Root of 108 is between and

1 answer:

5 0

Without using a calculator I would say the root is between 10 and 11

How I found the answer:
I know 10 * 10 = 100
I know 11 * 11 = 121
So the root of 100 = 10 and the root of 121 = 11 so that means the root of 108 is somewhere between 10 and 11, which the root of 108 = 10.3923

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4 divided by 992 because my calculator isn't helping and i'm not trynna fail

Black_prince [1.1K]

Answer:

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Step-by-step explanation:

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4 0

3 years ago

Urgent. There are 12 coloured counters in a bag. The counters are black, white or grey A counter is chosen at random. The probab

Nastasia [14]

Answer:

1/12

Step-by-step explanation:

Needed information

$\sf Probability\:of\:an\:event\:occurring = \dfrac{Number\:of\:ways\:it\:can\:occur}{Total\:number\:of\:possible\:outcomes}$

The sum of the probabilities of all outcomes must equal 1

Solution

We are told that the probability that the counter is not black is 3/4.

As the sum of the probabilities of all outcomes must equal 1, we can work out the probability that the counter is black by subtracting 3/4 from 1:

$\sf P(counter\:not\:black)=\dfrac34$

$\implies \sf P(counter\:black)=1-\dfrac34=\dfrac14$

We are told that the probability that the counter is not white is 2/3.

As the sum of the probabilities of all outcomes must equal 1, we can work out the probability that the counter is white by subtracting 2/3 from 1:

$\sf P(counter\:not\:white)=\dfrac23$

$\implies \sf P(counter\:white)=1-\dfrac23=\dfrac13$

We are told that there are black, white and grey counters in the bag. We also know that the sum of the probabilities of all outcomes must equal 1. Therefore, we can work out the probability the counter is grey by subtracting the probability the counter is black and the probability the counter is white from 1:

$\begin{aligned}\sf P(counter\:grey) & = \sf1-P(counter\:black)-P(counter\:white)\\\\ & =\sf 1-\dfrac14-\dfrac13\\\\ & = \sf \dfrac{12}{12}-\dfrac{3}{12}-\dfrac{4}{12}\\\\ & = \sf \dfrac{12-3-4}{12}\\\\ & = \sf \dfrac{5}{12}\end{aligned}$