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Answer 1
Answer: Ok ask question.......................

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Help ASAP Add these expressions(3x+15) + (4x+13)

MO bisects ∠LMN, m∠NMO =6x-20,and 2x+36. Solve for x and find m∠LMO

Answers

Answer:

fwe

Step-by-step explanation:

One normally distributed with mean value 20 in. and standard deviation .5 in. The length of the second piece is a normal rv with mean and standard deviation 15 in. and .4 in., respectively. The amount of overlap is normally distributed with mean value 1 in. and standard deviation .1 in. Assuming that the lengths and amount of overlap are independent of each other, what is the probability that the total length after insertion is between 34.5 and 35 in.

Answers

Answer:

The probability that the the total length after insertion is between 34.5 and 35 inches is 0.1589.

Step-by-step explanation:

Let the random variable X represent the length of the first piece, Y represent the length of the second piece and Z represents the overlap.

It is provided that:

X\sim N(20,\ 0.50^(2))\nY\sim N(15,\ 0.40^(2))\nZ\sim N(1,\ 0.10^(2))

It is provided that the lengths and amount of overlap are independent of each other.

Compute the mean and standard deviation of total length as follows:

E(T)=E(X+Y-Z)\n=E(X)+E(Y)-E(Z)\n=20+15-1\n=34

SD(T)=√(V(X+Y-Z))\n=√(V(X)+V(Y)+V(Z))\n=\sqrt{0.50^(2)+0.40^(2)+0.10^(2)}\n=0.6480741\n\approx 0.65

Since X, Y and Z all follow a Normal distribution, the random variable T, representing the total length will also follow a normal distribution.

T\sim N(34, 0.65^(2))

Compute the probability that the the total length after insertion is between 34.5 and 35 inches as follows:

P(34.5<T<35)=P((34.5-34)/(0.65)<(T-\mu_(T))/(\sigma_(T))<(35-34)/(0.65))\n\n=P(0.77<Z<1.54)\n\n=P(Z<1.54)-P(Z<0.77)\n\n=0.93822-0.77935\n\n=0.15887\n\n\approx 0.1589

*Use a z-table.

Thus, the probability that the the total length after insertion is between 34.5 and 35 inches is 0.1589.

Select a discrete probability distribution and present a real-life application of that distribution. Interpret the expected value and the standard deviation of your selected distribution within the context of the real-life example that you have selected, and describe how these values can be used by enterprise decision-makers.

Answers

Answer:

If a new product wants to be tested by a company and decides to show 50 samples of this product to 50 selected customers. The company estimates that the probability that the customer buys the product is 0.67, the objective is to determine approximately how many people expect to buy the product.

Let X the random variable of interest "Number of people that will buy a selected product", on this case we now that:  

X \sim Binom(n=50, p=0.67)  

The expected value is given by this formula:

E(X) = np=50*0.67=33.50

And the standard deviation for the random variable is given by:

sd(X)=√(np(1-p))=√(50*0.67*(1-0.67))=3.32

So then they can conclude that for each group of 50 people they expect that about 33-34 peoploe will buy the product with a standard deviation of 3.32.

Step-by-step explanation:

Previous concepts

A Bernoulli trial is "a random experiment with exactly two possible outcomes, "success" and "failure", in which the probability of success is the same every time the experiment is conducted". And this experiment is a particular case of the binomial experiment.

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

The probability mass function for the Binomial distribution is given as:  

P(X)=(nCx)(p)^x (1-p)^(n-x)  

Where (nCx) means combinatory and it's given by this formula:  

nCx=(n!)/((n-x)! x!)  

Solution to the problem

If a new product wants to be tested by a company and decides to show 50 samples of this product to 50 selected customers. The company estimates that the probability that the customer buys the product is 0.67, the objective is to determine approximately how many people expect to buy the product.

Let X the random variable of interest "Number of people that will buy a selected product", on this case we now that:  

X \sim Binom(n=50, p=0.67)  

The expected value is given by this formula:

E(X) = np=50*0.67=33.50

And the standard deviation for the random variable is given by:

sd(X)=√(np(1-p))=√(50*0.67*(1-0.67))=3.32

So then they can conclude that for each group of 50 people they expect that about 33-34 peoploe will buy the product with a standard deviation of 3.32.

A. X=1
B. X=3
C. X= 1,3
D. X= 0,3

Answers

The answer is A, X=1
Hope this helps
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Please help.

2/3x - 4 = -2

A) -3
B) -4
C) -9
D) 3

Answers

Answer:

D) 3

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............ that is the correct options

Answer:

D) 3

be confident to answer any questions

A typical person has an average heart rate of 70.0 70.0 beats/min. Calculate the given questions. How many beats does she have in 6.0 6.0 years? How many beats in 6.00 6.00 years? And finally, how many beats in 6.000 6.000 years? Pay close attention to significant figures in this question.

Answers

Answer:

a) 2.2 × 10⁸ beats

b) 2.20 × 10⁸ beats

c) 2.207 × 10⁸ beats

Step-by-step explanation:

Data provided in the question:

Average heart rate of a typical person = 70.0 beats/min

Now,

In the given cases, the significance is on the significant figures after the decimal

Therefore,

the answer is will be provided accordingly

Now,

a) Time = 6.0 years

[since 1 significant figure after decimal. answer will be give in  1 significant figure after decimal ]

time in minutes = 6.0 × 365 × 24 × 60

= 3.1 × 10⁶ minutes

Total beats = Average heart rate × Time

= 70 × 3.1 × 10⁶

= 2.2 × 10⁸ beats

b)  Time = 6.00 years

[since 2 significant figure after decimal. answer will be give in 2 significant figure after decimal ]

time in minutes = 6.00 × 365 × 24 × 60

= 3.15 × 10⁶ minutes

Total beats = Average heart rate × Time

= 70 × 3.15 × 10⁶

= 2.20 × 10⁸ beats

c) Time = 6.000 years

[since 3 significant figure after decimal. answer will be give in 3 significant figure after decimal ]

time in minutes = 6.000 × 365 × 24 × 60

= 3.154 × 10⁶ minutes

Total beats = Average heart rate × Time

= 70 × 3.154 × 10⁶

= 2.207 × 10⁸ beats
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