1.

|-5c+5|=40 solve for c

GrogVix [38]

Answer:

C=9 or c=-7

Step-by-step explanation:

|-5c+5|=40

-5c+5=-40 or - 5c+5=40

-5c=-40-5. - 5c=40-5

-5c=-45. - 5c=35

C=9. C=-7

5 0

3 years ago

Read 2 more answers

What is the answer plz show work

lana66690 [7]

There will be three students left out

6 0

3 years ago

What is the probability of spinning the spinner below once and having it land on blue or an even number?

spayn [35]

A. 3/16 would be the answer

4 0

3 years ago

Read 2 more answers

Please i need your help in this question, i will mark you to brainliest

ICE Princess25 [194]

Answer:

$\frac{2520}{64} {ln}^{3}$

Step-by-step explanation:

V=L×W×H

=3½in×1⅞×2¾

=3×2+1/2in × 1×8+7/8in × 2×4+3/4in

=7/2in × 15/8in × 24/4in

=2520/64 incube

so it's you tern to change into mixed fraction

sorry for that

but if it's helpful ❤❤

THANK YOU

8 0

2 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