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3 years ago

Convert 12 1/2% into a decimal (It isn't 12.50)Convert 12 1/2% into a fraction (It isn't 25/2)

Mathematics

1 answer:

Hoochie [10]3 years ago

6 0

Answer:

0.125 and 1/8

Step-by-step explanation:

To convert a percentage to a decimal, divide by 100 to move 2 decimal places to the left.

12.50% / 100 = 0.125

To convert a percentage into a fraction, rewrite the decimal number you got and put it over 1. Multiply the numerator and denominator by 1000 to move 3 decimal places to the right and get your fraction.

$\frac{0.125}{1}$ · $\frac{1000}{1000}$ = 125/1000

Divide the numerator and denominator by the GCF, 125, to get your simplified fraction.

125/125 = 1

1000/125 = 8

125/1000 = 1/8

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4 + b^2 x c / 2

statuscvo [17]

Answer:

The answer is 80 just substitute the values for b and c

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3 years ago

Alondra says that (12 + 31) × 5 is equivalent to 12 + 31 × 5 because of the associative property. Determine if Alondra is correc

MAXImum [283]

Answer:

Step-by-step explanation:

(12 + 31) × 5 = 215

12 + 31 × 5 = 167

order of operations are the rules that state the sequence in which the multiple operations in an expression should be solved.

Order of Operations :

1. Parentheses

2. Exponents

3. Multiplication and Division

4. Addition and Subtraction

7 0

2 years ago

An airplane has 100 seats for passengers. Assume that the probability that a person holding a ticket appears for the flight is 0

zmey [24]

Answer:

96.33% probability that everyone who appears for the flight will get a seat

Step-by-step explanation:

I am going to use the binomial approximation to the normal to solve this question.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

$E(X) = np$

The standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore 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 pvalue, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that $\mu = E(X)$ , $\sigma = \sqrt{V(X)}$ .