ 2 are defective given that  ​at least ​ 2 purchased PS4s are defective is 0.3846.

Step-by-step explanation:

Let <em>X</em> = number of defective PS4s.

It is provided that 4 PS4s of 8 are defective.

The probability of selecting a defective PS4 is:

P(X)=p=\frac{4}{8}=0.50

A customer bought <em>n</em> = 5 PS4s.

The random variable <em>X</em> follows a Binomial distribution with parameters <em>n</em> = 5 and <em>p</em> = 0.50.

The probability function of a Binomial distribution is:

P(X=x)={n\choose x}p^{x}(1-p)^{n-x};\ x=0, 1, 2, 3...

(a)

Compute the probability of exactly 2 defective PS4s among them as follows:

P(X=2)={5\choose 2}(0.50)^{2}(1-0.50)^{5-2}=10\times0.25\times0.125=0.3125

Thus, the probability of exactly 2 defective PS4s among them is 0.3125.

(b)

Compute the probability that exactly ​ 2 are defective given that  ​at least ​ 2 purchased PS4s are defective as follows:

P(X=2|X\geq 2)=\frac{P(X=2\cap X\geq2)}{P(X\geq2)} =\frac{P(X=2)}{P(X\geq 2)}

The value of P (X = 2) is 0.3125.

The value of P (X ≥ 2) is:

P(X\geq 2)=1-P(X

Then the value of P (X = 2 | X ≥ 2) is:

P(X=2|X\geq 2)=\frac{P(X=2)}{P(X\geq 2)}=\frac{0.3125}{0.8125} =0.3846

Thus, the probability that exactly ​ 2 are defective given that  ​at least ​ 2 purchased PS4s are defective is 0.3846.

6 0
3 years ago
Read 2 more answers
Solve the system using an equivalent system
DerKrebs [107]
Add y to both sides in 1st equation
-2x=y
sub -2x for y in2nd equation
3x-(-2x)=10
3x+2x=10
5x=10
divide 5
x=2

sub back
-2x=y
-2(2)=y
-4=y

(x,y)
(2,-4)
C is ans
8 0
3 years ago
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