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3 years ago

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When e = 4, f = 2, and g = 8If e varies jointly with f and g, what is the constant of variation?

2 answers:

tatyana61 [14]3 years ago

8 0

Answer:

$e=\frac{1}{4} fg$

Step-by-step explanation:

Joint variation problem solve using the equation y = kxz.

e ∝ fg

e=kfg

now substitute the values

$4=k*2*8\\4=16k\\k=\frac{4}{16} \\k=\frac{1}{4}$

The relationship will be:

$e=kfg\\e=\frac{1}{4} fg$

suter [353]3 years ago

7 0

Answer:

K = 1/4

Step-by-step explanation:

Formula for joint variation is X = Kyz

Where k is the constant and x,z are any given variables.

The question here says that

When e = 4, f = 2, and g = 8If e varies jointly with f and g, what is the constant of variation?

It means that e= k×f×g

And e = 4, f= 2 and g= 8

4 = k × 2 × 8

4 = 16 × k (Now divide through by 16 to get k)

K = 1/4

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32 percent of the customers of a fast food chain order the Whopper, French fries and a drink. A random sample of 10 cash registe

weqwewe [10]

Answer: 0.0023

Step-by-step explanation:

Let X be the binomial variable that represents the number of receipts will show that the above three food items were ordered.

probability of success p = 32% = 0.32

Sample size : n= 10

Binomial probability function :

$P(X=x)= \ ^nC_xp^x(1-p)^{n-x}$

Now, the probability that eight receipts will show that the above three food items were ordered :

$P(X=8)=\ ^{10}C_8(0.32)^8(1-0.32)^2\\\\=\dfrac{10!}{8!2!}(0.32)^8(0.68)^2\\\\=5\times9(0.0000508414176684)\\\\=0.00228786379508\approx0.0023$

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3 years ago

What is the side length, in inches, of the pets

tatiyna

Candy draws a square design with a side length of x inches for the window at the pet shop. She takes the design to the printer and asks for a sign that has an area of 16x2 – 40x + 25 square inches. What is the side length, in inches, of the pet shop sign?

Answer:

the length of the sign is $4x-5$ inches

Step-by-step explanation:

Given

Area of the square of design = $16x^{2} -40x+25$

First we find the roots of equation $16x^{2} -40x+25=0$

The roots of the quadratic equation $ax^{2} +bx^{2} +c=0$ are given by

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

where $a=16, b=-40, c=25$

$x=\frac{40\pm\sqrt{(-40)^2-4\times 16\times 25}}{2\times 16}$

$x=\frac{40\pm\sqrt{1600-1600}}{32}$

$x=\frac{40\pm\sqrt{0}}{32}$

$x=\frac{40}{32}$

$x=\frac{5}{4}$

$4x=5\\4x-5=0$

That is, the factors of the polynomial $16x^{2} -40x+25$ are $4x-5$ and $4x-5$ .

So, Area of the square design = $16x^{2} -40x+25$ = $(4x-5)^{2}$

Area of a square = Length^2

Thus, the length of the sign is $4x-5$ inches

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