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3 years ago

The probability that a randomly selected point on the grid below is in the blue area is 9/16

Mathematics

2 answers:

DochEvi [55]3 years ago

8 0

Answer:

The correct option is 1.

Step-by-step explanation:

it is given that the probability of a randomly selected point on the grid below is in the blue area is 9/16.

In the given grid we have only two colors that are blue and white.

Let A be the event of a randomly selected point on the given grid is in the blue area.

$P(A)=\frac{9}{16}$

If the randomly selected point on the given grid does not lie in the blue area, it means it lies on white area.

We will calculate P(A'), to find the probability that a randomly selected point is in the white on the grid.

We know that the sum of the probability is 1.

$P(A)+P(A')=1$

$P(A')=1-P(A)$

It is given that $P(A)=\frac{9}{16}$

$P(A')=1-\frac{9}{16}$

Therefore option 1 is correct.

Arturiano [62]3 years ago

5 0

the answer is 1 + 9/16!!!1

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Original price of a CD: $20.00 <br> Discount 42% <br> Tax 6%

vodomira [7]

First, we can subtract the discount from the original price. Divide the discount (42%) over 100.

42/100 = 0.42

Next, we need to multiply. It's basically asking what 42% of 20 is.

0.42 × $20 = $8.40

Now, we know that we will have $8.40 on the CD because of the discount. We can subtract the saved amount from the original amount to get the price.

$20 - $8.40 = $11.60

You will pay $11.60 for the CD without taxes.

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Now to add the taxes.... To do this, do the same first step we did when finding the discount price.

6/100 = 0.06

Multiply. Like I said, you can think of it as, '6% of 11.60'.

0.06 × 11.60 = $0.70

Last step is to ADD. Since taxes add more money onto the current price, we will have to add.

$11.60 + $0.70 = $12.30

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The total price of the CD will be: $12.30.

3 0

3 years ago

<img src="https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Csf%5Clim_%7Bx%20%5Cto%200%20%7D%20%5Cfrac%7B1%20-%20%5Cprod%20%5Climits_%

xxTIMURxx [149]

To demonstrate a method for computing the limit itself, let's pick a small value of n. If n = 3, then our limit is

$\displaystyle \lim_{x \to 0 } \frac{1 - \prod \limits_{k = 2}^{3} \sqrt[k]{\cos(kx)} }{ {x}^{2} }$

Let a = 1 and b the cosine product, and write them as

$\dfrac{a - b}{x^2}$

with

$b = \sqrt{\cos(2x)} \sqrt[3]{\cos(3x)} = \sqrt[6]{\cos^3(2x)} \sqrt[6]{\cos^2(3x)} = \left(\cos^3(2x) \cos^2(3x)\right)^{\frac16}$

Now we use the identity

$a^n-b^n = (a-b)\left(a^{n-1}+a^{n-2}b+a^{n-3}b^2+\cdots a^2b^{n-3}+ab^{n-2}+b^{n-1}\right)$

to rationalize the numerator. This gives

$\displaystyle \frac{a^6-b^6}{x^2 \left(a^5+a^4b+a^3b^2+a^2b^3+ab^4+b^5\right)}$

As x approaches 0, both a and b approach 1, so the polynomial in a and b in the denominator approaches 6, and our original limit reduces to

$\displaystyle \frac16 \lim_{x\to0} \frac{1-\cos^3(2x)\cos^2(3x)}{x^2}$

For the remaining limit, use the Taylor expansion for cos(x) :

$\cos(x) = 1 - \dfrac{x^2}2 + \mathcal{O}(x^4)$

where $\mathcal{O}(x^4)$ essentially means that all the other terms in the expansion grow as quickly as or faster than x⁴; in other words, the expansion behaves asymptotically like x⁴. As x approaches 0, all these terms go to 0 as well.

Then

$\displaystyle \cos^3(2x) \cos^2(3x) = \left(1 - 2x^2\right)^3 \left(1 - \frac{9x^2}2\right)^2$

$\displaystyle \cos^3(2x) \cos^2(3x) = \left(1 - 6x^2 + 12x^4 - 8x^6\right) \left(1 - 9x^2 + \frac{81x^4}4\right)$

$\displaystyle \cos^3(2x) \cos^2(3x) = 1 - 15x^2 + \mathcal{O}(x^4)$

so in our limit, the constant terms cancel, and the asymptotic terms go to 0, and we end up with

$\displaystyle \frac16 \lim_{x\to0} \frac{15x^2}{x^2} = \frac{15}6 = \frac52$

Unfortunately, this doesn't agree with the limit we want, so n ≠ 3. But you can try applying this method for larger n, or computing a more general result.

Edit: some scratch work suggests the limit is 10 for n = 6.

6 0

3 years ago

-(4k^2-13k-12)+5k^2-8k How would I be able to write this in standard form with the greatest exponent being first.

aliina [53]

Answer:

Step-by-step explanation:

Negative sign is outside the parenthesis. So, each term of (4k² - 13k -12) should be multiplied by (-1)

-(4k² - 13k -12) + 5k² - 8k = 4k²*(-1) - 13k*(-1) - 12*(-1) + 5k² - 8k

= -4k² + 13k + 12 +5k² - 8k

Combine like terms,

= -4k² + 5k² + 13k - 8k + 12

= k² + 5k + 12

Like terms are the terms with same variable with same exponent.

(-4k² ) and 5k² are like terms

13k and (-8k) are like terms

3 0

3 years ago

Matt has 50 apples and gave 20 to his brother jhon how many apples does he have now

alexdok [17]

Answer:

30 apples

Explanation:

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hammer [34]

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