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Aleks [24]
2 years ago
9

How do you evaluate 1.5x + 120 ≥ 270

Mathematics
1 answer:
Y_Kistochka [10]2 years ago
5 0

1.5x + 120 ≥ 270

1.5x ≥ 150

x ≥ 100

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good job on increasing our team sales from 90 to 130 units per day.” wow i never thought that we could increase our sales by
artcher [175]

The sales was increased by 44.4%.

Step-by-step explanation:

  • Lets do this cumulative
  • 90 of 10 % =  9  which is 99
  • 90 of 10 % = 9 which is 108
  • 90 of 10% = 9 which is 117
  • 90 of 10% = 9 which is 126
  • 90 of 4 % = 3.6 which is 129.6
  • 90 of 0.044% = 0.0396 which is 129.99
  • This comes to nearly 44.44 to be closer there was an increase.
  • Always rule 1 approximation use 50% of 90 if it is above 130.
  • Down it to 40% of 90 see if the sales is lower than 130.
  • It should always start with 50,40,30....10 percents.
  • Alternative should be 10,1 or decimals.
  • Approximation using decimals rounding up becomes simple.

7 0
1 year ago
At a financial institution, a fraud detection system identifies suspicious transactions and sends them to a specialist for revie
labwork [276]

Answer:

a. E(X) = 54.4

b. E(X) = 2.5

c. P(Y=2) = .0116

Step-by-step explanation:

a.

    E(X) = np = .40 probability * 136 trials = 54.4 blocked transmissions

    To get the expected value, we simply multiply probability times number of trials. You can look at it in simple terms by thinking if there's a 50% chance of flipping heads and you flip a coin twice, in an ideal world you will have .5*2 = 1 head.

b.

    i. Let X represent the number of suspicious transmissions reviewed until finding the first blocked one. We will use a geometric distribution to model the "first" transmission. Whenever we're looking for the "first" time something happens, we use geometric.

   ii. E(X) = 1/p , according to the geometric model.

              = 1/.4 = 2.5.

       We expect that the specialist will review 2.5 suspicious transactions <em>on average </em>before finding the first transmission that will be blocked.

c.

    i. Let Y represent the exact number of blocked transmissions out of 10. We will use a binomial distribution to model the "fixed" number of transmissions. Whenever we're looking for a "fixed" number of times something happens, we use binomial.

    ii. P(Y=k) = (n choose k)(p^k)(q^n-k)

        P(Y=2) = (¹⁰₂)(.4^2)(.6^10-2)

                    = 45 (.4^2)(.6^10-2) = .0016

        As for calculator notation, the n choose k can be accessed on a TI-84 via MATH -> PRB -> nCr. It looks like 10 nCr 2 on the display.

        Hence the probability that two transactions out of ten will be blocked is .0016 by the binomial model.

5 0
2 years ago
A random sample of 50 units is drawn from a production process every half hour. the true fraction of nonconforming products manu
Naya [18.7K]

solution:

The probability mass function for binomial distribution is,

 

Where,

X=0,1,2,3,…..; q=1-p

find the probability that (p∧ ≤ 0.06) , substitute the values of sample units (n) , and the probability of nonconformities (p) in the probability mass function of binomial distribution.

Consider x   to be the number of non-conformities. It follows a binomial distribution with n   being 50 and p  being 0.03. That is,

binomial (50,0.02)

Also, the estimate of the true probability is,

p∧  = x/50

The probability mass function for binomial distribution is,

 

Where,

X=0,1,2,3,…..; q=1-p

The calculation is obtained as

P(p^ ≤ 0.06) = p(x/20 ≤ 0.06)

         = 50cx ₓ (0.03)x ₓ (1-0.03)50-x  

=    (50c0 ₓ (0.03)0 ₓ (1-0.03)50-0 + 50c1(0.03)1 ₓ (1-0.03)50-1 + 50c2 ₓ (0.03)2 ₓ (1-0.03)50-2 +50c3 ₓ      (0.03)3 ₓ (1- 0.03)50-3 )

=(   ₓ (0.03)0 ₓ (1-0.03)50-0 +  ₓ (0.03)1 ₓ (1-0.03)50-1 +   ₓ (0.03)2 ₓ (1-0.03)50-2   ₓ (0.03)3 ₓ (1-0.03)50-3 )



5 0
2 years ago
Which oh the following are among the five basic postulates of Euclidean geometry
igomit [66]

1. A straight line segment can be drawn joining any two points.

2. Any straight line segment can be extended indefinitely in a straight line.

3. Given any straight line segment, a circle can be drawn having the segment as radius and one endpoint as center.

4. All right angles are congruent.

5. If two lines are drawn which intersect a third in such a way that the sum of the inner angles on one side is less than two right angles, then the two lines inevitably must intersect each other on that side if extended far enough. This postulate is equivalent to what is known as the parallel postulate.

8 0
2 years ago
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crimeas [40]

Answer:

coffee is the best answer in my mind

4 0
2 years ago
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