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sattari [20]
3 years ago
13

A bag contains 3 red and 2 blue counters. A counter is picked at

Mathematics
1 answer:
vichka [17]3 years ago
7 0

Answer:

24.

Step-by-step explanation:

The fraction of red counters in the bag = 3/5.

So you'd expect 3/5 * 40

= 120/5

= 24.

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Fill in the blanks for the expression
Oduvanchick [21]

Answer:

Step-by-step explanation:

80

7 0
3 years ago
Solve : 11 (9-v)=0 a) -9 b) 9 c) 99 d)-99
garri49 [273]
B. V=9 Multiply it out, then subtract 99 from both sides, then divide by 11
7 0
3 years ago
Read 2 more answers
Find the four rational numbers and irrational numbers between 2/5 and 5/6
Savatey [412]
The four?

Of course there are an infinite number of rationals between any two different real numbers, as well as an even bigger infinite number of irrationals.

The average will be between the numbers:

\frac 1 2 (\frac 2 5 + \frac 5 6) = \dfrac{37}{60}

That's a lot of work to get a number in between.  We can just see

\frac 1 2
is between the two numbers.

The mediant, or freshman addition, will always be in between:

\dfrac{2 + 5}{5 + 6} = \dfrac{7}{11}

2/5=.4 and 5/6 is about .83, so

.6 = \frac{6}{10}

is in between as is .7, .41, .530940394 and as many rationals as we care to generate.

To get irrationals we could just add a teeny irrational to the ones we just generated, like \frac{6}{10} + \frac{\pi}{100}

We could just change that denominator a bit and get as many as we like.

But let's get some square roots.  The geometric mean will be between

\sqrt{ \dfrac 2 5 \cdot \dfrac 5 6 } = \dfrac{1}{\sqrt{3}}


That's the tangent from one of trig's biggest cliches, but I digress.  It's in between.

While we're on trig cliches, 

\dfrac{1}{\sqrt{2}}

is in between as well.

Keeping with the trig theme, also in between are

\dfrac{\pi} 6

and

\dfrac{\pi}{4}


the angles associated with the some of the above trig function values.

We could obviously go on as long as we cared to.



7 0
4 years ago
How modeling subtraction with fraction strips is different from modeling addition strips
Wewaii [24]
Instead of adding more fraction strips you would take some away to get a smaller number.
8 0
3 years ago
Suppose that A = PDP-1 . Prove that det(A) = det(D)
daser333 [38]

Answer:

Check.

Step-by-step explanation:

To prove it we need to know that  for two matrices A and B we have that:

det(AB) = det(A)*det(B) and det(A^{-1}) = \frac{1}{det(A)}. Now:

A = PDP^{-1}

det(A) = det(PDP^{-1})

det(A) = det(P)*det(D)*det(P^{-1})

det(A) = det(P)*det(D)*\frac{1}{det(P)}

det(A) = det(P)*\frac{1}{det(P)}*det(D)

det(A) = det(D).

5 0
4 years ago
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