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3 years ago

If x is direct proportional to y cube and x=32 when y=2

Mathematics

1 answer:

Setler [38]3 years ago

5 0

Answer:

1x=864,

2y=3

Step-by-step explanation:

y=6,x=?

if x is directly proportional to y^3 &if (when x=32, y is=2) we can understand that x=4y^3

1,x=4y^3

x=4(6^3)

x=4(216)

x=864

2,x=4y^3

108=4y^3

108/4=4y^3/4

27=y^3

√27=y

y=3

i think it is easy to understand,but if the answers are not clear you can ask me &i hope it helps you

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If u don’t get it right I <br> Will report you

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Answer:

1 A

2 C

Step-by-step explanation:

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3 years ago

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Factories A, B and C produce computers. Factory A produces 4 times as manycomputers as factory C, and factory B produces 7 times

Elina [12.6K]

Answer:

The probability is $P(A') = 0.485$

Step-by-step explanation:

Let assume that the number of computer produced by factory C is k = 1

So From the question we are told that

The number produced by factory A is 4k = 4

The number produced by factory B is 7k = 7

The probability of defective computers from A is $P(A) = 0.04$

The probability of defective computers from B is $P(B) = 0.02$

The probability of defective computers from C is $P(C) = 0.03$

Now the probability of factory A producing a defective computer out of the 4 computers produced is

$P(a) = 4 * P(A)$

substituting values

$P(a) = 4 * 0.04$

$P(a) = 0.16$

The probability of factory B producing a defective computer out of the 7 computers produced is

$P(b) = 7 * P(B)$

substituting values

$P(b) = 7 * 0.02$

$P(b) = 0.14$

The probability of factory C producing a defective computer out of the 1 computer produced is

$P(c) = 1 * P(C)$

substituting values

$P(c) = 1 * 0.03$

$P(b) = 0.03$

So the probability that the a computer produced from the three factory will be defective is

$P(t) = P(a) + P(b) + P(c)$

substituting values

$P(t) = 0.16 + 0.14 + 0.03$

$P(t) = 0.33$

Now the probability that the defective computer is produced from factory A is

$P(A') = \frac{P(a)}{P(t)}$

$P(A') = \frac{ 0.16}{0.33}$

$P(A') = 0.485$

3 0

3 years ago

Find the 4th term of the expansion of (a - sqrt(2)) ^ 8

charle [14.2K]

Answer:

Applying the binomial theorem

$(x+a)^{n}=\sum _{k=0}^{n}{n \choose k}x^{n-k}a^{k}$

Here we have x = - $\sqrt{2}$ , n = 8,

=> 4th term (k = 3) would be:

${8 \choose 3}x^{8-3}a^{3}} = {8 \choose 3}(-\sqrt{2})^{5}a^{3}} = -224a^{3}\sqrt{2}$

Hope this helps!

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3 years ago

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What is the sum of the infinite geometric series 9−6+4−83+...9-6+4-83+...??

vivado [14]

Successive terms have a common factor of $-\dfrac23$ , which means the $n$ th partial sum of the series can be written as

$S_n=9-6+4-\dfrac83+\cdots+9\left(-\dfrac23\right)^{n-1}$
$S_n=9\left(1+\left(-\dfrac23\right)^1+\left(-\dfrac23\right)^2+\left(-\dfrac23\right)^3+\cdots+\left(-\dfrac23\right)^{n-1}$

$\implies-\dfrac23S_n=9\left(\left(-\dfrac23\right)^1+\left(-\dfrac23\right)^2+\left(-\dfrac23\right)^3+\left(-\dfrac23\right)^4+\cdots+\left(-\dfrac23\right)^n\right)$

$\implies S_n-\left(-\dfrac23\right)S_n=9\left(1-\left(-\dfrac23\right)^n\right)$
$\dfrac53S_n=9\left(1-\left(-\dfrac23\right)^n\right)$
$S_n=\dfrac{27}5\left(1-\left(-\dfrac23\right)^n\right)$

As $n\to\infty$ , the geometric term vanishes, leaving you with

$S=\displaystyle\lim_{n\to\infty}S_n=\dfrac{27}5$

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3 years ago

7x-3=-3+7x show your work

Nuetrik [128]

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3 years ago