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3 years ago

May someone help? 1+pq use p=6 and q=5

Mathematics

2 answers:

lilavasa [31]3 years ago

6 0

Answer: 31

Step-by-step solution:

1 + (6•5)
1 + 30
31

I hope this helps!

larisa86 [58]3 years ago

4 0

Answer:

31 is the answer

Step-by-step explanation:

1+(6x5)

6x5 is 30 and then add 1

when 2 numbers are exactlky next to ach other you multiply them so therefore the answer is 30+1=31

Hope this helped :)

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They paid $24 each for the rental.

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2 years ago

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Ardem collected data from a class survey. He then randomly selected samples of five responses to generate four samples.

scoundrel [369]

Solution: The sample mean of sample 1 is:

$\bar{x}=\frac{4+5+2+4+3}{5}= \frac{18}{5}=3.6$

The sample mean of sample 2 is:

$\bar{x}=\frac{2+2+6+5+7}{5}= \frac{22}{5}=4.4$

The sample mean of sample 3 is:

$\bar{x}=\frac{4+6+3+4+1}{5}= \frac{18}{5}=3.6$

The sample mean of sample 4 is:

$\bar{x}=\frac{5+2+4+3+6}{5}= \frac{20}{5}=4$

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Therefore, using his four samples, between 3.6 and 4.4 will Ardem's actual population mean lie.

Hence the option 3.6 and 4.4 is correct

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3 years ago

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An angle measured 10 degrees less than the measure of its complementary angle. Find the measure of both angles

Mazyrski [523]

Answer:

The measure of the angles are 40 degrees and 50 degrees

Step-by-step explanation:

Let

x ----> the measure of one angle in degrees

y ----> the measure of the complementary angle in degrees

we know that

If two angles are complementary, then their sum is equal to 90 degrees

$x+y=90$ ----> equation A

we have that

$x=y-10$ ----> equation B

Substitute equation B in equation A and solve for y

$(y-10)+y=90\\2y-10=90\\2y=100\\y=50$

Find the value of x

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2 years ago

Suppose that 5 cards are drawn from a well-shuffled deck of 52 cards. What is the probability that all 5 are not black?

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Here's the way I see it: 5 cards are drawn, one by one, without replacement. Half (or 26) of the original deck are black and half (26) are white.

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(26/52) * (25/51) * (24/50) * (23/49) * (22/48) = 0.025 (answer)

We start with 52 cards. We draw one, leaving 51 cards, 25 of which are red. And so on.

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2 years ago

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$y = e^x\\\\\displaystyle y = \sum_{k=1}^{\infty}\frac{x^k}{k!}\\\\\displaystyle y= 1+x+\frac{x^2}{2!} + \frac{x^3}{3!}+\ldots\\\\\displaystyle y' = \frac{d}{dx}\left( 1+x+\frac{x^2}{2!} + \frac{x^3}{3!}+\frac{x^4}{4!}+\ldots\right)\\\\$

$\displaystyle y' = \frac{d}{dx}\left(1\right)+\frac{d}{dx}\left(x\right)+\frac{d}{dx}\left(\frac{x^2}{2!}\right) + \frac{d}{dx}\left(\frac{x^3}{3!}\right) + \frac{d}{dx}\left(\frac{x^4}{4!}\right)+\ldots\\\\\displaystyle y' = 0+1+\frac{2x^1}{2*1} + \frac{3x^2}{3*2!} + \frac{4x^3}{4*3!}+\ldots\\\\\displaystyle y' = 1 + x + \frac{x^2}{2!}+ \frac{x^3}{3!}+\ldots\\\\\displaystyle y' = \sum_{k=1}^{\infty}\frac{x^k}{k!}\\\\\displaystyle y' = e^{x}\\\\$

This shows that $y' = y$ is true when $y = e^x$

-----------------------

Note 1: A more general solution is $y = Ce^x$ for some constant C.
Note 2: It might be tempting to say the general solution is $y = e^x+C$ , but that is not the case because $y = e^x+C \to y' = e^x+0 = e^x$ and we can see that $y' = y$ would only be true for C = 0, so that is why $y = e^x+C$ does not work.

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