Onsider the function f(x)=−2/3x+2 . What is f(1/2) ?

Cookies are on sale! Today each cookie costs $0.75 less than the normal price. Right now if you buy 7 of them it will only cost

Anestetic [448]

Let's say that the price of each normal cookie is n.
The equation would then be 7(n - .75)=2.80.
Use distributive property, getting 7n - 5.25=2.80.
Add 5.25 to each side of the equation, getting 7n=8.05.
Divide 7 from both sides of the equation, getting n=1.15.

6 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 $\frac{9}{15}=0.6$

Since there are no replacement, the probability of taking a second non 75-W bulb is now $\frac{8}{14}=0.5714$

Following this procedure 5 times, we find the probabilities

$\frac{9}{15},\frac{8}{14},\frac{7}{13},\frac{6}{12},\frac{5}{11}$

which are

0.6, 0.5714, 0.5384, 0.5, 0.4545

As the events are independent, the probability of choosing 5 non 75-W bulbs is the product

0.6*0.5714*0.5384*0.5*0.4545 = 0.0419 = 4.19%

3 0

3 years ago

In 2007, there were 31 laptops

abruzzese [7]

Answer:

$l=31+20(y-2007)$

where $l$ is the number of laptops, and $y$ is the year.

in 2017: $l=231$

Step-by-step explanation:

I will define the variable $x$ as the number of years that passed since 2007.

Since the school buys 20 lapts each year, after a number $x$ of years, the school will have

$20*x$ more laptops.

and thus, since the school starts with 31 laptops, the equation to model this situation is

$l=31+20*x$

where $l$ is the number of laptops.

since x is the number of years that have passed since 2007, it can be represented like this:

$x=y-2007$

where $y$ can be any year, so the equation to model the situation using the year:

$l=31+20(y-2007)$

and this way we can find the number of laptos at the end of 2017:

$y=2017$

and

$l=31+20(y-2007)$

$l=31+20(2017-2007)\\l=31+20(10)\\l=31+200\\l=231$

3 0

3 years ago

How many people can seat a theater if there are 156 seats and each row of seat has 13 rows.how many have a seat?

Elan Coil [88]

How you wrote it is kind of confusing but if I'm reading what you're saying right. There are 2028 .

5 0

4 years ago

HELPPPPP!!!!!!!!!!Please explain how you got your answer. <3 (:

gayaneshka [121]

Step-by-step explanation:

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8 0

2 years ago