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lesantik [10]
3 years ago
13

I did this and turned it in, I got a C on it, could you tell me what to fix so I can get an A?

Mathematics
1 answer:
mart [117]3 years ago
8 0
In my opinion I think your response sounds great
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HeLp Me pLeAsE?
Bogdan [553]
One because it only has one number behind the decimal
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3 years ago
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Find the perimeter and area , Please help me on this will give brainlist
Ronch [10]

Answer:

Area: x^2+x-6

Perimeter: 4x+2

Step-by-step explanation:

Area: multiply x+3 and x-2 and combine the like terms

Perimeter: multiply the length and width by two, then combine the like terms.

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 <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%

3 0
3 years ago
<img src="https://tex.z-dn.net/?f=%5Ctext%7BI%20was%20eating%20cookies%20and%20had%20some%20thoughts.%20If%20I%20wanted%20to%20c
andrew-mc [135]

In the attachement, there is what I came up with so far. I think that finding 'a' is non-trivial, if possible at all.

A_c - the area of a circle

A_{cs} - the area of a circular segment

5 0
3 years ago
Which of the following proportions could be used to convert 6 days to hours?
nekit [7.7K]

Answer:

Step-by-step explanation:

24 hours---------->1 day

x hours------------>6 days

Cross multiplying:

x=24x6

x=144 hrs

4 0
3 years ago
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