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MArishka [77]
4 years ago
13

Write an expression that shows how you can multiply 9×475 using expanded form and distributive form

Mathematics
1 answer:
Dima020 [189]4 years ago
3 0
Expanded form:

9(400+70+5)
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mariarad [96]

Answer:69

Step-by-step explanation:

4 0
3 years ago
Can someone help me with this math question i really don’t know how to do it!
julia-pushkina [17]

Answer:

1260

Step-by-step explanation:

He asked 180 people and 63 said cookies so I think you have to do 63 out of each 180 people so you would do 3600 ÷ 180 = 20, 20 x 63 =1260

Not sure if this is right, hope this helps

8 0
4 years ago
I need help with this polynomial question: (5x^5-2x)-(4x^4+3x^2). This ^ means to the power of.
jeyben [28]

Answer:

Step-by-step explanation:

5x^5 - 2x - 4x^4 - 3x^2

5x^5 - 4x^4 - 3x^2 - 2x

6 0
4 years ago
Make f the subject of the formula c=<img src="https://tex.z-dn.net/?f=%5Cfrac%7B5%28f-32%29%7D%7B9%7D" id="TexFormula1" title="\
lara31 [8.8K]

Answer:

f=\frac{9c}{5}+32

Step-by-step explanation:

c=\frac{5\left(f-32\right)}{9}

Switch sides:

\frac{5\left(f-32\right)}{9}=c

Multiply both sides by 9:

\frac{9\times5\left(f-32\right)}{9}=9c

5\left(f-32\right)=9c

Divide both sides by 5:

\frac{5\left(f-32\right)}{5}=\frac{9c}{5}

f-32=\frac{9c}{5}

Add 32 to both sides:

f-32+32=\frac{9c}{5}+32

f=\frac{9c}{5}+32

8 0
4 years ago
Suppose that two teams play a series of games that ends when one of them has won ???? games. Also suppose that each game played
Musya8 [376]

Answer:

(a) E(X) = -2p² + 2p + 2; d²/dp² E(X) at p = 1/2 is less than 0

(b) 6p⁴ - 12p³ + 3p² + 3p + 3; d²/dp² E(X) at p = 1/2 is less than 0

Step-by-step explanation:

(a) when i = 2, the expected number of played games will be:

E(X) = 2[p² + (1-p)²] + 3[2p² (1-p) + 2p(1-p)²] = 2[p²+1-2p+p²] + 3[2p²-2p³+2p(1-2p+p²)] = 2[2p²-2p+1] + 3[2p² - 2p³+2p-4p²+2p³] =  4p²-4p+2-6p²+6p = -2p²+2p+2.

If p = 1/2, then:

d²/dp² E(X) = d/dp (-4p + 2) = -4 which is less than 0. Therefore, the E(X) is maximized.

(b) when i = 3;

E(X) = 3[p³ + (1-p)³] + 4[3p³(1-p) + 3p(1-p)³] + 5[6p³(1-p)² + 6p²(1-p)³]

Simplification and rearrangement lead to:

E(X) = 6p⁴-12p³+3p²+3p+3

if p = 1/2, then:

d²/dp² E(X) at p = 1/2 = d/dp (24p³-36p²+6p+3) = 72p²-72p+6 = 72(1/2)² - 72(1/2) +6 = 18 - 36 +8 = -10

Therefore, E(X) is maximized.

6 0
3 years ago
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