How do you show the remainder of 166/6 as a fraction

Mathematics

2 answers:

Sedaia [141]3 years ago

8 0

The remainder of 166 is:

4.

You put the remainder over the divisor and simplify.

Easy!
27 4/6
or
27 2/3

podryga [215]3 years ago

4 0

I think It's. 1 4/6.

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. Tommy gives baseball cards to four of his friends. He gives Kiana 3 cards less than he gives Gary.

Licemer1 [7]

We need to write system of equations for this problem. Reading text we write 1 equation by 1.
first lets name our variables
x - Kiana cards
y - Gary cards
z - Monty cards
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x = y-3 which we can write as y = x+3
z = x + 7
m = 2z
x + y + z + m = 74

I think easiest way to solve this is to express all variables in last equation and solve for x.
x + x+3 + x + 7+ 2x + 14= 74
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y = 13
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3 years ago

A patio in the shape of a regular octagon (8-sided figure) has a side length of 24 ft and an apothem length of 60 ft. What is th

Damm [24]

Given:
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3 years ago

At a new exhibit in the Museum of Science, people are asked to choose between 94 or 220 random draws from a machine. The machine

Tresset [83]

Answer:

0.0869 = 8.69% probability of getting more than 61% green balls.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean $\mu = p$ and standard deviation $s = \sqrt{\frac{p(1-p)}{n}}$