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3 years ago

7

Which equations are related equations to x+14=12x+14=12?

1 answer:

Nataly_w [17]3 years ago

4 0

X + 14 = 12
x + 14 - 14 = 12 - 14
x = 12 - 14

Options B and C are related to the given equation.

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What is the focus of the parabola? <img src="https://tex.z-dn.net/?f=y%3D%20%5Cfrac%7B1%7D%7B4%7D%20x%5E%7B2%7D%20-x%2B3" id="Te

OlgaM077 [116]

We know that
the equation of the parabola is of the form
y=ax²+bx+c
in this problem
y=1/4x²−x+3
where
a=1/4
b=-1
c=3
the coordinates of the focus are
(-b/2a,(1-D)/4a)
where D is the discriminant b²-4ac
D=(-1)²-4*(1/4)*3-----> D=1-3---> D=-2
therefore
x coordinate of the focus
-b/2a----> 1/[2*(-1/4)]----> 2

y coordinate of the focus
(1-D)/4a------> (1+2)/(4/4)---> 3

the coordinates of the focus are (2,3)

3 0

3 years ago

Read 2 more answers

How to convert 43 cm into a fraction?<br><br> .

Alex787 [66]

Since a cm is 1/100th of a meter, 43 cm would simply be 43/100. Hope that helps!

5 0

4 years ago

What is the volume of a rectangular prism with length 12 in., height 16 in., and width 13 in.?

Lilit [14]

Answer:

2496in3

Step-by-step explanation:

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3 years ago

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A new security system needs to be evaluated in the airport. The probability of a person being a security hazard is 4%. At the ch

ivolga24 [154]

Answer:

(a) 0.9412

(b) 0.9996 ≈ 1

Step-by-step explanation:

Denote the events a follows:

$P$ = a person passes the security system

$H$ = a person is a security hazard

Given:

$P (H) = 0.04,\ P(P^{c}|H^{c})=0.02\ and\ P(P|H)=0.01$

Then,

$P(H^{c})=1-P(H)=1-0.04=0.96\\P(P|H^{c})=1-P(P|H)=1-0.02=0.98\\$

(a)

Compute the probability that a person passes the security system using the total probability rule as follows:

The total probability rule states that: $P(A)=P(A|B)P(B)+P(A|B^{c})P(B^{c})$

The value of P (P) is:

$P(P)=P(P|H)P(H)+P(P|H^{c})P(H^{c})\\=(0.01\times0.04)+(0.98\times0.96)\\=0.9412$

Thus, the probability that a person passes the security system is 0.9412.

(b)

Compute the probability that a person who passes through the system is without any security problems as follows:

$P(H^{c}|P)=\frac{P(P|H^{c})P(H^{c})}{P(P)} \\=\frac{0.98\times0.96}{0.9412} \\=0.9996\\\approx1$

Thus, the probability that a person who passes through the system is without any security problems is approximately 1.

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3 years ago

One container holds 1 7/8 quarts of water and A second container holds 5 3/4 quarts of water how many more quarts of water does

BartSMP [9]

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4 0

3 years ago