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3 years ago

13

NEEED HELP ASSSSAAAAPPPPP PLEASE

1 answer:

boyakko [2]3 years ago

4 0

Answer:

Option 2 - square root 5, 5 over 2, 2.71 repeating 71, 2 and 3 over 4.

Step-by-step explanation:

Given : Numbers 2.71 repeating 71, 2 and 3 over 4, square root 5, 5 over 2

To find : Order the set of numbers from least to greatest ?

Solution :

First we write numbers in one form i.e. in decimals,

2.71 repeating 71 = $2.7171....$

2 and 3 over 4 = $2\frac{3}{4}=\frac{11}{4}=2.75$

square root 5 = $\sqrt{5}=2.2360....$

5 over 2 = $\frac{5}{2}=2.5$

Arranging from least to greatest,

$2.2360.. < 2.5 < 2.7171.. < 2.75$

or $\sqrt{5}< \frac{5}{2} < 2.7171.. < 2\frac{3}{4}$

or square root 5, 5 over 2, 2.71 repeating 71, 2 and 3 over 4.

Therefore, Option 2 is correct.

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Random error is: A. an outside influence that affects the true score in a consistent way. B. the extent to which a measure will

k0ka [10]

D is the answer, i just done this one

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2 years ago

Slove.<br> -5 3/4 - 3 1/2 =<br> A. -9 1/4 <br> B. -2 1/4 <br> C. -8 2/3<br> D. 2 1/4

Komok [63]

Answer:

A

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

Find the absolute maximum and minimum values of f(x,y)=xy−4x in the region bounded by the x-axis and the parabola y=16−x2.

skad [1K]

Answer:

The absolute maximum and minimum is $20\; \text{and} -20.$

Step-by-step explanation:

We first check the critical points on the interior of the domain using the

first derivative test.

$f_x=y-4=0$

$f_y=x=0$

The only solution to this system of equations is the point (0, 4), which lies in the domain.

$f_{xx}=0, \;f_{yy}=0\; \text{and}\; f_{xy}=-1$

$\Rightarrow f_{xx}f_{yy}-f_{xy}=o-1=-1$

$\therefore (0,4)$ is a saddle point.

Boundary points - $(4,0), (-4,0), (0,16)$

Along boundary $y=16-x^2$

$f=x(16-x^2)-4x$

$=16x-x^3-4x$

$\Rightarrow f^'=16-3x^2-4=0$

$\Rightarrow 3x^2=12$

$\Rightarrow x=\pm2,\;\;y=14$

Values of f(x) at these points.

$\begin{array}{}(4,0)=-16\\(-4,0)=16\\(0,16)=0\\(2,14)=20\\(-2,14)=-20\end{array}$

Therefore, the absolute maximum and minimum is $20\; \text{and} -20.$

6 0

3 years ago

For Restaurant A, the owner wants to advertise a good cleanliness score to his customers. Which measure would make his score see

Semenov [28]

Answer:

median

Step-by-step explanation:

the median would make the score as good as possible, because there could be possible outliers. this really depends on if there are outliers or not. here is an example of what i need:

say you are trying to show your parents your grades:

here are your scores:

38, 88, 90, 91, 93, 95, 100

the median would be 91, which looks better.

the mean would be 85.

even though a <em>majority</em> of your grades are good, the one 38 makes the average go down. therefore, the median is a better advertiser.

8 0

4 years ago

Please help fassssst

Zarrin [17]

Answer:

$\dfrac{9}{2}$

Step-by-step explanation:

You can do this using the quadratic equation:

$x =\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}$

Let's first setup our expression:

4x² + 12x = 135

the quadratic formula is:

ax² + bx + c = 0

4x² + 12x - 135 = 0

Then:

a = 4

b = 12

c = -135

Now we plug in our coefficients and solve:

$x =\dfrac{-12\pm\sqrt{12^{2}-4(4)(-135)}}{2(4)}$

We solve for both to determine the positive one:

$x =\dfrac{-12+ \sqrt{12^{2}-4(4)(-135)}}{2(4)}$ $x =\dfrac{-12- \sqrt{12^{2}-4(4)(-135)}}{2(4)}$

$x =\dfrac{-12+ \sqrt{144+2160}}{8}$ $x =\dfrac{-12- \sqrt{144+2160}}{8}$

$x =\dfrac{-12+ \sqrt{2304}}{8}$ $x =\dfrac{-12- \sqrt{2304}}{8}$

$x =\dfrac{-12+ 48}{8}$ $x =\dfrac{-12- 48}{8}$

$x =\dfrac{36}{8} = \dfrac{9}{2}$ $x =\dfrac{-60}{8} = \dfrac{-15}{2}$

So if you are looking for a positive solution, just take the positive one as x.

8 0

3 years ago