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miss Akunina [59]
3 years ago
12

RT has the endpoints (0, 2) and (4, 2). What is its midpoint?

Mathematics
2 answers:
Nimfa-mama [501]3 years ago
3 0
R=(0,2), T=(4,2)
vector:
v=(4-0,2-2)=(4,0)
v/2=(2,0)

(0,2)+v/2=(2,2)

so (2,2) is the midpoint
Oksanka [162]3 years ago
3 0
I used the midpoint formula.

Hopefully this helped.

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According to the last census (2010), the mean number of people per household in the United States is LaTeX: \mu = 2.58 Assume a
Veseljchak [2.6K]

Answer:

P(2.50 < Xbar < 2.66) = 0.046

Step-by-step explanation:

We are given that Population Mean, \mu = 2.58 and Standard deviation, \sigma = 0.75

Also, a random sample (n) of 110 households is taken.

Let Xbar = sample mean household size

The z score probability distribution for sample mean is give by;

             Z = \frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } } ~ N(0,1)

So, probability that the sample mean household size is between 2.50 and 2.66 people = P(2.50 < Xbar < 2.66)

P(2.50 < Xbar < 2.66) = P(Xbar < 2.66) - P(Xbar \leq 2.50)

P(Xbar < 2.66) = P( \frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } } < \frac{2.66-2.78}{\frac{0.75}{\sqrt{110} } } ) = P(Z < -1.68) = 1 - P(Z  1.68)

                                                              = 1 - 0.95352 = 0.04648

P(Xbar \leq 2.50) = P( \frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } } \leq \frac{2.50-2.78}{\frac{0.75}{\sqrt{110} } }  ) = P(Z \leq  -3.92) = 1 - P(Z < 3.92)

                                                              = 1 - 0.99996 = 0.00004  

Therefore, P(2.50 < Xbar < 2.66) = 0.04648 - 0.00004 = 0.046

7 0
3 years ago
A) 5<br> 11 + 4<br> 3<br> add the following rational number
mel-nik [20]

Step-by-step explanation:

the question is not correct

edit it and I will answer u in comments

5 0
2 years ago
Determine whether each integral is convergent or divergent. If it is convergent evaluate it. (a) integral from 1^(infinity) e^(-
AVprozaik [17]

Answer:

a) So, this integral is convergent.

b) So, this integral is divergent.

c) So, this integral is divergent.

Step-by-step explanation:

We calculate the next integrals:

a)

\int_1^{\infty} e^{-2x} dx=\left[-\frac{e^{-2x}}{2}\right]_1^{\infty}\\\\\int_1^{\infty} e^{-2x} dx=-\frac{e^{-\infty}}{2}+\frac{e^{-2}}{2}\\\\\int_1^{\infty} e^{-2x} dx=\frac{e^{-2}}{2}\\

So, this integral is convergent.

b)

\int_1^{2}\frac{dz}{(z-1)^2}=\left[-\frac{1}{z-1}\right]_1^2\\\\\int_1^{2}\frac{dz}{(z-1)^2}=-\frac{1}{1-1}+\frac{1}{2-1}\\\\\int_1^{2}\frac{dz}{(z-1)^2}=-\infty\\

So, this integral is divergent.

c)

\int_1^{\infty} \frac{dx}{\sqrt{x}}=\left[2\sqrt{x}\right]_1^{\infty}\\\\\int_1^{\infty} \frac{dx}{\sqrt{x}}=2\sqrt{\infty}-2\sqrt{1}\\\\\int_1^{\infty} \frac{dx}{\sqrt{x}}=\infty\\

So, this integral is divergent.

4 0
3 years ago
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vazorg [7]
Use distributive property
5k - 6k -8 = 5x + 4
Combine like terms
-k - 8 = 5x + 4
add 8 to both sides
-k = 5x + 12
k = -5x- 12
4 0
3 years ago
Read 2 more answers
Lilah completed the division problem below.
victus00 [196]
A - False

B - True

C - True

D - False

E - True

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7.84 / 2 = 3.92
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