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2 years ago

20 POINTS

Mathematics

1 answer:

Nata [24]2 years ago

7 0

Answer:

1, 3, 7.

Step-by-step explanation:

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Suppose that Y and Z are points on a number line if YZ=20 and Y lies at 3 where could Z be located

creativ13 [48]

17? probably I think

5 0

2 years ago

An article suggests the uniform distribution on the interval (6.5, 19) as a model for depth (cm) of the bioturbation layer in se

Rama09 [41]

Answer:

The mean of depth is 12.75cm.

The variance of depth is of 13.02 cm².

Step-by-step explanation:

An uniform probability is a case of probability in which each outcome is equally as likely.

For this situation, we have a lower limit of the distribution that we call a and an upper limit that we call b.

The mean of the uniform distribution is:

$M = \frac{a+b}{2}$

The variance of the uniform distribution is given by:

$V = \frac{(b-a)^{2}}{12}$

Uniform distribution on the interval (6.5, 19)

This means that: $a = 6.5, b = 19$

Mean:

$M = \frac{6.5+19}{2} = 12.75$

The mean of depth is 12.75cm.

Variance:

$V = \frac{(19 - 6.5)^{2}}{12} = 13.02$

The variance of depth is of 13.02 cm².

4 0

3 years ago

Can someone help me please?!?

Vesna [10]

Answer:

Step-by-step explanation:

angle K and angle U are supplementary meaning when they are added they equal 180. In order to find angle K, you have to subtract 180-85 and it will get you 95.

4 0

2 years ago

If you place a 34-foot ladder against the top of a 30-foot building, how many feet will the bottom of the ladder be from the bot

aksik [14]

Answer:

It is 16 feet from the bottom of the building

Step-by-step explanation:

Let the bottom of the ladder be x feet from the bottom of the building.

Note that the length of the ladder becomes the hypotenuse.

Then using Pythagoras theorem,

${34}^{2} = {x}^{2} + {30}^{2}$

We solve for x,

${34}^{2} - {30}^{2} = {x}^{2}$

$256 = {x}^{2}$

$x = \sqrt{256} = 16$

Therefore 16 feet from the bottom of the building

5 0

3 years ago

Read 2 more answers

Derive the equation of the parabola with a focus at (2, −1) and a directrix of y = −one half.

liubo4ka [24]

Vertical parabola opens down so
(y-k)^2=4P(x-h)
(h,k) is center
P is distance from directix to center and from center to focus (it is 1/2 distance between focus and directix)
-1 to -1/2 is a distance of 1/2
1/2 of 1/2=1/4
P=1/4

so move 1/4 down from -1/2
-3/4
center is (2,-3/4)

so
y+3/4=(x-2)^2
minus 3/4
y=(x-2)^2-3/4
f(x)=(x-2)^2-3/4

you made an error, either the last or first option doesn't have -(x+2)^2

7 0

2 years ago