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3 years ago

Let f(x) = 5x + 6 and g(x) = -7x - 4. Find f(x) + g(x)

Mathematics

1 answer:

sattari [20]3 years ago

4 0

5x + 6 + (-7x - 4)

5x + 6 - 7x - 4

-2x + 2

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What the value of r that makes the value true? r+(-0.14)=0.47

siniylev [52]

$r+(-0.14)=0.47\\\\r-0.14=0.47\qquad|\text{add 0.14 to both sides}\\\\r-0.14+0.14=0.47+0.14\\\\r=0.61$

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Brian, Chris, and Damien took a math test that had 20 questions. The number of questions Brian got right is 14 more than 1/4 the

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shusha [124]

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Step-by-step explanation:

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2 years ago

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Use the fact that the mean of a geometric distribution is μ= 1 p and the variance is σ2= q p2. A daily number lottery chooses th

butalik [34]

Answer:

a). The mean = 1000

The variance = 999,000

The standard deviation = 999.4999

b). 1000 times , loss

Step-by-step explanation:

The mean of geometric distribution is given as , $\mu = \frac{1}{p}$

And the variance is given by, $\sigma ^2=\frac{q}{p^2}$

Given : $p=\frac{1}{1000}$

= 0.001

The formulae of mean and variance are :

$\mu = \frac{1}{p}$

$\sigma ^2=\frac{q}{p^2}$

$\sigma ^2=\frac{1-p}{p^2}$

a). Mean = $\mu = \frac{1}{p}$

= $\mu = \frac{1}{0.001}$

= 1000

Variance = $\sigma ^2=\frac{1-p}{p^2}$

= $\sigma ^2=\frac{1-0.001}{0.001^2}$

= 999,000

The standard deviation is determined by the root of the variance.

$\sigma = \sqrt{\sigma^2}$

= $\sqrt{999,000}$ = 999.4999

b). We expect to have play lottery 1000 times to win, because the mean in part (a) is 1000.

When we win the profit is 500 - 1 = 499

When we lose, the profit is -1

Expected value of the mean μ is the summation of a product of each of the possibility x with the probability P(x).

$\mu=\Sigma\ x\ P(x)= 499 \times 0.001+(-1) \times (1-0.001)$

= $ 0.50

Since the answer is negative, we are expected to make a loss.

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2 years ago

Find the equation of locus of a point which moves such that its distance from (0,2) is one third distance from (-2,3). ( I WILL

scoray [572]

Answer:

$8(x^2+y^2)-4x-30y+23=0$

Step-by-step explanation:

<u />

<u>Distance formula</u>

$\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Let P(x, y) = any point on the locus

Let A = (0, 2)

Let B = (-2, 3)

If a point moves such that its distance from (0, 2) is one third distance from (-2, 3):

$PA=\dfrac{1}{3}PB$

Therefore, using the distance formula:

$\implies \sqrt{(x-0)^2+(y-2)^2}=\dfrac{1}{3}\sqrt{(x-(-2))^2+(y-3)^2}$

Square both sides:

$\implies x^2+(y-2)^2=\dfrac{1}{9}[(x+2)^2+(y-3)^2]$

$\implies x^2+y^2-4y+4=\dfrac{1}{9}(x^2+4x+4+y^2-6y+9)$

Multiply both sides by 9:

$\implies 9x^2+9y^2-36y+36=x^2+4x+4+y^2-6y+9$

$\implies 8x^2+8y^2-4x-30y+23=0$

$\implies 8(x^2+y^2)-4x-30y+23=0$

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2 years ago