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2 years ago

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The World Health Organization lists the following overall literacy rates per 100 people for countries. Which answer choice inclu

des all countries with literacy rates above 90%?

2 answers:

IRISSAK [1]2 years ago

7 0

The second one lists all six countries with literacy rates above 90%.

Lynna [10]2 years ago

6 0

The second option. All those countries have literacy rates higher than 90%

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Please anyone can you guys help me it would mean the world

lapo4ka [179]

It says 404 not found. :(

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4 years ago

1/20 in dicimal please help

KonstantinChe [14]

0.05 is the answer for your question.

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4 years ago

Create a linear function that has a rate of change of -4 and a solution at (-5,2)

san4es73 [151]

You would use the point-slope equation of y-y₁=mx(x-x₁)

So the rate of change is the slope and equals m, so m=-4

and has a solution at (-5,2) which would be the (x₁, y₁) and then it is just plugging in the numbers

y - (2) = (-4)(x - (-5))

y - 2 = -4 (x+5)

y - 2 = -4x - 20

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3 years ago

Find the median<br> 23 ,15, 38 ,34 ,22, 2, 21 ,27 ,26 ,14, 32, 30, 19

xxTIMURxx [149]

First put them in ascending numerical order
2, 14, 15, 19, 22, 21, 26, 27, 23, 30, 32, 34, 38
Then start to eliminate numbers by pair (the farthest number from each side) until you get to 1 or two numbers left (in the very middle)
in this case you will be left with 26, which means that the median is 26

3 0

3 years ago

Read 2 more answers

Because of their connection with secant lines, tangents, and instantaneous rates, limits of the form ModifyingBelow lim With h

Gre4nikov [31]

Answer:

$\dfrac{1}{2\sqrt{x}}$

Step-by-step explanation:

$f(x) = \sqrt{x} = x^{\frac{1}{2}}$

$f(x+h) = \sqrt{x+h} = (x+h)^{\frac{1}{2}}$

We use binomial expansion for $(x+h)^{\frac{1}{2}}$

This can be rewritten as

$[x(1+\dfrac{h}{x})]^{\frac{1}{2}}$

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

From the expansion

$(1+x)^n=1+nx+\dfrac{n(n-1)}{2!}+\ldots$

Setting $x=\dfrac{h}{x}$ and $n=\frac{1}{2}$ ,

$(1+\dfrac{h}{x})^{\frac{1}{2}}=1+(\dfrac{h}{x})(\dfrac{1}{2})+\dfrac{\frac{1}{2}(1-\frac{1}{2})}{2!}(\dfrac{h}{x})^2+\tldots$

$=1+\dfrac{h}{2x}-\dfrac{h^2}{8x^2}+\ldots$

Multiplying by $x^{\frac{1}{2}}$ ,

$x^{\frac{1}{2}}(1+\dfrac{h}{x})^{\frac{1}{2}}=x^{\frac{1}{2}}+\dfrac{h}{2x^{\frac{1}{2}}}-\dfrac{h^2}{8x^{\frac{3}{2}}}+\ldots$

$x^{\frac{1}{2}}(1+\dfrac{h}{x})^{\frac{1}{2}}-x^{\frac{1}{2}}=\dfrac{h}{2x^{\frac{1}{2}}}-\dfrac{h^2}{8x^{\frac{3}{2}}}+\ldots$

$\dfrac{x^{\frac{1}{2}}(1+\dfrac{h}{x})^{\frac{1}{2}}-x^{\frac{1}{2}}}{h}=\dfrac{1}{2x^{\frac{1}{2}}}-\dfrac{h}{8x^{\frac{3}{2}}}+\ldots$

The limit of this as $h\to 0$ is

$\lim_{h\to0} \dfrac{f(x+h)-f(x)}{h}=\dfrac{1}{2x^{\frac{1}{2}}}=\dfrac{1}{2\sqrt{x}}$ (since all the other terms involve $h$ and vanish to 0.)

8 0

3 years ago