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2 years ago

Write the quadratic equation 3x ^ 2 - 9x = 5 standard form ax²+bx+c=0? in the

Mathematics

1 answer:

sleet_krkn [62]2 years ago

8 0

The answer to your question is B 3x^2-9x-5=0.

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Need some help with this. hope someone can.help

GrogVix [38]

$\pi = 3.14\\r=40cm\\h=1.3m=130cm\\V=\pi r^{2}h\\V=(3.14)(40^{2})(130)\\V=(3.14)(1600)(130)\\V=653120cm^{3}$

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2 years ago

Read 2 more answers

2x+8 find x pleaaasee hurry i need help

castortr0y [4]

Answer:

x = 4

Step-by-step explanation:

fFs

4 0

2 years ago

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You have a biased coin for which p(h)=pp(h)=p. you toss the coin 2020 times. what is the probability that you observe 88 heads a

Nat2105 [25]

The question describes a binomial probability with p(h) = p, then p(t) = 1 - p and number of trials (n) = 20

The probability of a binomial distribution is given by

$P(x)=\, ^nC_xp^x(1-p)^{n-x}$

Part A:

The probability of observing 8 heads and 12 tails is given by:

$P(8)=\, ^{20}C_8p^8(1-p)^{20-8}=\, ^{20}C_8p^8(1-p)^{12}$

Part B:

You observe more than 8 heads and more than 8 tails, when you observe 9 heads and 11 tails, 10 heads and 10 tails, and 11 heads and 9 tails.

Therefore, the probability of observing more than 8 heads and more than 8 tails is given by:

$P(9)+P(10)+P(11) \\ \\ =\, ^{20}C_9p^9(1-p)^{20-9}+\, ^{20}C_{10}p^{10}(1-p)^{20-10}+\, ^{20}C_{11}p^{11}(1-p)^{20-11} \\ \\ =\, ^{20}C_9p^9(1-p)^{11}+\, ^{20}C_{10}p^{10}(1-p)^{10}+\, ^{20}C_{11}p^{11}(1-p)^{9}$

7 0

2 years ago

Which of the following shows the equation after both sides have been distributed correctly?

gogolik [260]

Answer:

B. -4x + 8 = 15x + 40

Hope this helps!

8 0

2 years ago

How do I do this? *Look at the directions in the photo*

lora16 [44]

Answer:

$Area\ of\ material\ required\ for\ the\ first\ box=384\ inches^2\\Area\ of\ material\ required\ for\ the\ second\ box=486\ inches^2\\Area\ of\ material\ required\ for\ the\ first\ box=600\ inches^2\\Total\ Area\ of\ material\ required=1470\ inches^2$

Step-by-step explanation:

$We\ are\ given:\\Diameter\ of\ the\ first\ volleyball=8\ inches \\Diameter\ of\ the\ second\ volleyball=9\ inches\\Diameter\ of\ the\ third\ volleyball= 10\ inches.\\Hence,\\We\ know\ that,\\If\ the\ side\ of\ the\ cube\ box\ is\ s, it's\ Total\ Surface\ Area\ =No.\ of\\ faces\ in\ a\ regular\ polyhedron\ *Area\ of\ each\ face\ of\ the\ polyhedron=6*s^2=6s^2\\Hence,\\Lets\ apply\ this\ equation\ in\ finding\ the\ area\ of\ material\ required\ for\ the\\ three\ cases.\\$

$As\ the\ volleyball\ should\ wholly\ fit\ into\ the\ box,\ the\ diameter\ of\ the\\ volleyballs\ would\ be\ the\ side\ of\ the\ cube\ box.\\Hence,\\For\ the\ first\ volleyball,\\Diameter\ of\ the\ first\ volleyball=8\ inches\\Hence,\\Side\ of\ the\ cubical\ box\ for\ the\ first\ volleyball=8\ inches.\\Hence,\\The\ Total\ Surface\ Area\ of\ the\ first\ box=6s^2=6*8*8=384\ inches^2$

$For\ the\ second\ volleyball,\\Diameter\ of\ the\ second\ volleyball=9\ inches\\Hence,\\Side\ of\ the\ cubical\ box\ for\ the\ second\ volleyball=9\ inches.\\Hence,\\The\ Total\ Surface\ Area\ of\ the\ second\ box=6s^2=6*9*9=486\ inches^2$

$For\ the\ third\ volleyball,\\Diameter\ of\ the\ third\ volleyball=8\ inches\\Hence,\\Side\ of\ the\ cubical\ box\ for\ the\ third\ volleyball=10\ inches.\\Hence,\\The\ Total\ Surface\ Area\ of\ the\ third\ box=6s^2=6*10*10=600\ inches^2$

$Hence,\\If\ you\ are\ asked\ the\ Total\ Area\ to\ make\ all\ the\ boxes,\\ you\ just\ add\ them\ together.\\Hence,\\Total\ Area\ of\ Material\ required\ to\ make\ the\ three\ boxes=384+486+600=1470\ inches^2$

7 0

2 years ago