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3 years ago

Complete the table. then describe a pattern in the decimal forms o the fractions.

Mathematics

1 answer:

DerKrebs [107]3 years ago

3 0

It's going down by five but I'm sure on the last one..but to help add a zero like if it 0.5 it would be 0.50 that should help.

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An environmentalist wants to find out the fraction of oil tankers that have spills each month.Step 1 of 2:Suppose a sample of 47

torisob [31]

Answer: The proportion of oil tankers that had spills is $\dfrac{156}{474}$ or 0.329.

Step-by-step explanation:

Since we have given that

Number of tankers is drawn = 474

Number of tankers did not have spills = 318

Number of tankers have spills = 474 - 318 = 156

Proportion of oil tankers that had spills is given by

$\dfrac{Containing\ spill}{Total}=\dfrac{156}{474}=0.329$

Hence, the proportion of oil tankers that had spills is $\dfrac{156}{474}$ or 0.329.

3 0

3 years ago

what is the quotient 5-x/x^2 3x-4 divided by x^2-2x-15/x^2 5x 4 in simplifed form state any restrictions on the varible

zheka24 [161]

The quotient when $\frac{5-x}{x^2+3x-4} /\frac{x^2-2x - 15}{x^2+5x+4}$ in simplified form is $\frac{-(x+1)}{(x-1)(x+3)}$

<h3>What is an equation?</h3>

An equation is an expression that shows the relationship between two or more numbers and variables.

Given that equation:

$\frac{5-x}{x^2+3x-4} /\frac{x^2-2x - 15}{x^2+5x+4}$

$=\frac{5-x}{(x+4)(x-1)} /\frac{(x-5)(x+3)}{(x+4)(x+1)} \\\\=\frac{5-x}{(x+4)(x-1)} * \frac{(x+4)(x+1)}{(x-5)(x+3)}\\\\\frac{-(x-5)}{(x+4)(x-1)} * \frac{(x+4)(x+1)}{(x-5)(x+3)}\\\\=\frac{-(x+1)}{(x-1)(x+3)}$

The quotient when $\frac{5-x}{x^2+3x-4} /\frac{x^2-2x - 15}{x^2+5x+4}$ in simplified form is $\frac{-(x+1)}{(x-1)(x+3)}$

Find out more on equation at: brainly.com/question/2972832

#SPJ1

8 0

2 years ago

Stan The Boyz<br>Stream Stealer

pentagon [3]

Ummmmmm whatttt I’m confuseddd

5 0

3 years ago

Read 2 more answers

Determining Whether a Difference Is Statistically

SCORPION-xisa [38]

Using the z-distribution, it is found that:

The 95% confidence interval is of -1.38 to 1.38.

The value of the sample mean difference is of 1.74, which falls outside the 95% confidence interval.

<h3>What is the z-distribution confidence interval?</h3>

The confidence interval is:

$\overline{x} \pm zs$

In which:

$\overline{x}$ is the difference between the population means.

s is the standard error.

In this problem, we have a 95% confidence level, hence $\alpha = 0.95$ , z is the value of Z that has a p-value of $\frac{1+0.95}{2} = 0.975$ , so the critical value is z = 1.96.