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4 years ago

How do I show my work for this ?

Mathematics

1 answer:

Zarrin [17]4 years ago

3 0

First, you multiply all terms by 2.8 to get rid of the denominator:
m-13.72 = -19.908
then, you isolate the m on one side of the equation and all other terms on the other side:
m = -19.908+13.72
m = -6.188

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Evaluate the expression n ÷ 5 for n = 3.2. A. 0.064 B. 0.604 C. 0.64 D. 6.4

Licemer1 [7]

Answer:

n/5

substitution for n = 3.2

3.2/5

0.64

3 0

3 years ago

4|n|-5=7 please help me solve

DedPeter [7]

I think you meant to type

4(n)-5=7.

4n = 7 + 5

4n = 12

n = 12/4

n = 3

3 0

4 years ago

Read 2 more answers

katrin2010 [14]

Answer:

A ≈ 35.3 units²

Step-by-step explanation:

Calculate the radius CB using Pythagoras' identity in the right triangle.

CB² + AB² = AC²

CB² + 6² = 9²

CB² + 36 = 81 ( subtract 36 from both sides )

CB² = 45 = r²

Then area of quarter circle is

A = $\frac{1}{4}$ × πr² = $\frac{1}{4}$ × π × 45 ≈ 35.3

6 0

3 years ago

Use a proof by contradiction to show that the square root of 3 is national You may use the following fact: For any integer kirke

Ierofanga [76]

Answer:

1. Let us proof that √3 is an irrational number, using reductio ad absurdum. Assume that $\sqrt{3}=\frac{m}{n}$ where $m$ and $n$ are non negative integers, and the fraction $\frac{m}{n}$ is irreducible, i.e., the numbers $m$ and $n$ have no common factors.

Now, squaring the equality at the beginning we get that

$3=\frac{m^2}{n^2}$ (1)

which is equivalent to $3n^2=m^2$ . From this we can deduce that 3 divides the number $m^2$ , and necessarily 3 must divide $m$ . Thus, $m=3p$ , where $p$ is a non negative integer.

Substituting $m=3p$ into (1), we get

$3= \frac{9p^2}{n^2}$

which is equivalent to

$n^2=3p^2$ .

Thus, 3 divides $n^2$ and necessarily 3 must divide $n$ . Hence, $n=3q$ where $q$ is a non negative integer.

Notice that

$\frac{m}{n} = \frac{3p}{3q} = \frac{p}{q}$ .

The above equality means that the fraction $\frac{m}{n}$ is reducible, what contradicts our initial assumption. So, $\sqrt{3}$ is irrational.

2. Let us prove now that the multiplication of an integer and a rational number is a rational number. So, $r\in\mathbb{Q}$ , which is equivalent to say that $r=\frac{m}{n}$ where $m$ and $n$ are non negative integers. Also, assume that $k\in\mathbb{Z}$ . So, we want to prove that $k\cdot r\in\mathbb{Z}$ . Recall that an integer $k$ can be written as

$k=\frac{k}{1}$ .

Then,

$k\cdot r = \frac{k}{1}\frac{m}{n} = \frac{mk}{n}$ .

Notice that the product $mk$ is an integer. Thus, the fraction $\frac{mk}{n}$ is a rational number. Therefore, $k\cdot r\in\mathbb{Q}$ .

3. Let us prove by reductio ad absurdum that the sum of a rational number and an irrational number is an irrational number. So, we have $x$ is irrational and $p\in\mathbb{Q}$ .

Write $q=x+p$ and let us suppose that $q$ is a rational number. So, we get that

$x=q-p$ .

But the subtraction or addition of two rational numbers is rational too. Then, the number $x$ must be rational too, which is a clear contradiction with our hypothesis. Therefore, $x+p$ is irrational.

7 0

4 years ago

The fuel efficiency in miles per gallon of all bmw 320i's is (approximately) normally distributed with a mean of 25 and a standa

grigory [225]

Given:
μ = 25 mpg, the population mean
σ = 2 mpg, the population standard deviation

If we select n samples for evaluation, we should calculate z-scores that are based on the standard error of the mean.
That is,
$z= \frac{x-\mu}{\sigma / \sqrt{n} }$

The random variable is x = 24 mpg.

Part (i): n = 1
σ/√n = 2
z = (24 -25)/2 = -0.5
From standard tables,
P(x < 24) = 0.3085

Part (ii): n = 4
σ/√n = 1
z = (24 -25)/1 = -1
P(x < 24) = 0.1587

Part (iii): n=16
σ/√n = 0.5
z = (24 - 25)/0.5 = -2
P(x < 24) = 0.0228

Explanation:
The larger the sample size, the smaller the standard deviation.
Therefore when n increases, we are getting a result which is closer to that of the true mean.

6 0

4 years ago