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Natasha_Volkova [10]
3 years ago
11

A water tank contains 12 1/2 liters of water. Two- fifth of it was consumed. How much of it was left?​

Mathematics
1 answer:
seraphim [82]3 years ago
6 0
12500mlx0.4=5000ml

12500-5000=7500ml

7500ml=7.5L
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NEED HELP IMMEDIATELY. I WILL BRAINLIEST. NEED TO SOLVE FOR X INTERCEPT
Kay [80]

Answer:

(1+√7,0),(1−√7,0)

Step-by-step explanation:

You can't factor the expression evenly, so use the quadratic formula.

a = 2

b= -4

c= -12

\frac{-b+\sqrt{b^{2}-4ac } }{2a}

\frac{4+\sqrt{16{}-4(2)(-12)} }{2(2)}

\frac{4+\sqrt{16-4(-24)} }{4}

\frac{4+\sqrt{16+96} }{4}

\frac{4+\sqrt{112} }{4}

\sqrt{112} = 4\sqrt{7}

End result: (1+√7,0),(1−√7,0)

5 0
3 years ago
Find (f/g) (x) for the functions provided: ƒ(x) = x3 − 27, g(x) = 3x − 9
AnnZ [28]

Answer:

(\frac{f}{g})(x)=\frac{1}{3}(x^2+3x+9)

Step-by-step explanation:

We have been given that

f(x)=x^3-27,g(x)=3x-9

We can use the formula for difference of cubes to simplify the function f(x)

difference of cubes -  a^3-b^3=(a-b)(a^2+ab+b^2)

f(x)=x^3-27\\\\=x^3-3^3\\\\=(x-3)(x^2+3x+9)

And g(x) can be written as

g(x)=3x-9\\=3(x-3)

Thus, we have

(\frac{f}{g})(x)=\frac{(x-3)(x^2+3x+9)}{3(x-3}

On cancelling the common factors, we get

(\frac{f}{g})(x)=\frac{1}{3}(x^2+3x+9)

5 0
3 years ago
Lmno is a kite find the length of ln
Llana [10]
Where are the numbers so that people can help you
4 0
3 years ago
I got 77 for my total answer. I correct? if not can some help​
mote1985 [20]
Yes you got it correct.
8 0
3 years ago
Read 2 more answers
The​ life, in​ years, of a certain type of electrical switch has an exponential distribution with an average life β=44. If 100 o
Bond [772]

Answer:

0.9999

Step-by-step explanation:

Let X be the random variable that measures the time that a switch will survive.

If X has an exponential distribution with an average life β=44, then the probability that a switch will survive less than n years is given by

\bf P(X

So, the probability that a switch fails in the first year is

\bf P(X

Now we have 100 of these switches installed in different systems, and let Y be the random variable that measures the the probability that exactly k switches will fail in the first year.

Y can be modeled with a binomial distribution where the probability of “success” (failure of a switch) equals 0.0225 and  

\bf P(Y=k)=\binom{100}{k}(0.02247)^k(1-0.02247)^{100-k}

where  

\bf \binom{100}{k} equals combinations of 100 taken k at a time.

The probability that at most 15 fail during the first year is

\bf \sum_{k=0}^{15}\binom{100}{k}(0.02247)^k(1-0.02247)^{100-k}=0.9999

3 0
3 years ago
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