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PSYCHO15rus [73]
3 years ago
6

Mustafa’s horse ate 6,400 pounds of hay last year. How could Mustafa determine the number of tons of hay his horse ate last year

? by dividing 6,400 by 16 by multiplying 6,400 by 16 by dividing 6,400 by 2,000 by multiplying 6,400 by 2,000
Mathematics
1 answer:
Alona [7]3 years ago
3 0

Answer:

by dividing 6,400 by 2,000

Step-by-step explanation:

We know that 2000 pounds  equals 1 ton

To convert 6400 pounds to tons, we need to

6400 pounds * 1 ton/2000 pounds

Divide 6400 by 2000

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Find the mean for the data set. 7, 6, 7, 7, 7, 4, 4, 6, 5, 5, 7
saul85 [17]

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5.909

Step-by-step explanation:

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Suppose that diameters of a new species of apple have a bell-shaped distribution with a mean of 7.25cm and a standard deviation
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2 years ago
Y=x^2+bx+c;(-2,3)
Maksim231197 [3]
<h3>Answer:  b = 4 and c = 7.</h3>

===============================================

Explanation:

Comparing y = x^2+bx+c to y = ax^2+bx+c, we see that a = 1.

The vertex given is (-2,3). In general, the vertex is (h,k). So h = -2 and k = 3.

Plug those three values into the vertex form below

y = a(x-h)^2 + k

y = 1(x-(-2))^2 + 3

y = (x+2)^2 + 3

Then expand everything out and simplify

y = x^2+4x+4 + 3

y = x^2+4x+7

We see that b = 4 and c = 7.

8 0
2 years ago
Charlie know that 3 blocks are needed to make a tower that is 0.5 foot high. How many blocks will he need to make a tower that i
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6 0
3 years ago
Find a compact form for generating functions of the sequence 1, 8,27,... , k^3
pantera1 [17]

This sequence has generating function

F(x)=\displaystyle\sum_{k\ge0}k^3x^k

(if we include k=0 for a moment)

Recall that for |x|, we have

\displaystyle\frac1{1-x}=\sum_{k\ge0}x^k

Take the derivative to get

\displaystyle\frac1{(1-x)^2}=\sum_{k\ge0}kx^{k-1}=\frac1x\sum_{k\ge0}kx^k

\implies\dfrac x{(1-x)^2}=\displaystyle\sum_{k\ge0}kx^k

Take the derivative again:

\displaystyle\frac{(1-x)^2+2x(1-x)}{(1-x)^4}=\sum_{k\ge0}k^2x^{k-1}=\frac1x\sum_{k\ge0}k^2x^k

\implies\displaystyle\frac{x+x^2}{(1-x)^3}=\sum_{k\ge0}k^2x^k

Take the derivative one more time:

\displaystyle\frac{(1+2x)(1-x)^3+3(x+x^2)(1-x)^2}{(1-x)^6}=\sum_{k\ge0}k^3x^{k-1}=\frac1x\sum_{k\ge0}k^3x^k

\implies\displaystyle\frac{x+4x^3+x^3}{(1-x)^4}=\sum_{k\ge0}k^3x^k

so we have

\boxed{F(x)=\dfrac{x+4x^3+x^3}{(1-x)^4}}

5 0
3 years ago
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