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3 years ago

What Is The Average Of 4,4 And 5

Mathematics

2 answers:

velikii [3]3 years ago

6 0

Answer:

The answer is B, 4 and 1/3

Step-by-step explanation:

4+4+5 divided by three is 4 and 1/3.

ruslelena [56]3 years ago

5 0

Answer:

4 1/3

Step-by-step explanation:

First, add all three numbers together. Then, divide that number by three, because there were three numbers to begin with. If you put this into a calculator, you will get 4.33333333... because 1/3 is equal to 0.333333333. This is useful to know because then when you get 0.6666666 you know that it is 2/3. Anyway, your answer is 4 1/3.

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PLZ HELP WILL GIVE BRAINLIEST WORTH 10 PTS THX!!

antiseptic1488 [7]

Answer:

64/49

Step-by-step explanation:

The applicable rules of exponents are ...

(a^b)^c = a^(bc)

(a^b)(a^c) = a^(b+c)

a^-b = 1/a^b

a^0 = 1

_____

Using these rules, we can simplify the given expression to ...

$\left(\dfrac{8\cdot 4\cdot 2}{8\cdot 7}\right)^2\times\left(\dfrac{8^0}{7^{-3}}\right)^3\times 7^{-9}\\\\=\dfrac{8^2}{7^2}\times(8^{0\cdot 3}\cdot 7^{3\cdot 3})\times 7^{-9}\\\\=\dfrac{64}{49}\times(8^07^{9-9})=\boxed{\dfrac{64}{49}}$

6 0

3 years ago

Read 2 more answers

A bag contains different colored candies. There are 50 candies in the bag, 28 are red, 10 are blue, 8 are green and 4 are yellow

Mrrafil [7]

Answer:

$\displaystyle \frac{54}{5405}$ .

Step-by-step explanation:

How many unique combinations are possible in total?

This question takes 5 objects randomly out of a bag of 50 objects. The order in which these objects come out doesn't matter. Therefore, the number of unique choices possible will the sames as the combination

$\displaystyle \left(50\atop 5\right) = 2,118,760$ .

How many out of that 2,118,760 combinations will satisfy the request?

Number of ways to choose 2 red candies out a batch of 28:

$\displaystyle \left( 28\atop 2\right) = 378$ .

Number of ways to choose 3 green candies out of a batch of 8:

$\displaystyle \left(8\atop 3\right)=56$ .

However, choosing two red candies out of a batch of 28 red candies does not influence the number of ways of choosing three green candies out of a batch of 8 green candies. The number of ways of choosing 2 red candies and 3 green candies will be the product of the two numbers of ways of choosing

$\displaystyle \left( 28\atop 2\right) \cdot \left(8\atop 3\right) = 378\times 56 = 21,168$ .

The probability that the 5 candies chosen out of the 50 contain 2 red and 3 green will be:

$\displaystyle \frac{21,168}{2,118,760} = \frac{54}{5405}$ .

3 0

3 years ago

What is the answer for number 15 16 17

stira [4]

15.

6 pachages
each pcage has 30
6*30=180 total

17 left over
how many used
used=total-left over
used=180-17
used=163

16. xy is 2 digit number
multiplules of 10 (whole multipules) always end with 0 so y=0

and has 3 as one of its factors
factors of 3 is 3,6,9,12
30,60,90 is answer

17.
30 in 9 parks
30 times 9
30=10 times 3
3*10*9
group the 3 and 9
10*(3*9)
3*9=27
10*(27)
270

15. 163
16. 30,60,90
17. see above, 270

6 0

3 years ago

If f(x)=x/2 and g(x) = x-3,what is the value of g(f(8))

pochemuha

F(x) = x/2
g(x) = x - 3

f(8) = 8/2 = 4

g(f(x)) = 4 - 3 = 1

Answer g(f(8)) = 1

7 0

3 years ago

Read 2 more answers

Please help me What is the midpoint of a line segment with the endpoints (4, -3) and (7,-5)?

liq [111]

Answer:

(11/2, -1)