Help! Please! Let x=a+bi and y=c+di 2x+3y

Help help help help please please

N76 [4]

You would move each coordinate 6 left and 1 up

A(2,5)
B(0,2)
C(-1,3)

hope this helps!

7 0

3 years ago

What is 24.051 rounded to the nearest hundredth? How did you solve it?

4vir4ik [10]

This is like late elementary/early middle school stuff where you learn this.
First off, what number is in the position of the hundereth place? 0? No. 1? No. 5 yes! Now how i learned how to round is you take the number to the right of whatever you are being asked about and round it up or down. Now if the number (in this case it is 1) is lower than 5, you round down (or keep it the same) but if it is 5 or higher you round up. since 1 is lower then 5, that 5 in the hundereths place stays the same.

7 0

3 years ago

If 9 pencils cost $4.95, what is the unit price of the pencils?

kondaur [170]

About $1.8 per pencil

5 0

3 years ago

A survey conducted by the Consumer Reports National Research Center reported, among other things, that women spend an average of

Nookie1986 [14]

Answer:

(a) The probability that a randomly selected woman shop exactly two hours online is 0.217.

(b) The probability that a randomly selected woman shop 4 or more hours online is 0.0338.

(c) The probability that a randomly selected woman shop less than 5 hours online is 0.9922.

Step-by-step explanation:

Let X = time spent per week shopping online.

It is provided that the random variable X follows a Poisson distribution.

The probability function of a Poisson distribution is:

$P (X=x)=\frac{e^{-\lambda}\lambda^{x}}{x!} ;\ x=0,1,2,...$

The average time spent per week shopping online is, λ = 1.2.

(a)

Compute the probability that a randomly selected woman shop exactly two hours online over a one-week period as follows:

$P (X=2)=\frac{e^{1.2}(1.2)^{2}}{2!} =0.21686\approx0.217$

Thus, the probability that a randomly selected woman shop exactly two hours online is 0.217.

(b)

Compute the probability that a randomly selected woman shop 4 or more hours online over a one-week period as follows:

P (X ≥ 4) = 1 - P (X < 4)

= 1 - P (X = 0) - P (X = 1) - P (X = 2) - P (X = 3)

$=1-\frac{e^{1.2}(1.2)^{0}}{0!}-\frac{e^{1.2}(1.2)^{1}}{1!}-\frac{e^{1.2}(1.2)^{2}}{2!}-\frac{e^{1.2}(1.2)^{2}}{3!}\\=1-0.3012-0.3614-0.2169-0.0867\\=0.0338$

Thus, the probability that a randomly selected woman shop 4 or more hours online is 0.0338.

(c)

Compute the probability that a randomly selected woman shop less than 5 hours online over a one-week period as follows:

P (X < 5) = P (X = 0) + P (X = 1) + P (X = 2) + P (X = 3) + P (X = 4)

$=\frac{e^{1.2}(1.2)^{0}}{0!}+\frac{e^{1.2}(1.2)^{1}}{1!}+\frac{e^{1.2}(1.2)^{2}}{2!}+\frac{e^{1.2}(1.2)^{3}}{3!}+\frac{e^{1.2}(1.2)^{4}}{4!}\\=0.3012+0.3614+0.2169+0.0867+0.0260\\=0.9922$

Thus, the probability that a randomly selected woman shop less than 5 hours online is 0.9922.

8 0

4 years ago

Can anyone please help me with these

Readme [11.4K]

Answer:

4(x+2) = 4*x + 4*2 = 4x+8

3(a+2) = 3*a + 3*2 = 3a + 6

2(x+3) = 2*x + 2*3 = 2x +6

Hope this help you :3

6 0

4 years ago

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