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3 years ago

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Harper works at an electronics store as a salesperson. Harper earns a 6% commission on the total dollar amount of all phone sale

s she makes, and earns a 5% commission on all computer sales. Harper made a total of $2600 in sales and earned $137 in commission. Write a system of equations that could be used to determine the dollar amount of phone sales Harper made and the dollar amount of computer sales she made. Define the variables that you use to write the system.

1 answer:

agasfer [191]3 years ago

7 0

<u>Answer:</u>

The system of equations are

x + y = 2600 …….equation (1)

and, 6x + 5y = 13700 ……equation (2)

<u>Explanation:</u>

Let x = amount of sales of phone;

y = amount of sales of computer

For phone, 6% of sales is 6% of x, or 6% * x = 0.06x.

Similarly for computer,

5% of y = 0.05y

Given the total sales; x + y = 2600

And commission; 0.06x + 0.05y = 137; or 6x + 5y = 13700

Therefore the system of equations are

x + y = 2600 …….equation (1)

6 x + 5 y = 13700 ……equation (2)

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For firework launched from height 100ft with initial velocity 150ft/sec, equation made is correct

(a) equation will be $h(t) = -16t^2+150t+100$

(b) Now we have to see when it will land. At land or ground level height h will be equal to 0. So simply plug 0 in h place in equation made in part (a)

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Plugging these values in quadratic formula we get

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(c) To make table simply plug various value for t like t =0, 2, 4, 6, 8 till 10. Plug values in equation mad in part (a) and find h value for each t as shown

For t =0 seconds, $h = -16(0)^2+150(0)+100 = 100 feet$

For t =2 seconds, $h = -16(2)^2+150(2)+100 =336 feet$

For t =4 seconds, $h = -16(4)^2+150(4)+100 = 444 feet$

For t =6 seconds, $h = -16(6)^2+150(6)+100 = 424 feet$

For t =8 seconds, $h = -16(8)^2+150(8)+100 = 276 feet$

For t =10 seconds, $h = -16(10)^2+150(10)+100 = 0 feet$

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$x = \frac{-b}{2a}$

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$x = \frac{-b}{2a}$

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Answer:

(a) 0.343

(b) 0.657

(c) 0.189

(d) 0.216

(e) 0.353

Step-by-step explanation:

Let P(a vehicle passing the test) = p

$p = \frac{70}{100} = 0.7$

Let P(a vehicle not passing the test) = q

q = 1 - p

q = 1 - 0.7 = 0.3

(a) P(all of the next three vehicles inspected pass) = P(ppp)

= 0.7 × 0.7 × 0.7

= 0.343

(b) P(at least one of the next three inspected fails) = P(qpp or qqp or pqp or pqq or ppq or qpq or qqq)

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= 0.147 + 0.063 + 0.147 + 0.063 + 0.147 + 0.063 + 0.027

= 0.657

(c) P(exactly one of the next three inspected passes) = P(pqq or qpq or qqp)

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= 0.063 + 0.063 + 0.063

= 0.189

(d) P(at most one of the next three vehicles inspected passes) = P(pqq or qpq or qqp or qqq)

= (0.7 × 0.3 × 0.3) + (0.3 × 0.7 × 0.3) + (0.3 × 0.3 × 0.7) + (0.3 × 0.3 × 0.3)

= 0.063 + 0.063 + 0.063 + 0.027

= 0.216

(e) Given that at least one of the next 3 vehicles passes inspection, what is the probability that all 3 pass (a conditional probability)?

P(at least one of the next three vehicles inspected passes) = P(ppp or ppq or pqp or qpp or pqq or qpq or qqp)

= (0.7 × 0.7 × 0.7) + (0.7 × 0.7 × 0.3) + (0.7 × 0.3 × 0.7) + (0.3 × 0.7 × 0.7) + (0.7 × 0.3 × 0.3) + (0.3 × 0.7 × 0.3) + (0.3 × 0.3 × 0.7)

= 0.343 + 0.147 + 0.147 + 0.147 + 0.063 + 0.063 + 0.063

= 0.973

With the condition that at least one of the next 3 vehicles passes inspection, the probability that all 3 pass is,

$= \frac{P(all\ of\ the\ next\ three\ vehicles\ inspected\ pass)}{P(at\ least\ one\ of\ the\ next\ three\ vehicles\ inspected\ passes)}$

$= \frac{0.343}{0.973}$

= 0.353

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