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Find the z-score for which 12% of the distribution's area lies between-z and z.

Bezzdna [24]

Since the normal distribution is symmetrical, we only need to consider the distribution for 56% or 0.56, because 6% is half of 12%.
The Z score is 0.15 approximately.
Z score of -0.15 is 1-0.56=0.44 and the difference between 0.56 and 0.44=0.12 which is 12%.
Therefore 12% of the distribution lies between -0.15 and 0.15.

3 0

3 years ago

Samantha has 52 jelly beans. She decides that she will eat 4 jelly beans per minute, and she wants to create an expression that

IrinaK [193]

Answer:

-4x+52

Step-by-step explanation:

So we know that Samantha has 52 jelly beans and that she'll eat 4 jelly beans per minute in X minutes.

It'd be 4*x=4x

Usually, it'd be 52-4x

But you can create an equivalent

Lets say x=5

So she'd eat 20 jelly beans

52-20=32

BUT

-20+52=32

So, the answer would be D

3 0

3 years ago

Enter the function for me y=5x+4 reflected across both axes

likoan [24]

Answer:

if you "reflect" the equation your cenceling out the percentage of the ABZ formula meaning there is no solution

Step-by-step explanation:

4 0

3 years ago

PLS HELP WILL MARK YOU BRAINLIEST!!!

qaws [65]

I believe it’s D because the notation should be outside the parenthesis

8 0

3 years ago

Read 2 more answers

Find an equation of the plane with the given characteristics. The plane passes through the points (4, 3, 1) and (4, 1, -7) and i

Minchanka [31]

Answer:

$3x - 4y + z = 1$

Step-by-step explanation:

Given

$Point\ 1 = (4,3,1)$

$Point\ 2 = (4,1,-7)$

Perpendicular to $8x + 7y + 4z = 18$

Required

Determine the plane equation

The general equation of a plane is:

$a(x-x_1) + b(y - y_1) + c(z-z_1) = 0$

For $n =$

$(x_1,y_1,z_1) = (4,3,1)$

$(x_2,y_2,z_2) = (4,1,-7)$

First, we need to determine parallel vector $V_1$

$V_1 =$

$V_1$ is parallel to the required plane

From the question, the required plane is perpendicular to $8x + 7y + 4z = 18$

Next, we determine vector $V_2$

$V_2 =$

This implies that the required plane is parallel to $V_2$

Hence: $V_1$ and $V_2$ are parallel.

So, we can calculate the cross product $V_1 * V_2$

$V_1 =$

$V_2 =$

$n = V_1 * V_2$

$V_1 * V_2 =\left[\begin{array}{ccc}i&j&k\\0&2&8\\8&7&4\end{array}\right]$

The product is always of the form + - +

So:

$V_1 * V_2 = i\left[\begin{array}{cc}2&8\\7&4\end{array}\right]$ $-j\left[\begin{array}{cc}0&8\\8&4\end{array}\right]$ $+k\left[\begin{array}{cc}0&2\\8&7\end{array}\right]$