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zheka24 [161]
3 years ago
10

Evaluate the given expression. 5 P 2

Mathematics
1 answer:
Mamont248 [21]3 years ago
7 0

Answer:

20

Step-by-step explanation:

nPr = n!/(n-r)!

5P2 = 5!/(5-2)!

= 5!/3!

= 5×4×3!/3!

= 5×4

= 20

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Simplify. -10m+2m^4-13m-20m^4
Ulleksa [173]
First one:

you can add -10m and -13m but you can't add -10m and 2m^4 becuase the powers aren't the same so

when adding the like terms
look at the:
powers, (x^3 adds with x^3)
placehloder letter (x adds with x and y adds with y and so on)

-10m+2m^4-13m-20m^4
powers: m^1 and M^4
placeholders: all m


add
-10m-13m+2m^4-20m^4
-23m-18m^4




second one:
when multiplying exponents, you add with like
so if you multipliy
x^2yz^3 times x^4y^2z^2 thne you would get x^6y^3z^5
when multiply with coeficients
2x^2yz^3 times 4x^4y^2z^2=8x^6y^3z^5
so using associative property a(bc)=(ab)c
2/3 times p^4 times y^3 times y^4 times s^5 times 6 times p^2 times s^3
group like terms

(2/3 times 6) times (p^4 times p^2) times (y^3 times y^4) times (s^5 times s^3)
(4) times (p^6) times (y^7) times (s^8)
4p^6y^7s^8

8 0
3 years ago
there are 6 twenty pound boxes and 8 sixty pound boxes of produce on a truck. how many pounds of produce are on the truck
Klio2033 [76]

6 *20 = 120

8 *60 = 480


120 + 480 = 600 pounds total

6 0
4 years ago
A wrestling match has eliminations each round. The table below shows the number of wrestlers in each of the first 4 rounds of th
AlladinOne [14]
\bf \begin{array}{lllll}
round(x)&\boxed{1}&2&3&\boxed{4}\\\\
wrestlers[f(x)]&\boxed{64}&32&18&\boxed{9}
\end{array}
\\\\\\
slope = {{ m}}= \cfrac{rise}{run} \implies 
\cfrac{{{ f(x_2)}}-{{ f(x_1)}}}{{{ x_2}}-{{ x_1}}}\impliedby 
\begin{array}{llll}
average\ rate\\
of\ change
\end{array}\\\\
-------------------------------\\\\
f(x)=   \qquad 
\begin{cases}
x_1=1\\
x_2=4
\end{cases}\implies \cfrac{f(4)-f(1)}{4-1}\implies \cfrac{9-64}{4-1}\implies \cfrac{-55}{3}

55 over 3, or 55 wrestlers for every 3 rounds, but the wrestlers value is negative, thus 55 "less" wrestlers for every 3 rounds on average.
6 0
3 years ago
Read 2 more answers
Suppose the sediment density (g/cm3 ) of a randomly selected specimen from a certain region is known to have a mean of 2.80 and
Irina18 [472]

Answer:

0.918 is the probability that the sample average sediment density is at most 3.00

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 2.80

Standard Deviation, σ = 0.85

Sample size,n = 35

We are given that the distribution of sediment density is a bell shaped distribution that is a normal distribution.

Formula:

z_{score} = \displaystyle\frac{x-\mu}{\sigma}

Standard error due to sampling:

=\dfrac{\sigma}{\sqrt{n}} = \dfrac{0.85}{\sqrt{35}} = 0.1437

P(sample average sediment density is at most 3.00)

P( x \leq 3.00) = P( z \leq \displaystyle\frac{3.00 - 2.80}{0.1437}) = P(z \leq 1.3917)

Calculation the value from standard normal z table, we have,  

P(x \leq 3.00) = 0.918

0.918 is the probability that the sample average sediment density is at most 3.00

4 0
3 years ago
Which of the following are factors of 6x^2+30x-36
attashe74 [19]
I think the answer is 3,2, x+6, and x+2 because  the common factor is 6. Hope this helps.
5 0
3 years ago
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