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3 years ago

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A random variable is normally distributed with a mean of 25 and a standard deviation of 5. if an observation is randomly selecte

d from the distribution, determine two values of which the smallest has 25% of the values below it and the largest has 25% of the values above it.

1 answer:

Naddika [18.5K]3 years ago

7 0

<span>A random variable is normally distributed with a mean of 25 and a standard deviation of 5. if an observation is randomly selected from the distribution, determine two values of which the smallest has 25% of the values below it and the largest has 25% of the values above it.</span>

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Jill and Marcy go to an ice cream store where they have the option of getting a smoothie in either a cylindrical or rectangular

k0ka [10]

Answer: Jill got the larger smoothie and its volume is 1609.14 cubic cm.

Step-by-step explanation:

Since we have given that

Jill purchased the cylindrical container.

Height of container = 8 cm

Radius = 8 cm

So, volume of cylindrical container would be

$\pi r^2h\\\\=\dfrac{22}{7}\times 8\times 8\times 8\\\\=1609.14\ cm^3$

Marcy purchased the rectangular container.

Height of container = 8 cm

Width = 8 cm

Length = 8 cm

so, volume of rectangular container would be

$l\times b\times h\\\\=8\times 8\times 8\\\\=512\ cm^3$

Hence, Jill got the larger smoothie and its volume is 1609.14 cubic cm.

3 0

3 years ago

the two lines represent the amount of water, over time, in two tanks that are the same size. Which container is filling more qui

uranmaximum [27]

The missing figure is attached down

Answer:

Container A is filling more quickly

Step-by-step explanation:

<em>Let us explain the linear relation</em>

The linear equation is y = m x + b, where

m is the slope of the line which represents the rate of change (change in y/change in x)
b is the y-intercept which means the initial amount of y (at x = 0)

<em>If a line represents the amount of water over time, then it represents the rate of filling of the water</em>

∵ There are two containers A and B that have the same size

∵ The two lines represent the amount of water over time in

containers A and B

∴ The slopes of the lines represent the rat of water in the tanks

→ That means the greater slope represents the greater rate

∵ The greater the slope, the steeper the line

∴ The container that represented by the steeper the line is filling quickly

→ The line represents container A is steeper more than the line

represents container B

∴ Container A is filling more quickly

4 0

3 years ago

<img src="https://tex.z-dn.net/?f=%5Cleft%20%5C%7B%20%7B%7Bx%2By%3D1%7D%20%5Catop%20%7Bx-2y%3D4%7D%7D%20%5Cright.%20%5C%5C%5Clef

brilliants [131]

Answer:

<em>(a) x=2, y=-1</em>

<em>(b) x=2, y=2</em>

<em>(c)</em> $\displaystyle x=\frac{5}{2}, y=\frac{5}{4}$

<em>(d) x=-2, y=-7</em>

Step-by-step explanation:

<u>Cramer's Rule</u>

It's a predetermined sequence of steps to solve a system of equations. It's a preferred technique to be implemented in automatic digital solutions because it's easy to structure and generalize.

It uses the concept of determinants, as explained below. Suppose we have a 2x2 system of equations like:

$\displaystyle \left \{ {{ax+by=p} \atop {cx+dy=q}} \right.$

We call the determinant of the system

$\Delta=\begin{vmatrix}a &b \\c &d \end{vmatrix}$

We also define:

$\Delta_x=\begin{vmatrix}p &b \\q &d \end{vmatrix}$

And

$\Delta_y=\begin{vmatrix}a &p \\c &q \end{vmatrix}$

The solution for x and y is

$\displaystyle x=\frac{\Delta_x}{\Delta}$

$\displaystyle y=\frac{\Delta_y}{\Delta}$

(a) The system to solve is

$\displaystyle \left \{ {{x+y=1} \atop {x-2y=4}} \right.$

Calculating:

$\Delta=\begin{vmatrix}1 &1 \\1 &-2 \end{vmatrix}=-2-1=-3$

$\Delta_x=\begin{vmatrix}1 &1 \\4 &-2 \end{vmatrix}=-2-4=-6$

$\Delta_y=\begin{vmatrix}1 &1 \\1 &4 \end{vmatrix}=4-3=3$

$\displaystyle x=\frac{\Delta_x}{\Delta}=\frac{-6}{-3}=2$

$\displaystyle y=\frac{\Delta_y}{\Delta}=\frac{3}{-3}=-1$

The solution is x=2, y=-1

(b) The system to solve is

$\displaystyle \left \{ {{4x-y=6} \atop {x-y=0}} \right.$

Calculating:

$\Delta=\begin{vmatrix}4 &-1 \\1 &-1 \end{vmatrix}=-4+1=-3$

$\Delta_x=\begin{vmatrix}6 &-1 \\0 &-1 \end{vmatrix}=-6-0=-6$

$\Delta_y=\begin{vmatrix}4 &6 \\1 &0 \end{vmatrix}=0-6=-6$

$\displaystyle x=\frac{\Delta_x}{\Delta}=\frac{-6}{-3}=2$

$\displaystyle y=\frac{\Delta_y}{\Delta}=\frac{-6}{-3}=2$

The solution is x=2, y=2

(c) The system to solve is

$\displaystyle \left \{ {{-x+2y=0} \atop {x+2y=5}} \right.$

Calculating:

$\Delta=\begin{vmatrix}-1 &2 \\1 &2 \end{vmatrix}=-2-2=-4$

$\Delta_x=\begin{vmatrix}0 &2 \\5 &2 \end{vmatrix}=0-10=-10$

$\Delta_y=\begin{vmatrix}-1 &0 \\1 &5 \end{vmatrix}=-5-0=-5$

$\displaystyle x=\frac{\Delta_x}{\Delta}=\frac{-10}{-4}=\frac{5}{2}$

$\displaystyle y=\frac{\Delta_y}{\Delta}=\frac{-5}{-4}=\frac{5}{4}$

The solution is

$\displaystyle x=\frac{5}{2}, y=\frac{5}{4}$

(d) The system to solve is

$\displaystyle \left \{ {{6x-y=-5} \atop {4x-2y=6}} \right.$

Calculating:

$\Delta=\begin{vmatrix}6 &-1 \\4 &-2 \end{vmatrix}=-12+4=-8$

$\Delta_x=\begin{vmatrix}-5 &-1 \\6 &-2 \end{vmatrix}=10+6=16$

$\Delta_y=\begin{vmatrix}6 &-5 \\4 &6 \end{vmatrix}=36+20=56$

$\displaystyle x=\frac{\Delta_x}{\Delta}=\frac{16}{-8}=-2$

$\displaystyle y=\frac{\Delta_y}{\Delta}=\frac{56}{-8}=-7$

The solution is x=-2, y=-7

4 0

3 years ago

I need help with numbers 1-4 please if someone could help me

laila [671]

You can use the slope formula or just calculate
$\frac{rise}{run}$
So for the first problem you run 3 units right and rise 5 units so
$\frac{rise}{run} = \frac{5}{3}$
2. Notice the y axis goes by 2 units.
$\frac{ - 10}{4} = \frac{ - 5}{4}$
3.
$\frac{0}{12} = 0$
4.
$\frac{2}{0} = undefined$
Please let me know if you have questions

3 0

3 years ago

Need help please

zlopas [31]

Answer:

B

Step-by-step explanation:

6x5=30

30x2/3=20

8 0

1 year ago