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oee [108]
3 years ago
8

Randy has two 28-pound blocks of ice for his snow cone stand. If each snow cone requires 2/3 pound of ice, how many snow cones c

an Randy make?
Mathematics
1 answer:
irga5000 [103]3 years ago
3 0

Answer:

The answer would be 28 divided by 2/3.

Step-by-step explanation:

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From families with four children a family is chosen at random. Let X be the number of boys in the family. Calculate and sketch t
alisha [4.7K]

Solution :

Given :

X = the number of boys in a family of four children

Families having four children are chosen randomly.

The gender distribution in the four child family are equally probable.

Thus,

 X                P(X)                                          CDF

0             ^4C_0 (1/2)^{4}        = 1/16                     \frac{1}{16}

1               ^4C_1 (1/2)^{4}       = 1/4                       \frac{5}{16}

2                ^4C_2 (1/2)^{4}      = 3/8                      \frac{11}{16}

3                ^4C_3 (1/2)^{4}      = 1/4                       \frac{15}{16}

4                  ^4C_4 (1/2)^{4}    = 1/16                      1

4 0
3 years ago
Si 9/11, 5/m+3, 15/N-7, 18/p-1 son fracciones homogeneas como calculo m+n+p
notsponge [240]
I’m confused, can you explain
5 0
3 years ago
Two friends share 76 blueberries. To count the blueberries they put them
Usimov [2.4K]
In there mouth and they ate them
6 0
3 years ago
Calculate the total area of the shaded region.
LUCKY_DIMON [66]

so hmmm seemingly the graphs meet at -2 and +2 and 0, let's check

\stackrel{f(x)}{2x^3-x^2-5x}~~ = ~~\stackrel{g(x)}{-x^2+3x}\implies 2x^3-5x=3x\implies 2x^3-8x=0 \\\\\\ 2x(x^2-4)=0\implies x^2=4\implies x=\pm\sqrt{4}\implies x= \begin{cases} 0\\ \pm 2 \end{cases}

so f(x) = g(x) at those points, so let's take the integral of the top - bottom functions for both intervals, namely f(x) - g(x) from -2 to 0 and g(x) - f(x) from 0 to +2.

\stackrel{f(x)}{2x^3-x^2-5x}~~ - ~~[\stackrel{g(x)}{-x^2+3x}]\implies 2x^3-x^2-5x+x^2-3x \\\\\\ 2x^3-8x\implies 2(x^3-4x)\implies \displaystyle 2\int\limits_{-2}^{0} (x^3-4x)dx \implies 2\left[ \cfrac{x^4}{4}-2x^2 \right]_{-2}^{0}\implies \boxed{8} \\\\[-0.35em] ~\dotfill

\stackrel{g(x)}{-x^2+3x}~~ - ~~[\stackrel{f(x)}{2x^3-x^2-5x}]\implies -x^2+3x-2x^3+x^2+5x \\\\\\ -2x^3+8x\implies 2(-x^3+4x) \\\\\\ \displaystyle 2\int\limits_{0}^{2} (-x^3+4x)dx \implies 2\left[ -\cfrac{x^4}{4}+2x^2 \right]_{0}^{2}\implies \boxed{8} ~\hfill \boxed{\stackrel{\textit{total area}}{8~~ + ~~8~~ = ~~16}}

7 0
2 years ago
X-2y=5 2x-4y=10
Nitella [24]

Step-by-step explanation:

X-2y=5 ; 2x-4y=10

• x - 2y = 5 • 2x -4y = 10

-2y = -x+5 -4y= -2x +10

y1 = ½x - 5/2 y2 = ½x -5/2

• => y1 = y2

½x -5/2 = ½x -5/2

the systems have no solution

• graph as attached

8 0
2 years ago
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