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3 years ago

3/9 can also simplify to..... 3 2/6 1/3

Mathematics

2 answers:

shepuryov [24]3 years ago

5 0

Answer:

1/3

Step-by-step explanation: 3/9 can be divided by 3 on top and bottom so 3/9=1/3 because 1 times 3 is 3 and 3 times 3 is 9 which would equal 3/9.

Aloiza [94]3 years ago

4 0

Answer:

1/3

Step-by-step explanation:

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Researchers for a company that manufactures batteries want to test the hypothesis that the mean battery life of their new batter

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(b) one-sample t-test for a population mean

ur welcome :D

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3 years ago

Suppose quantity s is a length and quantity t is a time. Suppose the quantities v and a are defined by v = ds/dt and a = dv/dt.

finlep [7]

Answer:

a) $v = \frac{[L]}{[T]} = LT^{-1}$

b) $a = \frac{[L}{T}^{-1}]}{{T}}= L T^{-1} T^{-1}= L T^{-2}$

c) $\int v dt = s(t) = [L]=L$

d) $\int a dt = v(t) = [L][T]^{-1}=LT^{-1}$

e) $\frac{da}{dt}= \frac{[L][T]^{-2}}{T} = [L][T]^{-2} [T]^{-1} = LT^{-3}$

Step-by-step explanation:

Let define some notation:

[L]= represent longitude , [T] =represent time

And we have defined:

s(t) a position function

$v = \frac{ds}{dt}$

$a= \frac{dv}{dt}$

Part a

If we do the dimensional analysis for v we got:

$v = \frac{[L]}{[T]} = LT^{-1}$

Part b

For the acceleration we can use the result obtained from part a and we got:

$a = \frac{[L}{T}^{-1}]}{{T}}= L T^{-1} T^{-1}= L T^{-2}$

Part c

From definition if we do the integral of the velocity respect to t we got the position:

$\int v dt = s(t)$

And the dimensional analysis for the position is:

$\int v dt = s(t) = [L]=L$

Part d

The integral for the acceleration respect to the time is the velocity:

$\int a dt = v(t)$

And the dimensional analysis for the position is:

$\int a dt = v(t) = [L][T]^{-1}=LT^{-1}$

Part e

If we take the derivate respect to the acceleration and we want to find the dimensional analysis for this case we got:

$\frac{da}{dt}= \frac{[L][T]^{-2}}{T} = [L][T]^{-2} [T]^{-1} = LT^{-3}$

7 0

3 years ago

A box contains four 10 nf and eight 100 nf capacitors. (a) what is the probability of obtaining at least two 10 nf capacitors if

grigory [225]

there are 4 of the 10-nf and 8 of the 100-nf capacitors which is a total of 12 items.

the probability of drawing a 10-nf is: 4 out of 12 = $\frac{1}{3}$

the probability of drawing a 100-nf is: 8 out of 12 = $\frac{2}{3}$

(a) "at least two" means: 2 or 3 or 4

Probability of 2 (10's) and 2 (100's): $\frac{1}{3}$ x $\frac{1}{3}$ x $\frac{2}{3}$ x $\frac{2}{3}$ = $\frac{4}{81}$

Probability of 3 (10's) and 1 (100's): $\frac{1}{3}$ x $\frac{1}{3}$ x $\frac{1}{3}$ x $\frac{2}{3}$ = $\frac{2}{81}$

Probability of 4 (10's) and 0 (100's): $\frac{1}{3}$ x $\frac{1}{3}$ x $\frac{1}{3}$ x $\frac{1}{3}$ = $\frac{1}{81}$

2 or 3 or 4: $\frac{4}{81}$ + $\frac{2}{81}$ + $\frac{1}{81}$ = $\frac{7}{81}$

(b) without replacement

Probability of 2 (10's) and 2 (100's): $\frac{4}{12}$ x $\frac{3}{11}$ x $\frac{8}{10}$ x $\frac{7}{9}$ = $\frac{4 x 3 x 8 x 7}{12 x 11 x 10 x 9}$

Probability of 3 (10's) and 1 (100's): $\frac{4}{12}$ x $\frac{3}{11}$ x $\frac{2}{10}$ x $\frac{8}{9}$ = $\frac{4 x 3 x 2 x 8}{12 x 11 x 10 x 9}$

Probability of 4 (10's) and 0 (100's): $\frac{4}{12}$ x $\frac{3}{11}$ x $\frac{2}{10}$ x $\frac{1}{9}$ = $\frac{4 x 3 x 2 x 1}{12 x 11 x 10 x 9}$

2 or 3 or 4: $\frac{4 x 3 x 8 x 7}{12 x 11 x 10 x 9}$ + $\frac{4 x 3 x 2 x 8}{12 x 11 x 10 x 9}$ + $\frac{4 x 3 x 2 x 1}{12 x 11 x 10 x 9}$ = $\frac{672 + 192 + 24}{12 x 11 x 10 x 9}$ = $\frac{888}{12 x 11 x 10 x 9}$ = $\frac{111}{1485}$

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3 years ago

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Leona [35]

G(F(2))
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3 years ago

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Answer:

Step-by-step explanation:

SEE THE IMAGE FOR SOLUTION

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3 years ago