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IceJOKER [234]
3 years ago
9

If u can do this your a real mathematics like OMG

Mathematics
2 answers:
Semenov [28]3 years ago
7 0

Answer: z+x x y

2 4/5 + (3/5 × 1/3) use BODMAS

14/5 + 1/5

15/5

3

Step-by-step explanation:

Svetllana [295]3 years ago
6 0
2(4/5) + (3/5) • (1/3)

(14/5) + (3/5) • (1/3)

(42/15) + (9/15) • (5/15)

(42/15) + (45/225)

(42/15) + (1/5)

(14/5) + (1/5)

(15/5) = 3 [I believe this is it]
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Please help ASAP!!!!Determine the total number of outcomes.
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Answer:

I believe it would be A

Step-by-step explanation:

A deck of cards has 52 cards in it, so it cant be C or D, and with all the possibilities it could not be B

8 0
3 years ago
At the beginning of a semester an anonymous survey was conducted on students in a statistics class. Two of the questions on the
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Answer:

a. Chi-square test of independence

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The chi square statistics is also used to test the hypothesis about the independence of two variables each of which is classified into a number of categories or attributes.

In the given problem the Equal , more or less are the attributes.

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4 0
4 years ago
If two lemons cost 15 cents, how many can be bought for 60 cents?
KiRa [710]

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3 years ago
Read 2 more answers
What is expression for 1/4y+3/8 factored
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3/2
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3 years ago
a) Estimate the volume of the solid that lies below the surface z = 7x + 5y2 and above the rectangle R = [0, 2]⨯[0, 4]. Use a Ri
SSSSS [86.1K]

In the x direction we consider the m=2 subintervals [0, 1] and [1, 2] (each with length 1), while in the y direction we consider the n=2 subintervals [0, 2] and [2, 4] (with length 2). Then the lower right corners of the cells in the partition of R are (1, 0), (2, 0), (1, 2), (2, 2).

Let f(x,y)=7x+5y^2. The volume of the solid is approximately

\displaystyle\iint_Rf(x,y)\,\mathrm dx\,\mathrm dy\approx f(1,0)\cdot1\cdot2+f(2,0)\cdot1\cdot2+f(1,2)\cdot1\cdot2+f(2,2)\cdot1\cdot2=\boxed{164}

###

More generally, the lower-right-corner Riemann sum over m=\mu and n=\nu subintervals would be

\displaystyle\sum_{m=1}^\mu\sum_{n=1}^\nu\left(7\frac{2m}\mu+5\left(\frac{4n-4}\nu\right)^2\right)\frac{2-0}\mu\frac{4-0}\nu=\frac83\left(101+\frac{21}\mu+\frac{40}{\nu^2}-\frac{120}\nu\right)

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7 0
3 years ago
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