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rosijanka [135]
3 years ago
10

TRANSLATE THE INDICATED PHRASE... the diameter of a circle is twice the radius. if c represents the radius, then write an expres

sion for the diameter.
Mathematics
1 answer:
KonstantinChe [14]3 years ago
6 0

Answer:

the expression for the diameter with radius c is 2c

Step-by-step explanation:

Translate the given statement

the diameter of a circle is twice the radius

Given that the radius of the circle is 'c'

Diameter = 2 times the radius

Diameter = 2c

So the expression for the diameter with radius c is 2c

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4. What is the volume of the cereal box?<br> cereal<br> 18 in<br> 3 in<br> 12 in<br> 135
marusya05 [52]

Answer:

12 in

Step-by-step explanation:

i don't know, you didn't show me the picture.

5 0
3 years ago
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Makayla has 2 yards of ribbon to use for a craft project she cuts the ribbon into pieces that are 1/4 yard long how many pieces
olga55 [171]

Answer:

8

Step-by-step explanation:

2/ 1/4 = 8

6 0
3 years ago
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. 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
Somebody’s please help me with this it’s due tomorrow and I don’t have any idea how to do it
svet-max [94.6K]

Answer:

1) They will stop 10 times for a break

2) A baker can make 10 cakes

3) Tammy will distrubute 36 bags with 2 1/2 pounds of candy

Step-by-step explanation:

Let me know if you have any questions !

Hope it helps

3 0
2 years ago
Please help ive been stuck!!
Kobotan [32]
I would say no, if you follow the lines on the graph, 12 on the x axis and 5 on the Y axis they are not within the range indicated in gray.
8 0
3 years ago
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