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Free_Kalibri [48]
3 years ago
7

SOMEONE DO NUMBER 1 FOR ME IM RUSHING. GIVING BRAINLIEST!!

Mathematics
1 answer:
MrRissso [65]3 years ago
6 0

Answer:

Step-by-step explanation:

                            Mixed number          Improper fraction

a) 1 + 1 + 1/3 =            2\frac{1}{3}                                7/3

b) 1 + 1 + 1 + 5/8 =     3\frac{5}{8}                                 29/8

c) 1 + 1 + 1 +2/4 =      3\frac{2}{4}                                  14/4

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the point that the function intercepts the x axis are the zeros of the function, where the products set equal to zero give you the x coordinate in which it intercepts.

x=-15 and x=9

set both equal to 0

x+15 and x-9

These are your two products

(x+15)(x-9)

or

x^2+6x-135

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PLZ!!! I NEED AN ANSWER!!! What is the Vertex? y = 3x^ 2 − 6x − 2
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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 <em>4+5+6 = 15 bulbs</em>. If we want to select 3 randomly there are  K ways of doing this, where K is the<em> combination of 15 elements taken 3 at a time </em>

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 <em>the events are disjoint</em>, 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, <em>supposing there is no replacement</em>, 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%

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Step-by-step explanation:

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