All categories

3 years ago

Enter a related multiplication sentence to solve 1/6 divided by 3

Mathematics

1 answer:

goldfiish [28.3K]3 years ago

6 0

Answer:

0.05555555555

Step-by-step explanation:

usless you flip it around then it would be 2 cause 1/6 / 3 = .0555555555555 because the 1/6 butif they want u to flip it then it would be 6/3=2

You might be interested in

Need help here guys.

Julli [10]

Answer:

they will pay 656 more dollars

Step-by-step explanation:

you multiply 42 by 48 and then subtract it from the full payment

5 0

2 years ago

Write each fraction as a percent!!~!! 7/20 3/5

grin007 [14]

Answer:

35%

60%

Step-by-step explanation:

7/20= 35% . 7 divided by 20 equals .35

3/5=60% 3 divided by 5 equals .6

4 0

3 years ago

Zero product property: (x-4)(-5x+1)=0 lesser x= greater x=

GaryK [48]

Answer:

lesser x= (x-4)greater x=(-5x+1)

5 0

3 years ago

40 packs of baseball cards for discounted price of 64 he sells 30 packs of baseball cards to A friend at cost much should he cha

Trava [24]

Answer: $48

Step by step:
1. 30 packs times 64 dollars is 1920
2. 1920 divided by 40 packs is 48
because if you pay 64 for 40 packs and your trying to figure out how much for 30 packs you will want to set it up like this:

$\frac{40}{64} \times \frac{30}{x}$
then you will use fishy method Google fish method in math if this doesn't help

5 0

3 years ago

The lifetime of two light bulbs are modeled as independent and exponential random variables X and Y, with parameters lambda and

Scorpion4ik [409]

Answer:

$\bf h(x)=max(\lambda,\mu) e^{-max(\lambda,\mu) x}\;(x\geq0)$

Step-by-step explanation:

The PDF of X is

$\bf f(x)=\lambda e^{-\lambda x}\;(x\geq0)$

The PDF of Y is

$\bf g(x)=\mu e^{-\mu x}\;(x\geq0)$

The means of X and Y are respectively,

$\bf \displaystyle\frac{1}{\lambda}\;,\displaystyle\frac{1}{\mu}$

so we can see that the larger the parameter, the smaller the mean. Hence the PDF of Z = min(X, Y) is an exponential with the largest parameter of the two.

Therefore, the PDF of Z is

$\bf h(x)=max(\lambda,\mu) e^{-max(\lambda,\mu) x}\;(x\geq0)$

7 0

2 years ago