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4 years ago

Solve the equation for c. d=a+b+c/ac

1 answer:

7 0

Verrry interesting !
In the fraction at the end of the equation ... c/ac ... the numerator and denominator can both be divided by 'c', and the fraction becomes 1/a .

Then the equation becomes d = a + b + 1/a

The equation can't be solved for 'c', because there is no relationship between 'c' and any of the other terms. 'c' can be anything at all, and it has no effect on anything else in the equation.

Or, to put a finer point on it, 'c' is not a function of 'a', 'b', or 'd' .

You might be interested in

How many solutions does the following system have?

Allisa [31]

No solution since the two equations do not intersect

7 0

3 years ago

Please help when there are two x's i get confused. answer #3 with an answer!

igomit [66]

Answer:

x = 1

Step-by-step explanation:

We know that the two angles must be equal to each other. So:

65x - 12 = 43x + 10

First add 12 to both sides:

65x - 12 + 12 = 43x + 10 + 12

65x = 43x + 22

Then subtract 43x from both sides:

65x - 43x = 43x - 43x + 22

22x = 22

x = 1

7 0

3 years ago

What is the value of X

zhannawk [14.2K]

Answer:

27=x

Step-by-step explanation:

We can use the triangle angle bisector theorem

x+8 2x-5

--------- = ------------

10 14

Using cross products

14(x+8) = 10 (2x-5)

Distribute

14x +112 = 20x -50

Subtract 14x from each side

14x-14x +112 = 20x-14x -50

112 = 6x-50

Add 50 to each side

112+50 = 6x-50+50

162 = 6x

Divide by 6

162/6 = 6x/6

27=x

3 0

3 years ago

Ship collisions in the Houston Ship Channel are rare. Suppose the number of collisions are Poisson distributed, with a mean of 1

alexandr1967 [171]

Answer:

$a) \simeq 0.3012$ $b) \simeq 0.0494$ c) $\simeq 0.2438$

Step-by-step explanation:

Rate of collision,

1.2 collisions every 4 months

or, $\frac{1.2}{4}$

= 0.3 collisions per month

So, the Poisson distribution for the random variable no. of collisions per month (X) is given by,

P(X =x) = $\frac{e^{-\lambda}\times {\lambda}^{x}}}{x!}
$

for x ∈ N ∪ {0}

= 0 otherwise --------------------------------------(1)

here, $\lambda = 0.3$ collision / month

No collision over a 4 month period means no collision per month or X =0

Putting X = 0 in (1) we get,

P(X = 0) = $\frac{e^{-0.3}\times {\0.3}^{0}}{0!}
$

$\simeq 0.7408182207$ ------------------------------------(2)

Now, since we are calculating this for 4 months,

so, P(No collision in 4 month period)

= $0.7408182207^{4}$

$\simeq 0.3012$ -----------------------------------------------------------(3)

2 collision in 2 month period means 1 collision per month or X =1

Putting X =1 in (1) we get,

P(X =1) = $\frac{e^{-0.3}\times {\0.3}^{1}}{1!}
$

$\simeq 0.2222454662$ ------------------------------------(4)

Now, since we are calculating this for 2 months, so ,

P(2 collisions in 2 month period)

= $0.2222454662^{2}$

$\simeq 0.0494$ -----------------------------------------(5)

1 collision in 6 months period means

$\frac{1}{6}$ collision per month

Now, P(1 collision in 6 months period)

= P( X = 1/6] (which is to be estimated)

= $\frac {P(X=0)\times 5 + P(X =1)\times 1}{6}$

= \frac {0.7408182207 \times 5 + 0.2222454662 \times 1}{6}[/tex]

$\simeq 0.6543894283$ -------------------------------------------(6)

So,

P(1 collision in 6 month period)

= $0.6543894283^{6}$

$\simeq 0.0785267444$ ------------------------------------------------(7)

So,

P(No collision in 6 months period)

= $(P(X =0)^{6}$

$\simeq 0.1652988882$ ---------------------------------(8)

so,

P(1 or fewer collision in 6 months period)

= (8) + (7 ) = 0.0785267444 +0.1652988882

$\simeq 0.2438$ ---------------------------------------------(9)

7 0

3 years ago

(doing a bunch of these so if you can't answer the first time, you will have another chance to)

posledela

Answer:

thanks your amazing

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers