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3 years ago

HELP ASAP PLEASE I NEED ANS N SOLUTION

Mathematics

1 answer:

Blababa [14]3 years ago

5 0

Given:

A bowl contains 25 chips numbered 1 to 25.

A chip is drawn randomly from the bowl.

To find:

The probability that it is

a. 9 or 10?

b. even or divisible by 3?

c. divisible by 5 and divisible by 10?

Solution:

a. We have,

Number of total chips = 25

Favorable out comes are either 9 or 10. So,

Number of favorable outcomes = 2

The probability that the selected chip is either 9 or 10 is:

$\text{Probability}=\dfrac{\text{Number of favorable outcomes}}{\text{Number of total outcomes}}$

$\text{Probability}=\dfrac{2}{25}$

Therefore, the probability that the selected chip is either 9 or 10 is $\dfrac{2}{25}$ .

b. The numbers that are even from 1 to 25 are 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24.

The numbers from 1 to 25 that are divisible by 3 are 3, 6, 9, 12, 15, 18, 21, 24.

The numbers that are either even or divisible by 3 are 2, 3, 4, 6, 8, 9, 10, 12, 14, 15, 16, 18, 20, 21, 22, 24.

Number of favorable outcomes = 16

The probability that the selected chip is either even or divisible by 3 is:

$\text{Probability}=\dfrac{16}{25}$

Therefore, the probability that the selected chip is either even or divisible by 3 is $\dfrac{16}{25}$ .

c. The numbers from 1 to 25 that are divisible by 5 are 5, 10, 15, 20, 25.

The numbers from 1 to 25 that are divisible by 10 are 10, 20.

The numbers that are divisible by both 5 and 10 are 10 and 20.

Number of favorable outcomes = 2

The probability that the selected chip is divisible by 5 and divisible by 10 is:

$\text{Probability}=\dfrac{2}{25}$

Therefore, the probability that the selected chip is divisible by 5 and divisible by 10 is $\dfrac{2}{25}$ .

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The annual inventory cost C for a manufacturer is given below, where Q is the order size when the inventory is replenished. Find

shtirl [24]

Answer:

Change in annual cost is 1.63 (decreasing).

Instantaneous rate of change is 1.65 (decreasing).

Step-by-step explanation:

Given that,

$C=\dfrac{1017000}{Q}+6.8\:Q$

The annual change in cost is given by,

$Change\:in\:C=C\left(348\right)-C\left(347\right)$

Calculate the value of cost at Q = 347 and Q =348 by substituting the value in cost function.

Calculating value of C at Q=347,

$C=\dfrac{1017000}{347}+6.8\left(347\right)$

$C=5290.44$

Calculating value of C at Q=348,

$C=\dfrac{1017000}{348}+6.8\left(348\right)}$

$C= 5288.81$

Substituting the value,

$Change\:in\:C=5288.81-5290.44$

$Change\:in\:C=-1.63$

Negative sign indicate that there is decrease in annual cost when Q is increased from 347 to 348

Therefore, change is annual cost is 1.63 (decreasing).

Instantaneous rate of change is given by the formula,

$f’\left(x\right)=\lim_{h\rightarrow 0}\dfrac{f\left(x+h\right)-f\left(x\right)}{h}$

Rewriting,

$C’\left(Q\right)=\lim_{h\rightarrow 0}\dfrac{C\left(Q+h\right)-C\left(Q\right)}{h}$

Now calculate \dfrac{C\left(Q+h\right) by substituting the value Q+h in cost function,

$C\left(Q+h\right)= \dfrac{1017000}{Q+h}+6.8\left(Q+h\right)$

Therefore,

$C’\left(Q\right)= \lim _{h\to 0}\:\dfrac{\dfrac{1017000}{Q+h}+6.8\left(Q+h\right)-\left(\dfrac{1017000}{Q}+6.8Q\right)}{h}$

By using distributive law,

$C’\left(Q\right) = \lim _{h\to 0}\:\dfrac{\dfrac{1017000}{Q+h}+6.8Q+6.8h-\dfrac{1017000}{Q}-6.8Q}{h}$

Cancelling out common factors,

$C’\left(Q\right) = \lim _{h\to 0}\:\dfrac{\dfrac{1017000}{Q+h}-\dfrac{1017000}{Q}+6.8h}{h}$

Now, LCD of $\left(Q+h\right)$ and $Q$ is $Q(Q+h)$ . So multiplying first term by $Q$ and second term by

Therefore,

$C’\left(Q\right) = \lim _{h\to 0}\:\dfrac{\dfrac{1017000}{Q+h}\times\dfrac{Q}{Q}-\dfrac{1017000}{Q}\times\dfrac{Q+h}{Q+h}+6.8h}{h}$

Simplifying,

$C’\left(Q\right) = \lim _{h\to 0}\:\dfrac{\dfrac{1017000Q}{\left(Q+h\right)\left(Q\right)}-\dfrac{1017000\left(Q+h\right)}{Q\left(Q+h\right)}+6.8h}{h}$

$C’\left(Q\right) = \lim _{h\to 0}\:\dfrac{\dfrac{1017000Q-1017000\left(Q+h\right)}{\left(Q+h\right)\left(Q\right)}+6.8h}{h}$

$C’\left(Q\right) = \lim _{h\to 0}\:\dfrac{\dfrac{1017000Q-1017000Q-1017000h}{\left(Q+h\right)\left(Q\right)}+6.8h}{h}$

$C’\left(Q\right) = \lim _{h\to 0}\:\dfrac{\dfrac{-1017000h}{\left(Q+h\right)\left(Q\right)}+6.8h}{h}$

Factoring out h from numerator,

$C’\left(Q\right) = \lim _{h\to 0}\:\dfrac{h\left(\dfrac{-1017000}{\left(Q+h\right)\left(Q\right)}+6.8\right)}{h}$

Cancelling out h,

$C'\left(Q\right) = \lim_{h\to 0}\: -\dfrac{1017000}{\left(Q+h\right)Q}+6.8$

Calculating the limit by plugging value h = 0,

$C'\left(Q\right) = -\dfrac{1017000}{\left(Q+0\right)Q}+6.8$

$C'\left(Q\right) = -\dfrac{1017000}{Q^{2}}+6.8$

Given that Q=347,

$C'\left(347\right) = -\dfrac{1017000}{347^{2}}+6.8$

$C'\left(347\right) = -1.65$

Negative sign indicate that there is decrease in instantaneous rate when Q is 347

Therefore, instantaneous rate of change at Q=347 is 1.65 (decreasing).

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