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3 years ago

Please answer/do 3-7!! Whoever does it right, i will give brainlest! :) also, PLEASE FILL IT OUT EITHER ON PAPER/ PHONE etc. !!!

Mathematics

2 answers:

meriva3 years ago

7 0

Answer:

3-7 is negitive 4 in your answer it's 3-7 = -4

Step-by-step explanation:

It's negitive four because if you subtract you will get 4 but since it's other way around. And if you do It, it will say that even if there is no negitive sign it won't change the problem because it's still the same problem.

1) 3 - 7

2) there are 3 ways to do this to get the answer:

1) 3/7

2) 7 + 3 - 4 + 3

3) now 7 + 3 is ten - 7 is 3 + 4 is 7, so then if you do 3 - 7 you get negitive four

4) now check your answer by subtracting 3 and 7

5) then you get -4

btw: ( hope this helped :) )

Klio2033 [76]3 years ago

6 0

3. 54.87 + 7.48 = 62.35 55 + 7 = 62

5. 0.215 + 3.74 = 3.955 0 + 3 = 3

7. 18.419 - 6.47 = 11.949 I’m assuming it should be 18 - 6 = 11 because of the decimal but I could be wrong about it

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We considered the differences between the temperature readings in January 1 of 1968 and 2008 at 51 locations in the continental

mr Goodwill [35]

Answer:

$1.1-2.02\frac{4.9}{\sqrt{50}}=-0.30$

$1.1+2.02\frac{4.9}{\sqrt{50}}=2.50$

So on this case the 90% confidence interval would be given by (-0.30;2.50)

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$\bar X=1.1$ represent the sample mean for the sample

$\mu$ population mean (variable of interest)

s=4.9 represent the sample standard deviation

n=51 represent the sample size

Solution to the problem

The confidence interval for the mean is given by the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

In order to calculate the critical value $t_{\alpha/2}$ we need to find first the degrees of freedom, given by:

$df=n-1=51-1=50$

Since the Confidence is 0.90 or 90%, the value of $\alpha=0.1$ and $\alpha/2 =0.05$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.05,50)".And we see that $t_{\alpha/2}=2.02$

Now we have everything in order to replace into formula (1):

$1.1-2.02\frac{4.9}{\sqrt{50}}=-0.30$

$1.1+2.02\frac{4.9}{\sqrt{50}}=2.50$

So on this case the 90% confidence interval would be given by (-0.30;2.50)

8 0

3 years ago

2.75 + .003 + .158 =

Alecsey [184]

2.75 Answer is 2.911

0.003 Make sure you line up the decimal points like shown in example

0.158

_________

2.911

6 0

3 years ago

Read 2 more answers

If f(x)=-2x^2+5x-3 and g(x) is the reflection of f(x) across the y-axis, then an equation of g is which of the following

Ymorist [56]

To reflect a function across the y-axis you have to change

$x\mapsto -x$

So, the function becomes

$f(x)=-2x^2+5x-3 \mapsto f(-x)=-2(-x)^2+5(-x)-3=-2x^2-5-3$

3 0

3 years ago

WHAT IS THE ANSWER PLS HELPPPPP!!!!!

bezimeni [28]

Answer:

5^x - 5^-x

Step-by-step explanation:

g(x) = 5^-x

h(x) = 5^x

We want h(x) - g(x)

h(x) - g(x) = 5^x - 5^-x

This cannot be simplified

6 0

3 years ago

Can u answer these for me with the work shown

babymother [125]

Answer:

$\frac{x^3 + 2x^2 -9x-18}{x^3-x^2-6x}= \frac{(x+3)}{x}$

$\frac{3x^2 - 5x - 2}{x^3 - 2x^2} = \frac{3x + 1}{x^2}$

$\frac{6 - 2x}{x^2 - 9} * \frac{15 + 5x}{4x}=-\frac{5}{2x}$

$\frac{x^2 -6x + 9}{5x - 15} / \frac{5}{3-x} = \frac{-(x-3)^2}{25}$

$\frac{x^3 - x^2 -x + 1}{x^2 - 2x+1}= x +1$

$\frac{9x^2 + 3x}{6x^2} = \frac{3x + 1}{2x}$

$\frac{x^2-3x+2}{4x} * \frac{12x^2}{x^2 - 2x} / \frac{x - 1}{x} = 3x$

Step-by-step explanation:

Required

Simplify

Solving (1):

$\frac{x^3 + 2x^2 -9x-18}{x^3-x^2-6x}$

Factorize the numerator and the denominator

$\frac{x^2(x + 2) -9(x+2)}{x(x^2-x-6)}$

Factor out x+2 at the numerator

$\frac{(x^2 -9)(x+2)}{x(x^2-x-6)}$

Express x^2 - 9 as difference of two squares

$\frac{(x^2 -3^2)(x+2)}{x(x^2-x-6)}$

$\frac{(x -3)(x+3)(x+2)}{x(x^2-x-6)}$

Expand the denominator

$\frac{(x -3)(x+3)(x+2)}{x(x^2-3x+2x-6)}$

Factorize

$\frac{(x -3)(x+3)(x+2)}{x(x(x-3)+2(x-3))}$

$\frac{(x -3)(x+3)(x+2)}{x(x+2)(x-3)}$

Cancel out same factors

$\frac{(x+3)}{x}$

Hence:

$\frac{x^3 + 2x^2 -9x-18}{x^3-x^2-6x}= \frac{(x+3)}{x}$

Solving (2):

$\frac{3x^2 - 5x - 2}{x^3 - 2x^2}$

Expand the numerator and factorize the denominator

$\frac{3x^2 - 6x + x - 2}{x^2(x- 2)}$

Factorize the numerator

$\frac{3x(x - 2) + 1(x - 2)}{x^2(x- 2)}$

Factor out x - 2

$\frac{(3x + 1)(x - 2)}{x^2(x- 2)}$

Cancel out x - 2

$\frac{3x + 1}{x^2}$

Hence:

$\frac{3x^2 - 5x - 2}{x^3 - 2x^2} = \frac{3x + 1}{x^2}$

Solving (3):

$\frac{6 - 2x}{x^2 - 9} * \frac{15 + 5x}{4x}$

Express x^2 - 9 as difference of two squares

$\frac{6 - 2x}{x^2 - 3^2} * \frac{15 + 5x}{4x}$

Factorize all:

$\frac{2(3 - x)}{(x- 3)(x+3)} * \frac{5(3 + x)}{2(2x)}$

Cancel out x + 3 and 3 + x

$\frac{2(3 - x)}{(x- 3)} * \frac{5}{2(2x)}$

$\frac{3 - x}{x- 3} * \frac{5}{2x}$

Express $3 - x$ as $-(x - 3)$

$\frac{-(x-3)}{x- 3} * \frac{5}{2x}\\$

$-1 * \frac{5}{2x}$

$-\frac{5}{2x}$

Hence:

$\frac{6 - 2x}{x^2 - 9} * \frac{15 + 5x}{4x}=-\frac{5}{2x}$

Solving (4):

$\frac{x^2 -6x + 9}{5x - 15} / \frac{5}{3-x}$

Expand x^2 - 6x + 9 and factorize 5x - 15

$\frac{x^2 -3x -3x+ 9}{5(x - 3)} / \frac{5}{3-x}$

Factorize

$\frac{x(x -3) -3(x-3)}{5(x - 3)} / \frac{5}{3-x}$

$\frac{(x -3)(x-3)}{5(x - 3)} / \frac{5}{3-x}$

Cancel out x - 3

$\frac{(x -3)}{5} / \frac{5}{3-x}$

Change / to *

$\frac{(x -3)}{5} * \frac{3-x}{5}$

Express $3 - x$ as $-(x - 3)$

$\frac{(x -3)}{5} * \frac{-(x-3)}{5}$

$\frac{-(x-3)(x -3)}{5*5}$

$\frac{-(x-3)^2}{25}$

Hence:

$\frac{x^2 -6x + 9}{5x - 15} / \frac{5}{3-x} = \frac{-(x-3)^2}{25}$

Solving (5):

$\frac{x^3 - x^2 -x + 1}{x^2 - 2x+1}$

Factorize the numerator and expand the denominator

$\frac{x^2(x - 1) -1(x - 1)}{x^2 - x-x+1}$

Factor out x - 1 at the numerator and factorize the denominator

$\frac{(x^2 - 1)(x - 1)}{x(x -1)- 1(x-1)}$

Express x^2 - 1 as difference of two squares and factor out x - 1 at the denominator

$\frac{(x +1)(x-1)(x - 1)}{(x -1)(x-1)}$

$x +1$

Hence:

$\frac{x^3 - x^2 -x + 1}{x^2 - 2x+1}= x +1$

Solving (6):

$\frac{9x^2 + 3x}{6x^2}$

Factorize:

$\frac{3x(3x + 1)}{3x(2x)}$

Divide by 3x

$\frac{3x + 1}{2x}$

Hence:

$\frac{9x^2 + 3x}{6x^2} = \frac{3x + 1}{2x}$

Solving (7):

$\frac{x^2-3x+2}{4x} * \frac{12x^2}{x^2 - 2x} / \frac{x - 1}{x}$

Change / to *

$\frac{x^2-3x+2}{4x} * \frac{12x^2}{x^2 - 2x} * \frac{x}{x-1}$

Expand

$\frac{x^2-2x-x+2}{4x} * \frac{12x^2}{x^2 - 2x} * \frac{x}{x-1}$

Factorize

$\frac{x(x-2)-1(x-2)}{4x} * \frac{12x^2}{x(x - 2)} * \frac{x}{x-1}$

$\frac{(x-1)(x-2)}{4x} * \frac{12x^2}{x(x - 2)} * \frac{x}{x-1}$

Cancel out x - 2 and x - 1

$\frac{1}{4x} * \frac{12x^2}{x} * \frac{x}{1}$

Cancel out x

$\frac{1}{4x} * \frac{12x^2}{1} * \frac{1}{1}$

$\frac{12x^2}{4x}$

$3x$

Hence:

$\frac{x^2-3x+2}{4x} * \frac{12x^2}{x^2 - 2x} / \frac{x - 1}{x} = 3x$

8 0

3 years ago