True or false? The value zc is a value from the standard normal distribution such that P(−zc < z < zc) = c.

Mathematics

1 answer:

rewona [7]3 years ago

5 0

Answer:

False

Step-by-step explanation:

The standard normal distribution is represented by P(-zc < z < zc) = c

From this, you see that -zc is less than the variable of interest z and z is less than zc.

The equation simply means that the probability that -zc is less than z and z is less than zc, is c.

The value zc is NOT a value from the above distribution. All values in the distribution are of the variable z. For example, z might be equal to 1, z might be equal to 6, but the variable of interest in the distribution is z, not zc.

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A box company is designing a new rectangular gift container. The marketing department has designed a box with a width 2 inches s

Sladkaya [172]

Volume of the box= 56 cubic inches
let x is the length, then
width =2 inches shorter than its length = x - 2
height = 3 inches taller than its length = x+3
Volume = length x width x height
56 = x x (x-2) x (x+3)
56 = (x² -2x)(x+3)
56 = x³ +3x² -2x² - 6x
56 = x³ + x² -6x
x³+x²-6x-56 = 0
using the rational root theorem and factoring the polynomial;
(x-4)(x² +5x +14) = 0
from here;
x-4 = 0
x = 4
So, length = 4 inches
width = x - 2 = 4 -2 = 2 inches
length = x + 3 = 4 + 3 = 7 inches
volume = l x w x h = 4 x 2 x 7 = 56

3 0

2 years ago

What is 2(7/10)*1/3+4/15=

NeTakaya

2(7÷10)×1÷3+4÷15=1.4×0.33+0.267
=0.462+0.267
=0.729(0.73)

4 0

2 years ago

Plz help me!!!!!!!!!

Over [174]

Answer: $\bold{\dfrac{4\pm \sqrt{6}}{2}}$

Step-by-step explanation:

$\dfrac{3}{y-2}-2=\dfrac{1}{y-1}\\\\\\\text{Multiply by the LCD (y-2)(y-1) to clear the denominator:}\\\\\dfrac{3}{y-2}(y-2)(y-1)-2(y-2)(y-1)=\dfrac{1}{y-1}(y-2)(y-1)\\\\\\3(y-1)-2(y-2)(y-1)=1(y-2)\\\\3y-3-2(y^2-3y+2)=y-2\\\\3y-3-2y^2+6y-4=y-2\\\\-2y^2+9y-7=y-2\\\\0=2y^2-8y+5\quad \rightarrow \quad a=2,\ b=-8,\ c=5\\\\\\\text{Quadratic formula is: }x=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}\\\\\\x=\dfrac{-(-8)\pm \sqrt{(-8)^2-4(2)(5)}}{2(2)}\\\\\\.\ =\dfrac{8\pm \sqrt{64-40}}{2(2)}$

$.\ =\dfrac{8\pm \sqrt{24}}{2(2)}\\\\\\.\ =\dfrac{8\pm 2\sqrt{6}}{2(2)}\\\\\\.\ =\dfrac{4\pm \sqrt{6}}{2}$