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1 year ago

CAN SOMEONE PLEASE HELP ME PLEASE!!!!!

Mathematics

1 answer:

OverLord2011 [107]1 year ago

7 0

Answer:

x4+7x3-2x2-9x-3 remainder -10

or we could write it as

x4+7x3-2x2-9x-3 - 10/(x+8)

Step-by-step explanation:

x+8 ) x5+15x4+54x3−25x2−75x−34 (x4+7x3-2x2-9x-3 <--- Quotient.

x5+ 8x4

7x4+54x3

7x4+56x3

-2x3-25x2

-2x3-16x2

-9x2-75x

-9x2-72x

-3x-34

-3x-24

-10 <--- Remainder.

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Answer:

$(x+4)^2 + (y-3)^2 + (z+2)^2 - 3^2 = 0$

Step-by-step explanation:

You have to find a, b, c, r so that the polynomial expands to match the given equation.

$(x-a)^2 + (y-b)^2 + (z-c)^2 - r^2 =\\x^2-2ax+a^2+\\y^2-2by+b^2+\\z^2-2cz+c^2-r^2 =\\ x^2 + y^2 + z^2 + 8x - 6y + 4z + 20$

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3 years ago

Divide. Round to the nearest tenth.<br> 8 divided by 6.403

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1.2

Step-by-step explanation:

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2 years ago

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Suppose a bag contains many marbles, 54% of which are purple. You draw 5 marbles from the bag with replacement.

dolphi86 [110]

Part A

Use the binomial probability distribution formula.

p = 0.54 = probability of getting a purple marble

n = 5 = sample size

x = 2 = number of purple we want to get

$P(x) = \frac{n!}{x!(n-x)!}*p^x*(1-p)^{n-x}\\\\P(2) = \frac{5!}{2!(5-2)!}*0.54^2*(1-0.54)^{5-2}\\\\P(2) = 10*0.54^2*0.46^3\\\\P(2) = 0.283831776\\\\$

The $\frac{n!}{x!(n-x)!}$ portion is from the nCr combination formula. The exclamation marks indicate a factorial.

Alternatively, you could use Pascal's Triangle for that portion.

<h3>Answer: 0.283831776</h3>

This decimal value is exact. Round it however you need to.

============================================================

Part B

To find the expected value, aka the mean, we multiply the sample size and probability of getting a purple marble on any single selection.

n*p = 5*0.54 = 2.7

<h3>Answer: 2.7</h3>

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