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melisa1 [442]
3 years ago
9

I NEED HELP ASAP 85 POINTS At a bargain store, Tanya bought 5 items that each cost the same amount. Tony bought 3 items that eac

h cost the same amount, but each was $2.5 less than the items that Tanya bought. Both Tanya and Tony paid the same amount of money. What was the individual cost of each person's items?
(A)Write an equation. Let x represent the cost of one of Tanya's items.
(B)Solve the equation. Show your work.
(C)Check your solution. Show your work.
(D)State the solution in complete sentences.
Mathematics
1 answer:
Andrei [34K]3 years ago
5 0

Answer:

Step-by-step explanation:

A) 5x = 3(x - 2.5)

-----------------------

B)

Distribute

5x = 3x - 7.5

Subtract 3x from both sides

2x = -7.5

Divde both sides by 2

x = -3.75

-----------------------

C)

5 * (-3.75) = 3(-3.75 - 2.5)

-18.75 = -18.75

D)

Tanya paid –$3.75 for each of 5 items

Tony paid –$6.25 for each of 3 items

There seems to be an error in the question since the cost of items is negative

Tony bought fewer items and paid less for each.

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Answer:

b) 0.0007

c) 0.4163

d) 0.2375

Step-by-step explanation:

We are given the following:

We treat securities lose value as a success.

P(Security lose value) = 70% = 0.7

Then the securities lose value follows a binomial distribution, where

P(X=x) = \binom{n}{x}.p^x.(1-p)^{n-x}

where n is the total number of observations, x is the number of success, p is the probability of success.

Now, we are given n = 20.

a) Assumptions

  • There are 20 independent trials.
  • Each trial have two possible outcome: security loose value or security does not lose value.
  • The probability for success of each trial is same, p = 0.7

b) P(all 20 securities lose value)

We have to evaluate:

P(x = 20)\\= \binom{20}{0}(0.7)^0(1-0.7)^{20}\\= 0.0007

0.0007 is the probability that all 20 securities lose value.

c) P(at least 15 of them lose value.)

P(x \geq 15)\\= P(x=15) + P(x = 16) + P(x=17) + P(x=18) +P(x = 19) +P(x = 20)\\ \binom{20}{15}(0.7)^{15}(1-0.7)^{5} +...+ \binom{20}{20}(0.7)^{20}(1-0.7)^{0} \\=0.4163

d) P(less than 5 of them gain value.)

P(gain value) = 1 - 0.7 = 0.3

P(x < 5)\\= P(x=0) + P(x = 1) + P(x=2) + P(x=3) +P(x = 4)\\ \binom{20}{0}(0.3)^{0}(1-0.3)^{20} +...+ \binom{20}{4}(0.3)^{4}(1-0.3)^{16} \\=0.2375

7 0
3 years ago
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Answer:

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Step-by-step explanation:

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Step-by-step explanation:

The y-axis of the table represents the range of the function.

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Factor  56−16 56−16  using the GCF. I TRIED IT AND ITS NOT 8
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Length of the base of the given triangle is (x+3) units and the perpendicular height is (x-1)units. If the area of the triangle
Lana71 [14]

Step-by-step explanation:

the area of a triangle is

baseline × height / 2

so, in our case we know this is 10 :

(x+3)(x-1)/2 = 10

(x+3)(x-1) = 20

x² - x + 3x - 3 = 20

x² + 2x - 23 = 0

that is exactly the provided equation. so, yes, x satisfies it.

the general solution to a quadratic equation is

x = (-b ± sqrt(b² - 4ac))/(2a)

in our case

a = 1

b = 2

c = -23

x = (-2 ± sqrt(2² - 4×1×-23))/(2×1) =

= (-2 ± sqrt(4 + 92))/2 = (-2 ± sqrt(96))/2 =

= (-2 ± sqrt(16×6))/2 = (-2 ± 4×sqrt(6))/2 =

= -1 ± 2×sqrt(6)

x1 = -1 + 2×sqrt(6) = 3.898979486... ≈ 4

or using sqrt(6) ≈ 2.45

= -1 + 2×2.45 = -1 + 4.9 = 3.9 ≈ 4

x2 = -1 - 2×sqrt(6) = -5.898979486... ≈ -6

or using sqrt(6) ≈ 2.45

= -1 - 2×2.45 = -1 - 4.9 = -5.9 ≈ -6

x2 would give us negative lengths for the triangle, which does not make sense.

so, x = 4 is our solution.

and that makes the height x-1 = 4-1 = 3 units.

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