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3 years ago

Easy Math Question: Will Upvote! (73) + (6x+4) + (8y-7) =180 Solve for X

Mathematics

2 answers:

sineoko [7]3 years ago

7 0

The answer is X=-4y/3+55/3

yan [13]3 years ago

3 0

Your answer will be x= -4y/3 + 55/3

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UHHHHH IM GONNA FAIL MATH YALL Just answer it ig-? I WILL MARK AS BRAINLIEST

tankabanditka [31]

Answer:

30.96 m^2

Step-by-step explanation:

Side length of the square is also the diameter of the semicircle so:

r = 0.5 * 12 m = 6 m

Since you want the area of the shaded area, you have to subtract the area of both semi circles from the area of the square:

Square Area - Semi Circle Area - Semi Circle Area = Shaded Region Area

(12 m * 12 m) - (0.5 * 3.14 * 6^2 m) - (0.5 * 3.14 * 6^2 m)

= 144 - 56.52 - 56.52

= 30.96 m^2

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