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2 years ago

WRITE AN COMPOUND INEQUALITY FOR THIS SENTENCE

1 answer:

7 0

Hello there!

Write a compound inequality for this sentence: 'w' is less than 4 and greater than or equal to -8
Only use 'w' once.

W is less than four ~ $4 \ \textgreater \ w$

W is greater than or equal to -8 ~ $w \geq -8$

If we were to put this together, it would look like:

$4 \ \textgreater \ w \geq -8$

I hope I helped!

Let me know if you need anything else!

~ Zoe

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Can someone explain division of polynomials to me please.

kykrilka [37]

Answer: Division of Polynomials is just like the long division that most of us despise but this division is with Variables

Step-by-step explanation:

Example

With whatever equation you have you will

First: set up the division putting the dividend inside the divisor outside and to the left
Second: Ignore everything past the leading terms and just focus on the leading _ of the divisor and the leading _ of the dividend (just like in regular long division.
Thirdly: Take whatever is on top and multiply is by the divisor {What is on the side} carry the result underneath put it exactly below the number from the dividend
Fourth: Multiply the number that is on top by the number that is on the side, carry what is on the side underneath putting it below the other dividend.
Fifth: Do the subtraction
Sixth To subtract change all the signs in the second line, then add down.
Next: Carry down that last term from the dividend
From there you multiply and then add down again and you should be left with the answer....

If this was to many words let me know and I will upload a picture and explain with a real equation.

3 0

2 years ago

Four cards are dealt from a standard fifty-two-card poker deck. What is the probability that all four are aces given that at lea

elena-s [515]

Answer:

The probability is 0.0052

Step-by-step explanation:

Let's call A the event that the four cards are aces, B the event that at least three are aces. So, the probability P(A/B) that all four are aces given that at least three are aces is calculated as:

P(A/B) = P(A∩B)/P(B)

The probability P(B) that at least three are aces is the sum of the following probabilities:

The four card are aces: This is one hand from the 270,725 differents sets of four cards, so the probability is 1/270,725

There are exactly 3 aces: we need to calculated how many hands have exactly 3 aces, so we are going to calculate de number of combinations or ways in which we can select k elements from a group of n elements. This can be calculated as:

$nCk=\frac{n!}{k!(n-k)!}$

So, the number of ways to select exactly 3 aces is:

$4C3*48C1=\frac{4!}{3!(4-3)!}*\frac{48!}{1!(48-1)!}=192$

Because we are going to select 3 aces from the 4 in the poker deck and we are going to select 1 card from the 48 that aren't aces. So the probability in this case is 192/270,725

Then, the probability P(B) that at least three are aces is:

$P(B)=\frac{1}{270,725} +\frac{192}{270,725} =\frac{193}{270,725}$

On the other hand the probability P(A∩B) that the four cards are aces and at least three are aces is equal to the probability that the four card are aces, so:

P(A∩B) = 1/270,725

Finally, the probability P(A/B) that all four are aces given that at least three are aces is:

$P=\frac{1/270,725}{193/270,725} =\frac{1}{193}=0.0052$

5 0

3 years ago

A child's sandbox is 5 feet wide, 4 feet long, and 3 feet deep. Grace fills the sandbox so that the sand is 2 ½ feet deep. What

8_murik_8 [283]

Answer:

the volume of sand is $50 ft^{3}$

Step-by-step explanation:

This problem bothers on the mensuration of solid shapes, rectangular prism.

Given data

$Length l= 4 ft Width w= 5ft\\Height h = 3ft$

if the sandbox is filled with sand of $2\frac{1}{2 } ft$ deep the volume of the sand in the box will be calculated based on the the depth of the sand

converting the depth of sand from mixed fraction to proper fraction we have $2\frac{1}{2} = \frac{5}{2}$

the expression for the volume of a rectangular prism is

$volume of sand = length * width* height$

substituting our data into the expression we have

$volume of sand= 5*4 * \frac{5}{2} \\volume of sand = \frac{100}{2} \\volume of sand = 50 ft^{3}$

5 0

2 years ago

B) Let g(x) =x/2sqrt(36-x^2)+18sin^-1(x/6)<br><br> Find g'(x) =

jolli1 [7]

I suppose you mean

$g(x) = \dfrac x{2\sqrt{36-x^2}} + 18\sin^{-1}\left(\dfrac x6\right)$

Differentiate one term at a time.

Rewrite the first term as

$\dfrac x{2\sqrt{36-x^2}} = \dfrac12 x(36-x^2)^{-1/2}$

Then the product rule says

$\left(\dfrac12 x(36-x^2)^{-1/2}\right)' = \dfrac12 x' (36-x^2)^{-1/2} + \dfrac12 x \left((36-x^2)^{-1/2}\right)'$

Then with the power and chain rules,

$\left(\dfrac12 x(36-x^2)^{-1/2}\right)' = \dfrac12 (36-x^2)^{-1/2} + \dfrac12\left(-\dfrac12\right) x (36-x^2)^{-3/2}(36-x^2)' \\\\ \left(\dfrac12 x(36-x^2)^{-1/2}\right)' = \dfrac12 (36-x^2)^{-1/2} - \dfrac14 x (36-x^2)^{-3/2} (-2x) \\\\ \left(\dfrac12 x(36-x^2)^{-1/2}\right)' = \dfrac12 (36-x^2)^{-1/2} + \dfrac12 x^2 (36-x^2)^{-3/2}$

Simplify this a bit by factoring out $\frac12 (36-x^2)^{-3/2}$ :

$\left(\dfrac12 x(36-x^2)^{-1/2}\right)' = \dfrac12 (36-x^2)^{-3/2} \left((36-x^2) + x^2\right) = 18 (36-x^2)^{-3/2}$

For the second term, recall that

$\left(\sin^{-1}(x)\right)' = \dfrac1{\sqrt{1-x^2}}$

Then by the chain rule,

$\left(18\sin^{-1}\left(\dfrac x6\right)\right)' = 18 \left(\sin^{-1}\left(\dfrac x6\right)\right)' \\\\ \left(18\sin^{-1}\left(\dfrac x6\right)\right)' = \dfrac{18\left(\frac x6\right)'}{\sqrt{1 - \left(\frac x6\right)^2}} \\\\ \left(18\sin^{-1}\left(\dfrac x6\right)\right)' = \dfrac{18\left(\frac16\right)}{\sqrt{1 - \frac{x^2}{36}}} \\\\ \left(18\sin^{-1}\left(\dfrac x6\right)\right)' = \dfrac{3}{\frac16\sqrt{36 - x^2}} \\\\ \left(18\sin^{-1}\left(\dfrac x6\right)\right)' = \dfrac{18}{\sqrt{36 - x^2}} = 18 (36-x^2)^{-1/2}$