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Thepotemich [5.8K]

3 years ago

10

Ahmed, Walid and Faisal made $1446 profit from selling used books. If Ahmed to receive double amount of Faisal and Walid to rece

ive triple amount of Faisal. How much does each one receive? any help pls urgent

1 answer:

Strike441 [17]3 years ago

7 0

I would say take that a divided

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Suppose xy=−4 and dy/dt=−3. Find dx/dt when x=−1.

user100 [1]

When x=-1: $\quad (-1)y=-4\qquad\to\qquad y=4\)$

Ok that gives us a little more information.
If we implicitly differentiate with respect to t, from the very start, then we can apply our product rule, ya?

$x'y+xy'=0$

The right side is zero, derivative of a constant is zero.
Where x' is dx/dt and y' is dy/dt.

From here, plug in all the stuff you know:
y' = -3
x = -1
y = 4

and solve for x'.

Hope that helps!

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3 years ago

One thirdOne third of a number is added to 77, giving a result of at least 33.

alexgriva [62]

Answer:

in my ipinion...

Step-by-step explanation:

let a no.be x

1/3*x+77=33

1/3*x=-44

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2 years ago

How are sequences whose consecutive terms differ by a constant modeled, and what can those models tell us?

alex41 [277]

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Step-by-step explanation:

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2 years ago

At a fabric store, the probability that a customer buys thread is 0.15. The probability that a customer buys a needle given that

Sveta_85 [38]

Answer:

The correct statement is:

C. Buying a needle and buying a thread are dependent events.

Explanation:

The first statement is incorrect because the given events are dependent on each other because the probability that a customer buys a needle given that the customer buys thread is 0.25.

The second statement is incorrect because we cannot say all those who buy a thread also buy a needle.

The fourth statement is incorrect because the probability that a customer buys a needle and thread is $0.25 \times 0.15=0.0375$

8 0

3 years ago

Suiting at 6 a.m., cars, buses, and motorcycles arrive at a highway loll booth according to independent Poisson processes. Cars

dem82 [27]

Answer:

Step-by-step explanation:

From the information given:

the rate of the cars = $\dfrac{1}{5} \ car / min = 0.2 \ car /min$

the rate of the buses = $\dfrac{1}{10} \ bus / min = 0.1 \ bus /min$

the rate of motorcycle = $\dfrac{1}{30} \ motorcycle / min = 0.0333 \ motorcycle /min$

The probability of any event at a given time t can be expressed as:

$P(event \ (x) \ in \ time \ (t)\ min) = \dfrac{e^{-rate \times t}\times (rate \times t)^x}{x!}$

∴

(a)

$P(2 \ car \ in \ 20 \ min) = \dfrac{e^{-0.20\times 20}\times (0.2 \times 20)^2}{2!}$

$P(2 \ car \ in \ 20 \ min) =0.1465$

$P ( 1 \ motorcycle \ in \ 20 \ min) = \dfrac{e^{-0.0333\times 20}\times (0.0333 \times 20)^1}{1!}$

$P ( 1 \ motorcycle \ in \ 20 \ min) = 0.3422$

$P ( 0 \ buses \ in \ 20 \ min) = \dfrac{e^{-0.1\times 20}\times (0.1 \times 20)^0}{0!}$

$P ( 0 \ buses \ in \ 20 \ min) = 0.1353$

Thus;

P(exactly 2 cars, 1 motorcycle in 20 minutes) = 0.1465 × 0.3422 × 0.1353

P(exactly 2 cars, 1 motorcycle in 20 minutes) = 0.0068

(b)

the rate of the total vehicles = 0.2 + 0.1 + 0.0333 = 0.3333

the rate of vehicles with exact change = rate of total vehicles × P(exact change)

$= 0.3333 \times \dfrac{1}{4}$

= 0.0833

∴

$P(zero \ exact \ change \ in \ 10 minutes) = \dfrac{e^{-0.0833\times 10}\times (0.0833 \times 10)^0}{0!}$

P(zero exact change in 10 minutes) = 0.4347

c)

The probability of the 7th motorcycle after the arrival of the third motorcycle is:

$P( 4 \ motorcyles \ in \ 45 \ minutes) =\dfrac{e^{-0.0333\times 45}\times (0.0333 \times 45)^4}{4!}$

$P( 4 \ motorcyles \ in \ 45 \ minutes) =0.0469$

Thus; the probability of the 7th motorcycle after the arrival of the third one is = 0.0469

d)

P(at least one other vehicle arrives between 3rd and 4th car arrival)

= 1 - P(no other vehicle arrives between 3rd and 4th car arrival)

The 3rd car arrives at 15 minutes

The 4th car arrives at 20 minutes

The interval between the two = 5 minutes

<u>For Bus:</u>

P(no other vehicle other vehicle arrives within 5 minutes is)

= $\dfrac{6}{12} = 0.5$

<u>For motorcycle:</u>

$= \dfrac{2 }{12} = \dfrac{1 }{6}$

∴

The required probability = $1 - \Bigg ( \dfrac{e^{-0.5 \times 0.5^0}}{0!} \times \dfrac{e^{-1/6}\times (1/6)^0}{0!} \Bigg)$

= 1- 0.5134

= 0.4866

6 0

3 years ago