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2 years ago

1/2(6+3p)-3/4(8-10p)Expand and simplify show all steps

1 answer:

7 0

$\frac{6+3p}{2} - \frac{3}{4} (8 - 10p) \\ \\ \frac{6 + 3p}{2} - \frac{3(8-10p)}{4} \\ \\ \frac{6+3p}{2} - \frac{3 \times 2(4 - 5p)}{4} \\ \\ \frac{6+3p}{2} - \frac{6(4-5p)}{4} \\ \\ \frac{6+3p}{2} - \frac{3(4-5p)}{2} \\ \\ \frac{6+3p - 3(4-5p)}{2} \\ \\ \frac{3(2 +p - 4 + 5p}{2} \\ \\ \frac{3(2 + (p + 5p) - 4)}{2} \\ \\ \frac{3(2+6p-4)}{2} \\ \\ \frac{3 \times 2 (1 + 3p-2)}{2} \\ \\ \frac{3 \times 2(3p - 1)}{2} \\ \\ \frac{6(3p -1 }{2} \\ \\ 3(3p - 1) \\ \\ Answer: \fbox {3(3p-1)}$

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23 security systems a home security system is designed to have a 99% reliability rate. suppose that nine homes equipped with thi

Bess [88]

C. eight or fewer

99% means 99/100

. with 9 alarms there is no way you can trigger 8 alarms with that 99% rate

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2 years ago

What is 2 divided 1 6/7 ? Please answer

Aliun [14]

If you would like to divide 2 by 1 6/7, you can do this using the following steps:

1 6/7 = 13/7

2 / 1 6/7 = 2 / 13/7 = 2/1 / 13/7 = 2/1 * 7/13 = 14/13 = 1 1/13

The correct result would be 1 1/13.

6 0

2 years ago

A freight train completes its journey of 150 miles 1 hour earlier if its original speed is increased by 5 miles/hour what is the

vlabodo [156]

<span>A freight train completes its journey of 150 miles 1 hour earlier if its original speed is increased by 5 miles/hour. What is the train’s original speed?
***
let x=original speed
x+5=increased speed
travel time=distance/speed
..
lcd:x(x+5)
150(x+5)-150x=x(x+5)
150x+750-150x=x^2+5x
x^2+5x-750=0
(x-25)(x+30)=0
x=25
What is the train’s original speed? 25 mph</span>

7 0

2 years ago

Assume that x and y are both differentiable functions of t and find the required values of dy/dt and dx/dt. x2 + y2 = 25 (a) Fin

Artemon [7]

Answer:

(a) $\frac{dy}{dt}=-3\frac{3}{4}$

(b) $\frac{dx}{dt}=3\frac{3}{4}$

Step-by-step explanation:

$x^{2} +y^{2}=25$

Take $\frac{d}{dt}$ of of each term.

$\frac{d}{dt}(x^{2})+\frac{d}{dt}(y^{2})=\frac{d}{dt}(25)\\\\(\frac{d}{dx}(x^{2})*\frac{dx}{dt}) +(\frac{d}{dy}(y^{2})*\frac{dy}{dt})=\frac{d}{dt}(25)\\\\2x\frac{dx}{dt} +2y\frac{dy}{dt} = 0\\\\$

For Question a

$2y\frac{dy}{dt}=-2x\frac{dx}{dt}\\\\\frac{dy}{dt}=\frac{-2x\frac{dx}{dt}}{2y} \\\\\frac{dy}{dt}=-\frac{x}{y}\frac{dx}{dt}$

Given that x = 3, y = 4, and dx/dt = 5.

$\frac{dy}{dt}=-\frac{3}{4}*5=-\frac{15}{4}\\ \\\frac{dy}{dt}=-3\frac{3}{4}$

For Question b

$2x\frac{dx}{dt}=-2y\frac{dy}{dt}\\\\\frac{dx}{dt}=\frac{-2y\frac{dy}{dt}}{2x} \\\\\frac{dx}{dt}=-\frac{y}{x}\frac{dy}{dt}$

Given that x = 4, y = 3, and dx/dt = -5.

$\frac{dx}{dt}=-\frac{3}{4}*-5=\frac{15}{4}\\ \\\frac{dx}{dt}=3\frac{3}{4}$

5 0

2 years ago

A bin of 5 transistors is known to contain 2 that are defective. The transistors are to be tested, one at a time, until the defe

xxMikexx [17]

Answer:

$P(N_1 = a , N_2 = b)= \frac{1}{5-a C 1} * \frac{5-a C 1}{5C2} = \frac{1}{5C2}=\frac{1}{10}$

Step-by-step explanation:

For the random variable $N_1$ we define the possible values for this variable on this case $[1,2,3,4,5]$ . We know that we have 2 defective transistors so then we have 5C2 (where C means combinatory) ways to select or permute the transistors in order to detect the first defective:

$5C2 = \frac{5!}{2! (5-2)!}= \frac{5*4*3!}{2! 3!}= \frac{5*4}{2*1}=10$

We want the first detective transistor on the ath place, so then the first a-1 places are non defective transistors, so then we can define the probability for the random variable $N_1$ like this:

$P(N_1 = a) = \frac{5-a C 1}{5C2}$

For the distribution of $N_2$ we need to take in count that we are finding a conditional distribution. $N_2$ given $N_1 =a$ , for this case we see that $N_2 \in [1,2,...,5-a]$ , so then exist $5-a C 1$ ways to reorder the remaining transistors. And if we want b additional steps to obtain a second defective transistor we have the following probability defined:

$P(N_2 =b | N_1 = a) = \frac{1}{5-a C 1}$

And if we want to find the joint probability we just need to do this:

$P(N_1 = a , N_2 = b) = P(N_2 = b | N_1 = a) P(N_1 =a)$

And if we multiply the probabilities founded we got:

$P(N_1 = a , N_2 = b)= \frac{1}{5-a C 1} * \frac{5-a C 1}{5C2} = \frac{1}{5C2}=\frac{1}{10}$

8 0

2 years ago