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Elina [12.6K]
3 years ago
13

How are 3 thirds and 6 sixths alike?

Mathematics
2 answers:
EleoNora [17]3 years ago
5 0
Answer: they both equal to one

Explanation:

3/3 can simplify to 1/1 and 6/6 can simplify to 1/1
UNO [17]3 years ago
4 0

Answer:

Brainliest

Step-by-step explanation:

The both equal one

three 1/3 = 3/3 = 1

six 1/6 = 6/6 = 1

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The answer should be: (2 - 1) = m(-3 - 2)
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For a normal distribution with μ=500 and σ=100, what is the minimum score necessary to be in the top 60% of the distribution?
STatiana [176]

Answer:

475.

Step-by-step explanation:

We have been given that for a normal distribution with μ=500 and σ=100. We are asked to find the minimum score that is necessary to be in the top 60% of the distribution.

We will use z-score formula and normal distribution table to solve our given problem.

z=\frac{x-\mu}{\sigma}

Top 60% means greater than 40%.

Let us find z-score corresponding to normal score to 40% or 0.40.

Using normal distribution table, we got a z-score of -0.25.

Upon substituting our given values in z-score formula, we will get:

-0.25=\frac{x-500}{100}

-0.25*100=\frac{x-500}{100}*100

-25=x-500

-25+500=x-500+500

x=475

Therefore, the minimum score necessary to be in the top 60% of the distribution is 475.

3 0
2 years ago
Express as -80/112 as a rational: numerator = -5​
Anastasy [175]

Answer:

-80\112 = -5\7

we divided both numbers by 16

8 0
2 years ago
Find the probability that a randomly generated bit string of length 10 does not contain a 0 if bits are independent and if a 0 b
jeka94

Answer:

Step-by-step explanation:

From the given information; Let's assume that  R should represent the set of all possible outcomes generated from  a bit string of length 10 .

So; as each place is fitted with either 0 or 1

\mathbf{|R|= 2^{10}}

Similarly; the event E signifies the randomly generated bit string of length 10 does not contain a 0

Now;

if  a 0 bit and a 1 bit are equally likely

The probability that a randomly generated bit string of length 10 does not contain a 0 if bits are independent and if a 0 bit and a 1 bit are equally likely is;

\mathbf{P(E) = \dfrac{|E|}{|R|}}

so ; if bits string should not contain a 0 and all other places should be occupied by 1; Then:

\mathbf{{|E|}=1 }   ; \mathbf{|R|= 2^{10}}

\mathbf{P(E) = \dfrac{1}{2^{10}}}

4 0
3 years ago
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