WHO WANTS BRANLIEST?!?!?!?!?!?!?!?

Answers

Answer 1

Answer:

Answer:

I Do

Step-by-step explanation:

Related Questions

Write an inequality for a game that allows 2 and not more than 4 players
I'm having trouble with finding the solutions.
Line A y= 2x + 3 is parallel to another line B, what is the slope of the line B?
The following figures are similar, find the value of x
This is a two-part question: A 12-pack of bottled mineral water sells for $3.85. For the following,state whether to use multiplication or division for the specified amount:a) The cost of one bottle of water:b) The cost of four 12-packs of water:

What are the zeros of the function below? Check all that apply

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Answer:

Step-by-step explanation:

Hope this helps man i have the same answer rn and im in middle school

Which is an example of a situation that is in equilibrium?A. The amount of air in a room increases quickly when the door is
opened.
B. The amount of money in a bank account never changes
C. The amount of water in a cup decreases as it evaporates
D. A flower slowly grows taller

Answers

Answer:B the amount of money in a bank account never changes.

Step-by-step explanation:

Answer:

B. The amount of money in a bank account never changes.

Step-by-step explanation:

Equilibrium is achieved when the state of a reversible reaction of opposing forces cancel each other out. While in a state of equilibrium, the competing influences are balanced out. Imagine a cup with a hole in it being filled with water from a tap. The level of water in this cup would stay the same if the rate at which the water that flows inside is the same as the water that flows outside. Option B will be the correct answer because the amount of money going into the account is at the same rate of money coming out of the account.

Find the seventh term of thegeometric sequence, given the
first term and common ratio.
a_=1 and r=-2/3
[?]

Answers

Answer:

$T_7 = (64)/(729)$

Step-by-step explanation:

Given

$a =1$

$r = (2)/(3)$

Required

Determine the 7th term

The nth term of a gp is:

$T_n = a * r^{n-1$

So, we have:

$T_7 = 1 * (2)/(3)^{7-1$

$T_7 = 1 * (2)/(3)^{6$

$T_7 = 1 * (2^6)/(3^6)$

$T_7 = 1 * (64)/(729)$

$T_7 = (64)/(729)$

Disk requests come in to the disk driver for cylinders 10, 22, 20, 2, 40, 6, and 38, in that order. A seek takes 6 msec per cylinder. How much seek time is needed for (a) First-come, first served. (b) Closest cylinder next. (c) Elevator algorithm (initially moving upward). In all cases, the arm is initially at cylinder 20.

Answers

Answer:

Step-by-step explanation:

The order for FCFS is: 20->10->22->20->2->40->6->38.

Distance is

10+12+2+18+38+34+32 = 146

cylinders, so time is

146* 6 = 876 sec.

The order for elevator is:

20->20->22->38->40->10->6->2.

Distance is

0+2+16+2+30+4+4 = 58

cylinders, so time is

58 * 6 =348 msec.

The order for CCN is:

20->20->22->10->6->2->38->40.

Distance is

0+2+12+4+4+36+2 = 60

cylinders, so time is 60 * 6 =360 msec.

A manufacturing company regularly conducts quality control checks at specified periods on the products it manufactures. Historically, the failure rate for LED light bulbs that the company manufactures is 5%. Suppose a random sample of 10 LED light bulbs is selected. What is the probability that a) None of the LED light bulbs are defective? b) Exactly one of the LED light bulbs is defective? c) Two or fewer of the LED light bulbs are defective? d) Three or more of the LED light bulbs are not defective?

Answers

Answer:

a) There is a 59.87% probability that none of the LED light bulbs are defective.

b) There is a 31.51% probability that exactly one of the light bulbs is defective.

c) There is a 98.84% probability that two or fewer of the LED light bulbs are defective.

d) There is a 100% probability that three or more of the LED light bulbs are not defective.

Step-by-step explanation:

For each light bulb, there are only two possible outcomes. Either it fails, or it does not. This means that we use the binomial probability distribution to solve this problem.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_(n,x).p^(x).(1-p)^(n-x)$

In which $C_(n,x)$ is the number of different combinatios of x objects from a set of n elements, given by the following formula.

$C_(n,x) = (n!)/(x!(n-x)!)$

And p is the probability of X happening.

In this problem we have that:

$n = 10, p = 0.05$

a) None of the LED light bulbs are defective?

This is P(X = 0).

$P(X = x) = C_(n,x).p^(x).(1-p)^(n-x)$

$P(X = 0) = C_(10,0)*(0.05)^(0)*(0.95)^(10) = 0.5987$

There is a 59.87% probability that none of the LED light bulbs are defective.

b) Exactly one of the LED light bulbs is defective?

This is P(X = 1).

$P(X = x) = C_(n,x).p^(x).(1-p)^(n-x)$

$P(X = 1) = C_(10,1)*(0.05)^(1)*(0.95)^(9) = 0.3151$

There is a 31.51% probability that exactly one of the light bulbs is defective.

c) Two or fewer of the LED light bulbs are defective?

This is

$P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)$

$P(X = 2) = C_(10,2)*(0.05)^(2)*(0.95)^(8) = 0.0746$

$P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.5987 + 0.3151 + 0.0746 0.9884$

There is a 98.84% probability that two or fewer of the LED light bulbs are defective.

d) Three or more of the LED light bulbs are not defective?

Now we use p = 0.95.

Either two or fewer are not defective, or three or more are not defective. The sum of these probabilities is decimal 1.

$P(X \leq 2) + P(X \geq 3) = 1$

$P(X \geq 3) = 1 - P(X \leq 2)$

In which

$P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)$

$P(X = 0) = C_(10,0)*(0.95)^(0)*(0.05)^(10)\cong 0$

$P(X = 1) = C_(10,1)*(0.95)^(1)*(0.05)^(9) \cong 0$

$P(X = 2) = C_(10,1)*(0.95)^(2)*(0.05)^(8) \cong 0$

$P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0$

$P(X \geq 3) = 1 - P(X \leq 2) = 1$

There is a 100% probability that three or more of the LED light bulbs are not defective.

Final answer:

The question relates to binomial distribution in probability theory. The probabilities calculated include those of none, one, two or less, and three or more LED bulbs being defective out of a random sample of 10.

Explanation:

This question relates to the binomial probability distribution. A binomial distribution is applicable because there are exactly two outcomes in each trial (either the LED bulb is defective or it's not) and the probability of a success remains consistent.

a) In this scenario, 'none of the bulbs being defective' means 10 successes. The formula for probability in a binomial distribution is p(x) = C(n, x) * [p^x] * [(1-p)^(n-x)]. Plugging in the values, we find p(10) = C(10, 10) * [0.95^10] * [0.05^0] = 0.5987 or 59.87%.

b) 'Exactly one of the bulbs being defective' implies 9 successes and 1 failure. Following the same formula, we get p(9) = C(10, 9) * [0.95^9] * [0.05^1] = 0.3151 or 31.51%.

c) 'Two or less bulbs being defective' means 8, 9 or 10 successes. We add the probabilities calculated in (a) and (b) with that of 8 successes to get this probability. Therefore, p(8 or 9 or 10) = p(8) + p(9) + p(10) = 0.95.

d) 'Three or more bulbs are not defective' means anywhere from 3 to 10 successes. As the failure rate is low, it's easier to calculate the case for 0, 1 and 2 successes and subtract it from 1 to find this probability. This gives us p(>=3) = 1 - p(2) - p(1) - p(0) = 0.98.

Learn more about Binomial Probability here:

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