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2 years ago

Solve the following equation by completing the square: 2x^2 – 2x – 4 = 0

Mathematics

1 answer:

Rus_ich [418]2 years ago

6 0

Answer:

the answer is B

Step-by-step explanation:

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ASAP, GEOMETRY PLEASE HELP ME

Black_prince [1.1K]

Answer: 405 yd squared

Step-by-step explanation:

V = (bh)/2

b = (15 x 9)/2 = 67.5

67.5 x 12 = 810

810/ 2 = 405

3 0

1 year ago

Please help me!! due tonight!

olasank [31]

Answer:

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Step-by-step explanation:

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2 years ago

Each calendar will sell for $5 each write an equatio to model the total income for selling calendars

Brrunno [24]

To answer the question above, I let x be the number of calendars sold. You may use any other letter as this is just for representation. The total income generated in selling calendars is calculated by multiplying the number of calendars with the price. That is,
total income = 5x
If we let total income be y, our equation is further simplified into,
y = 5x

4 0

2 years ago

Suppose X has an exponential distribution with mean equal to 23. Determine the following:

e-lub [12.9K]

Answer:

a) P(X > 10) = 0.6473

b) P(X > 20) = 0.4190

c) P(X < 30) = 0.7288

d) x = 68.87

Step-by-step explanation:

Exponential distribution:

The exponential probability distribution, with mean m, is described by the following equation:

$f(x) = \mu e^{-\mu x}$

In which $\mu = \frac{1}{m}$ is the decay parameter.

The probability that x is lower or equal to a is given by:

$P(X \leq x) = \int\limits^a_0 {f(x)} \, dx$

Which has the following solution:

$P(X \leq x) = 1 - e^{-\mu x}$

The probability of finding a value higher than x is:

$P(X > x) = 1 - P(X \leq x) = 1 - (1 - e^{-\mu x}) = e^{-\mu x}$

Mean equal to 23.

This means that $m = 23, \mu = \frac{1}{23} = 0.0435$

(a) P(X >10)

$P(X > 10) = e^{-0.0435*10} = 0.6473$

P(X > 10) = 0.6473

(b) P(X >20)

$P(X > 20) = e^{-0.0435*20} = 0.4190$

P(X > 20) = 0.4190

(c) P(X <30)

$P(X \leq 30) = 1 - e^{-0.0435*30} = 0.7288$

P(X < 30) = 0.7288

(d) Find the value of x such that P(X > x) = 0.05

$P(X > x) = e^{-\mu x}$

$0.05 = e^{-0.0435x}$

$\ln{e^{-0.0435x}} = \ln{0.05}$

$-0.0435x = \ln{0.05}$

$x = -\frac{\ln{0.05}}{0.0435}$

$x = 68.87$

5 0

2 years ago

Solve the following system of equations <br> x^2+2y^2=59<br> 2x^2+y^2=43

aivan3 [116]

Answer: x = (-3,3), y = (-5,5)

Step-by-step explanation:

Add both of the equations together, for $3x^{2}+3y^{2} = 102$ . Now we can divide both sides by 3, getting $x^2+y^2=34$ . we subtract the first equation from our equation we just got, getting $y^2=25$ y= (5,-5). once we plug that in, we get $50+x^2 = 59$ , $x^2 = 9$ x = (-3,3)

8 0

2 years ago