Tentukankata ganti nama

In the field of quality control, the science of statistics is often used to determine if a process is "out of control". Suppose

devlian [24]

Answer:

probability = 0.008

probability = 0.0256

Explanation:

we know here probability of defective is 0.2

so probability of not defective is 1 - 0.2 = 0.8

as we know 3 item is arrive off process line in succession

so The probability that an item is defective is

as P(defective) = 0.20

as all item are independent so

probability that all three items are defective is

probability = 0.20 × 0.20 × 0.20 = 0.008

and

probability that exactly 3 of next 4 are defective

so number of way that can choose 3 out of 4 is

= $\frac{4!}{3! ( 4-3)!}$

= 4

so as all are independent probability is

probability = ( the number of way to choose 3 out of 4 ) × ( 3 item defective ) × ( 1 item not defective )

probability = $_4 C_3$ × 0.2³ × ( 1- 0.2)

probability = 4 × 0.008 × 0.8

probability = 0.0256

4 0

3 years ago

Coaches can receive a bonus from the booster club at the conclusion of a season.

inna [77]

Answer:

The statement is true

Explanation:

Booster club is the organization which is formed in order to support the organization, associated club and sports team. The club support through coordinating events or raising the funds.

So, yes, the coaches are legally could receive the bonus from the booster club at the season conclusion as the booster clubs are not state funded, so they are intended to use the money for the school resources.

And there is no legal matter is involves with giving the coaches bonuses as the booster club raise the fund from the extra circular activities.

4 0

3 years ago

a bacteria population starts with 200 bacteriaa and grows at a rate of r(t) = (450.268)e 1.12567t bacteria per hour. How many ba

Virty [35]

Answer:

11,513

Explanation:

Data provided in the question:

Initial bacteria = 200

Growth rate r(t) = $(450.268)e^{1.12567t}$

Now,

Total growth after 3 hours = $\int\limits^3_0 {(450.268)e^{1.12567t}} \, dt$

or

Total growth after 3 hours = $450.268\int\limits^3_0 {e^{1.12567t}} \, dt$

or

Total growth after 3 hours = $450.268[\frac{e^{1.12567t}}{1.12567}]^3_0$

[ ∵ $\int{d(e^x)}{dt}=\frac{e^x}{\frac{dx}{dt}}$ ]

Thus,

Total growth after 3 hours = 400 × $[{e^{1.12567t}]^3_0$

or

Total growth after 3 hours = 400 × $[{e^{1.12567(3)}-e^{1.12567(0)}]$

or

Total growth after 3 hours ≈ 11313

Hence,

Total bacteria after 3 hours = Initial bacteria + Total growth after 3 hours

= 200 + 11313

= 11,513

4 0

3 years ago

An alternative to CFL bulbs and incandescent bulbs are light-emitting diode (LED) bulbs. A 100 W incandescent bulb can be replac

mariarad [96]

Answer:

It cost $5.84 to run the LED bulb for one year if it runs for five hours a day.

Explanation:

E = Pt

= (16W)(365*5)

= 29200Wh

= 29.2 kWh

cost of operation = E($0.2/kWh)

= (29.2 kWh)($0.2/kWh)

= $5.84

Therefore, It cost $5.84 to run the LED bulb for one year if it runs for five hours a day.

8 0

3 years ago

Given the cost formula, y = $7,000 + $1.80x, total cost for an activity level of 4,000 units would be:

zaharov [31]

The total cost for this certain activity level can be calculated by substituting 4000 to the x of the equation given above,
y = 7000 + 1.8(4000)
y = 14200

Thus, the total cost of the activity level is $14,200.

4 0

3 years ago

Tentukankata ganti nama​

Tentukankata ganti nama