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3 years ago

1 liter is equal to how many cubic centimeters?

Mathematics

2 answers:

Alika [10]3 years ago

7 0

Answer:

1000

Step-by-step explanation:

Kamila [148]3 years ago

4 0

The answer is 1,000 cubic centimeters

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I don't know how to solve this equation. Help please. 8!-3!(5!)

kakasveta [241]

The answer is 39600
Use calculator to evaluate it
8!-3!(5!) = 39600
Or
8!=1×2×3×4×5×6×7×8
3!=1×2×3
5!=1×2×3×4×5
Solve it this way by replacing 8!, 3! and 5!

3 0

3 years ago

many credit card companies charge a compound interest rate of 1.8% per month on a credit card balance. Miriam owes $750 on a cre

mash [69]

The answer is E.

First you need to find the rate. You must divide 1.8 by 100

1.8/100 = .018

Then add 1.

.018 + 1 = 1.018

Then times the starting amount with 1.018. The starting amount 750

750(1.018) = 763.5.
As you can see, in option E that is the 2nd number.

Now we must find the next number.

You will do 750(1.018)^x

x represents the time.

763.5 was the first month.

750(1.018)^2 is the 2nd month.

750(1.018)^2 = 777.244 which is the 3rd number in option E

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3 years ago

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I need help with 5 and 6

Vanyuwa [196]

I think it’s b but I’m not sure

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4 years ago

How do I write an inequality for this graph? Dint mind the other ones just number 5

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