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aev [14]
3 years ago
14

1. The sum of x and -45 is at most 35

Mathematics
1 answer:
Lorico [155]3 years ago
5 0

Answer:

The given set of equation are:   x+ (-45) ≤  35  ,  x - (-45) ≥  35

For the given equality to be true, x  ≤ 35

Step-by-step explanation:

Here, given the first number = x

Second number  = -45

Now, Sum of x  and - 45 is at most 35.

⇒ x+ (-45) ≤  35

Also,  The difference of x and -45 is at least 35.

⇒ x - (-45) ≥  35

Now, simplifying the given set of equations:

x  - 45  ≤  35  ⇒    -x - (-45) >   - 35   ( as 3  <  4  ⇒ -3  > -4)

or, -x  + 45  >  - 35

and second equation is x  + 45 ≥  35

Now, solving both the equations by not taking sign of inequality in to the consideration, we get

x - 45  = 35

x + 45 = 35  

Adding both equations,we get:   ⇒ 2x = 70

                                                          or x = 35

Hence for the given equality to be true, x  ≤ 35

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Anvisha [2.4K]

The answer would be 9v ^4 + 27v ^3 - 36v^2

Step-by-step explanation:

You first would multiply each term in the parenthesis by 9v^2

9v^2 (v^2) + 9v^2 (3v) - 9v^2 (4)

After this you combine the like terms

9v^2 * v^2 is 9v^4

9v^2  * 3v is 27v^3

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Because you can not simplify it any smaller, this is your final answer

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Step-by-step explanation:

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A construction company is considering submitting bids for contracts of three different projects. The company estimates that it h
julsineya [31]

Answer:

a.P(x)=\frac{n!}{x!(n-x)!}*p^{x}*(1-p)^{n-x}\\

b. E(x) = 0.3

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Step-by-step explanation:

The probability that the company won x bids follows a binomial distribution because we have n identical and independent experiments with a probability p of success and (1-p) of fail.

So, the PMF of X is equal to:

P(x)=\frac{n!}{x!(n-x)!}*p^{x}*(1-p)^{n-x}\\

Where p is 0.1 and it is the chance of winning. Additionally, n is 3 and it is the number of bids. So the PMF of X is:

P(x)=\frac{3!}{x!(3-x)!}*0.1^{x}*(1-0.1)^{n-x}\\

For binomial distribution:

E(x)=np\\S(x)=\sqrt{np(1-p)}

Therefore, the company can expect to win 0.3 bids and it is calculated as:

E(x) = np = 3*0.1 = 0.3

Additionally, the standard deviation of the number of bids won is:

S(x)=\sqrt{np(1-p)}=\sqrt{3(0.1)(1-0.1)}=0.5196

Finally, the probability to won 1, 2 or 3 bids is equal to:

P(1)=\frac{3!}{1!(3-1)!}*0.1^{1}*(1-0.1)^{3-1}=0.243\\P(2)=\frac{3!}{2!(3-2)!}*0.1^{2}*(1-0.1)^{3-2}=0.027\\P(3)=\frac{3!}{3!(3-3)!}*0.1^{3}*(1-0.1)^{3-3}=0.001

So, the expected profit for the company is equal to:

E=-10,000+50,000(0.243)+100,000(0.027)+150,000(0.001)\\E=5,000

Because there is a probability of 0.243 to win one bid and it will produce 50,000 of income, there is a probability of 0.027 to win 2 bids and it will produce 100,000 of income and there is a probability of 0.001 to win 3 bids and it will produce 150,000 of income.

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pychu [463]

Answer:

<h3>n = 1/4</h3>

Step-by-step explanation:

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