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Aleonysh [2.5K]
2 years ago
6

Which represents the solution set to the inequality 5.1(3 + 2.2x) > –14.25 – 6(1.7x + 4)? x < –2.5 x > 2.5 (–2.5, ∞) (–

∞, 2.5)

Mathematics
2 answers:
german2 years ago
4 0
The inequality simplifies to
  15.3 + 11.22x > -14.25 -10.2x -24
  21.42x > -53.55 . . . . . . . . . . . . . . . . add 10.2x-15.3
  x > -2.5 . . . . . . . . . . . . . . . . . . . . . . . divide by 21.42

The solution set is best represented by
  (-2.5, ∞)

Setler79 [48]2 years ago
3 0

Answer:

The set which represent the solution is (-2.5;+∞).

Step-by-step explanation:

Let's first expands each one of the parenthesis  in the inequality:

5.1(3 + 2.2x) > -14.25 - 6(1.7x + 4)

5.1*3 + 5.1*2.2x > -14.25 - 6*1.7x -6*4

15.3 + 11.22x > -14.25 - 10.2x - 24

Now we must to organize the equition, for dong this we will group the terms which contains the variable x to left member of the inequality and the rest of the term to the right member:

11.22x +10.2x > -14.25  - 24 - 15.3

21.42x >  - 53.55

Solving x:

x >  - 53.55/21.42

x >  - 2.5  

That mean that the solution set is  (-2.5;+∞)

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2 years ago
A customer visiting the suit department of a certain store will purchase a suit with probability .22, a shirt with probability .
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Answer:

a) The probability that he doesnt but any items is 0.49

b) He buys exactly 1 of those items with probability 0.28

Step-by-step explanation:

lets call su the event that the customer purchases a suit, sh the event that teh customer purchases a shirt and t the event that the customer purchases a tie.

Remembe that for events A, B and C we have that

P(A U B) = P(A) + P(B) - P(A ∩ B)

P(A U B U C) = P(A) + P(B) + P(C) - P(A ∩ B) - P(A ∩ C) - P(B ∩ C) + P(A ∩ B ∩ C)

Also, we are given that

P(su) = 0.22

P(sh) = 0.3

p(t) = 0.28

p(su ∩ sh) =  0.11

P(su ∩ t) = 0.14

P(sh ∩ t) = 0.1

P(sh ∩ t ∩ su) = 0.06

The event that he doesnt buy any item has as complementary event su ∪ sh ∪ t, therefore

P( he doesnt but any items) = 1-P(su U sh U t) =

1-( P(su) + p(sh) + p(t) - P(su ∩ sh) - p(su∩t) - p(sh∩t) + p(su∩sh∩t) ) =

1-(0.22+0.30+0.28-0.11-0.14-0.1+0.06) = 1-0.51 = 0.49

b) The probability that he buys at least 2 items is equal to

p(su ∩ t) + p(su ∩ sh) + p(sh ∩ t) -2 p(su ∩ t ∩ sh) (because we are counting the triple intersection 3 times, so we need to remove it twice)

This number is

0.14+0.11+0.1-2*0.06 = 0.23

Thus, the probability that he buys exactly one item can be computed by substracting from one the probability of the complementary event : she buys 2 or more or non items

P(he buys exactly one item) = 1- ( p(he buys none items) + p(he buys at least 2) ) = 1- 0.49-0.23 = 0.28

7 0
2 years ago
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In choice situations of this type, subjects often exhibit the "center stage effect," which is a tendency to choose the item in t
victus00 [196]

Complete Question

The complete question is shown on the first uploaded image

Answer:

a

The probability that he or she would choose the pair of socks in the center position is   p =\frac{1}{5}

The correct answer choice is

X has a binomial distribution with parameters n=100 and p=1/5  

b

The mean is  \mu = 20

The standard deviation is \sigma=4

c

The probability, P =0.0002

d

The correct answer is

The experiment supports the center stage effect. If participants were truly picking the socks at random, it would be highly unlikely for 34 or more to choose the center pair.

Using the R the probability Pe = 0.0003

The probabilities P \approx Pe

Step-by-step explanation:

Since the person selects his or her desired pair of socks at random , then the probability that the person would choose the pair of socks in the center position from all the five identical pair is mathematically evaluated as

                  p =\frac{1}{5}

                    =0.2

The mean of this distribution is mathematical represented as

           \mu = np

substituting the value

         \mu = 100 * 0.2

             \mu = 20

The standard deviation is mathematically represented as

         \sigma = \sqrt{np (1-p)}

substituting the value

           = \sqrt{100 * 0,2 (1-0.2)}

           \sigma=4

Applying normal approximation the probability that 34 or more subjects would choose the item in the center if each subject were selecting his or her preferred pair of socks at random would be mathematically represented as

               P=P(X \ge 34 )

By standardizing the normal approximation we have that

              P(X \ge 34) \approx P(Z \ge z)

Now z is mathematically evaluated as

               z = \frac{x-\mu}{\sigma }

Substituting values

             z = \frac{34-20}{4}

               =3.5

So  using the z table the P(Z \ge 3.5) is  0.0002

The probability P and Pe that 34 or more subject would choose the center pair is very small  So

The correct answer is

The experiment supports the center stage effect. If participants were truly picking the socks at random, it would be highly unlikely for 34 or more to choose the center pair.

 

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