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3 years ago

Find the sum and product of the roots. 2x^2 + 6x - 8 = 0

1 answer:

3 0

Suppose you can factorize:

$ax^2+bx+c=a(x-r_1)(x-r_2)$

Then by expanding the right side, you have

$ax^2+bx+c=a(x^2-(r_1+r_2)x+r_1r_2)=ax^2-a(r_1+r_2)x+ar_1r_2$

which means in this case, you have to have

$\begin{cases}6=-2(r_1+r_2)\\-8=2r_1r_2\end{cases}\implies\begin{cases}r_1+r_2=-3\\r_1r_2=-4\end{cases}$

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A bottler of drinking water fills plastic bottles with a mean volume of 1,000 milliliters (mL) and standard deviation The fill v

trapecia [35]

Answer:

84.13% of bottles will have volume greater than 994 mL

Step-by-step explanation:

Mean volume = u = 1000

Standard deviation = $\sigma$ = 6

We need to find the proportion of bottles with volume greater than 994. So our test value is 994. i.e.

x = 994

Since the data is normally distributed we will use the concept of z-score to find the required proportion. First we convert 994 to its equivalent z-score, then using the z-table we will find the corresponding value of proportion. The formula to calculate the z score is:

$z=\frac{x-u}{\sigma}$

Substituting the values, we get:

$z=\frac{994-1000}{6}=-1$

This means 994 is equivalent to a z score of -1. Now we will find the proportion of z values which are greater than -1 from the z table.

i.e. P(z > -1)

From the z-table this value comes out to be:

P(z >- 1) = 1 - P(z < -1) = 1 - 0.1587 = 0.8413

Since, 994 is equivalent to a z score of -1, we can write that proportion of values which will be greater than 994 would be:

P( X > 994 ) = P( z > -1 ) = 0.8413 = 84.13%

4 0

2 years ago

A Web ad can be designed from four different colors, three font types, five font sizes, three images, and five text phrases. A s

Varvara68 [4.7K]

Answer:

P(A∪B)=17/20 or 0.85

P(A∪B')=2/5 or 0.4

P(A'∪B')=4/5 or 0.8

Step-by-step explanation:

There are four font colors so each color had equal chance and thus,

P(A)=1/4

There are 5 font sizes and so not the smallest fonts are 4.Thus,

P(B)=4/5

P(A∪B)=P(A)+P(B)-P(A∩B)

The design is generated randomly so event A and event B are independent.

P(A∩B)=P(A)*P(B)

P(A∩B)=1/4(4/5)=1/5

P(A∪B)=P(A)+P(B)-P(A∩B)

P(A∪B)=1/4+4/5-1/5=1/4+3/5

P(A∪B)=17/20 or 0.85

P(A∪B')=P(A)+P(B')-P(A∩B')

P(B')=1-P(B)=1-4/5=1/5

P(A∩B')=P(A)*P(B')=1/4*1/5=1/20

P(A∪B')=P(A)+P(B')-P(A∩B')

P(A∪B')=1/4+1/5-1/20=9/20-1/20=8/20

P(A∪B')=2/5 or 0.4

P(A'∪B')=P(A∩B)'

P(A'∪B')=1-P(A∩B)

P(A'∪B')=1-1/5=4/5

P(A'∪B')=4/5 or 0.8

5 0

3 years ago

Write an equation in standard form for a line that passes through (4,1) & (5,7)

tatuchka [14]

Y = mx + b

First equation 1 = m(4) + b, bring everything to one side m(4) + b - 1 = 0
Second equation 7 = m(5) + b, bring everything to one side m(5) + b - 7 = 0

Set them equal to each other,

m(4) + b - 1 = m(5) + b - 7

If you bring the b over to the left hand side it becomes

m(4) + b - b - 1 = m(5) - 7

m(4) - 1 = m(5) - 7

Solve for m

6 = m

Plug m = 6 into either equation from the beginning,

m(4) + b - 1 = 0

6(4) + b - 1 = 0

24 + b - 1 = 0

b = -23

Knowing m and b we can now make an equation

y = mx + b

y = 6x -23 Final answer

3 0

2 years ago

Two airplanes leave an airport at the same time, one going NW bearing 135 degrees at 415mph and the other going east at 327 mph.

arsen [322]

Find the distances of each plane for the 4 hours, then use the law of cosine.

See attached picture: