Lesson 1 Homework

A tractor costs $16,900 and depreciates in value by 13% per year. How much will the tractor be worth after 8 years?

Assoli18 [71]

Answer: $-676

Step-by-step explanation:

Multiply 13% or 0.13 by the 16,900 to get 2,197. Multiply the 2,197 by 8 because that’s how many years there are and you get 17,576 then subtract that from the 16,900 to get -676 which means that the tractor would be worth pretty much nothing

8 0

3 years ago

Find the points of intersection of the graphs of the equations.

AURORKA [14]

Answer:

To find the point of intersection, just solve for x and y in the equations

so since x - y = -5

x = y - 5

now that we have a value of x, we will substitute x with in in the other equation

5x + 3y = -9

5(y-5) + 3y = -9

5y - 25 + 3y = -9

8y = 16

y = 2

now use this value of y in the equation we made for x

x = y - 5

x = 2 - 5

x = -3

Hence, the point of contact is (-3 , 2)

4 0

3 years ago

. A box in a certain supply room contains four 40-W lightbulbs, five 60-W bulbs, and six 75-W bulbs. Suppose that three bulbs ar

yaroslaw [1]

Answer:

a) 59.34%

b) 44.82%

c) 26.37%

d) 4.19%

Step-by-step explanation:

(a)

There are in total 4+5+6 = 15 bulbs. If we want to select 3 randomly there are K ways of doing this, where K is the combination of 15 elements taken 3 at a time

$K=\binom{15}{3}=\frac{15!}{3!(15-3)!}=\frac{15!}{3!12!}=\frac{15.14.13}{6}=455$

As there are 9 non 75-W bulbs, by the fundamental rule of counting, there are 6*5*9 = 270 ways of selecting 3 bulbs with exactly two 75-W bulbs.

So, the probability of selecting exactly 2 bulbs of 75 W is

$\frac{270}{455}=0.5934=59.34\%$

(b)

The probability of selecting three 40-W bulbs is

$\frac{4*3*2}{455}=0.0527=5.27\%$

The probability of selecting three 60-W bulbs is

$\frac{5*4*3}{455}=0.1318=13.18\%$

The probability of selecting three 75-W bulbs is

$\frac{6*5*4}{455}=0.2637=26.37\%$

Since the events are disjoint, the probability of taking 3 bulbs of the same kind is the sum 0.0527+0.1318+0.2637 = 0.4482 = 44.82%

(c)

There are 6*5*4 ways of selecting one bulb of each type, so the probability of selecting 3 bulbs of each type is

$\frac{6*5*4}{455}=0.2637=26.37\%$

(d)

The probability that it is necessary to examine at least six bulbs until a 75-W bulb is found, supposing there is no replacement, is the same as the probability of taking 5 bulbs one after another without replacement and none of them is 75-W.

As there are 15 bulbs and 9 of them are not 75-W, the probability a non 75-W bulb is $\frac{9}{15}=0.6$