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3 years ago

Convert 17 years into minutes

Mathematics

2 answers:

Setler79 [48]3 years ago

7 0

Answer:

8,941,136.4 Minutes/ 8,935,200 Minutes

Step-by-step explanation:

Ksenya-84 [330]3 years ago

3 0

Answer:

8935200

Step-by-step explanation:

multiply by 525600

1hour =525600minutes

so we say 17years x 525600minutes

which is 8935200

can I please get brainliest??

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USPshnik [31]

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2. -4/3

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3 years ago

What is the solution to this equation 1/2n^2+18=0

ololo11 [35]

Answer:

n = ±6 .

Step-by-step explanation:

A quadratic equation is given to us and we need to find out the solution of the given equation . The given equation is ,

$\rm\implies -\dfrac{1}{2}n^2+18=0$

Subtracting 18 both sides ,

$\rm\implies -\dfrac{1}{2}n^2 = -18$

Multiplying both sides by -2 ,

$\rm\implies -2\times\dfrac{1}{2}n^2 = -2\times -18$

On simplyfing , we get ,

$\rm\implies n^2= 36$

Putting squareroot both sides ,

$\rm\implies n= \sqrt{36}$

This equals to ,

$\rm\implies \boxed{\quad\blue{\rm n =\pm 6 }}$

Hence the value of n is ±6 .

6 0

3 years ago

Suppose that X has an exponential distribution with mean equal to 10. Determine the following: a. P(X > 10) b. P(X > 20) c

GrogVix [38]

Answer:

(a) The value of P (X > 10) is 0.3679.

(b) The value of P (X > 20) is 0.1353.

(d) The value of x is 30.

Step-by-step explanation:

The probability density function of an exponential distribution is:

$f(x)=\lambda e^{-\lambda x};\ x>0, \lambda>0$

The value of E (X) is 10.

The parameter λ is:

$\lambda=\frac{1}{E(X)}=\frac{1}{10}=0.10$

(a)

Compute the value of P (X > 10) as follows:

$P(X>10)=\int\limits^{\infty}_{10} {0.10 e^{-0.10 x}} \, dx \\=0.10\int\limits^{\infty}_{10} { e^{-0.10 x}} \, dx\\=0.10|\frac{e^{-0.10 x}}{-0.10} |^{\infty}_{10}\\=|e^{-0.10 x} |^{\infty}_{10}\\=e^{-0.10\times10}\\=0.3679$

Thus, the value of P (X > 10) is 0.3679.

(b)

Compute the value of P (X > 20) as follows:

$P(X>20)=\int\limits^{\infty}_{20} {0.10 e^{-0.10 x}} \, dx \\=0.10\int\limits^{\infty}_{20} { e^{-0.10 x}} \, dx\\=0.10|\frac{e^{-0.10 x}}{-0.10} |^{\infty}_{20}\\=|e^{-0.10 x} |^{\infty}_{20}\\=e^{-0.10\times20}\\=0.1353$

Thus, the value of P (X > 20) is 0.1353.

(c)

Compute the value of P (X < 30) as follows:

$P(X$

Thus, the value of P (X < 30) is 0.9502.

(d)

It is given that, P (X < x) = 0.95.

Compute the value of x as follows:

$P(X$

Take natural log on both sides.

$ln(e^{-0.10x})=ln(0.05)\\-0.10x=-2.996\\x=\frac{2.996}{0.10}\\ =29.96\\\approx30$

Thus, the value of x is 30.

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Vedmedyk [2.9K]

Answer:

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3 years ago

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hammer [34]

Answer:

$y ( \times + 6) 2 - 4$

7 0

3 years ago

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Convert 17 years into minutes​

Convert 17 years into minutes