Why are you allowed to move the decimal points before dividing with decimals? Explain your reasoning.

2 answers:

7 0

If you move the decimal point the same number of places in the dividend and the divisor, you are multiplying them both by the same power of 10. That does not change the quotient.

Goshia [24]3 years ago

6 0

Answer:

If you move the decimal point the same number of places in the dividend and the divisor, you are multiplying them both by the same power of 10. That does not change the quotient.

Step-by-step explanation:

it was what i got and it was right

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(3x² - 4x+8) + (-7² - 27-8)

iragen [17]

Answer:

3x^ - 4x - 76

Step-by-step explanation:

4 0

3 years ago

According to the National Bridge Inspection Standard (NBIS), public bridges over 20 feet in length must be inspected and rated e

slamgirl [31]

Answer:

1.80% probability that in a random sample of 12 major Denver bridges, at least 4 will have an inspection rating of 4 or below in 2020.

Step-by-step explanation:

For each bridge, there are only two possible outcomes. Either it has rating of 4 or below, or it does not. The probability of a bridge being rated 4 or below is independent from other bridges. So we use the binomial probability distribution to solve this problem.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

For the year 2020, the engineers forecast that 9% of all major Denver bridges will have ratings of 4 or below.

This means that $p = 0.09$

Use the forecast to find the probability that in a random sample of 12 major Denver bridges, at least 4 will have an inspection rating of 4 or below in 2020.

Either less than 4 have a rating of 4 or below, or at least 4 does. The sum of the probabilities of these events is 1.

$P(X < 4) + P(X \geq 4) = 1$