What is the inverse operation of n/4.3=9.4

How to the solve the equation by completing the square?

saw5 [17]

Https://m.quickmath.com/solve-equation-inequality type it in to heee

8 0

3 years ago

On each round, Ann and Bob each simultaneously toss a fair coin. Let Xn be the number of heads tossed in the 2n flips which occu

CaHeK987 [17]

Answer:

$P=\frac{2n!}{m!*(2n-m)!}*0.5^{2n}$

Step-by-step explanation:

In a coin toss the probability of tossing a head is 0.5 (50% head/50% tails)

If n is the number of rounds and 2n the number of coins tossed (one for each player), the probability of having m heads tossed is:

$R=\frac{2n!}{m!*(2n-m)!}$

R is the number of cases (combination of coins tossed) that gives a m number of heads. Each case has a probability of $P_{case}=0.5^{2n}$ so:

$P=\frac{2n!}{m!*(2n-m)!}*0.5^{2n}$

<u>For example, to toss 4 heads in 5 rounds: </u>

n=5
2n=10
m=4

$P=\frac{10!}{4!*(10-4)!}*0.5^{10}$

$P=\frac{10*9*8*7*6!}{4!*6!}*0.5^{10}$

$P=\frac{10*9*8*7}{4!}*0.5^{10}$

$P=\frac{10*9*8*7}{4!}*0.5^{10}=0.205$

8 0

3 years ago

Evaluate the iterated integral. $$ \int\limits_0^{2\pi}\int\limits_0^y\int\limits_0^x {\color{red}9} \cos(x+y+z)\,dz\,dx\,dy $$

KengaRu [80]

$\displaystyle\int_{y=0}^{y=2\pi}\int_{x=0}^{x=y}\int_{z=0}^{z=x}\cos(x+y+z)\,\mathrm dz\,\mathrm dx\,\mathrm dy=\int_{y=0}^{y=2\pi}\int_{x=0}^{x=y}\sin(x+y+z)\bigg|_{z=0}^{z=x}\,\mathrm dx\,\mathrm dy$
$\displaystyle=\int_{y=0}^{y=2\pi}\int_{x=0}^{x=y}\sin(2x+y)-\sin(x+y)\,\mathrm dx\,\mathrm dy$
$\displaystyle=\int_{y=0}^{y=2\pi}-\frac12\left(\cos(2x+y)-2\cos(x+y)\right)\bigg|_{x=0}^{x=y}\,\mathrm dx\,\mathrm dy$
$\displaystyle=\int_{y=0}^{y=2\pi}-\frac12\left((\cos3y-2\cos2y)-(\cos y-2\cos y)\right)\bigg|_{x=0}^{x=y}\,\mathrm dy$
$\displaystyle=-\frac12\int_{y=0}^{y=2\pi}(\cos3y-2\cos2y+\cos y)\,\mathrm dy$
$\displaystyle=-\frac12\left(\frac13\sin3y-\sin2y+\sin y\right)\bigg|_{y=0}^{y=2\pi}$
$=0$

4 0

3 years ago

Solve for y 2x+y=3 thank you.

xxMikexx [17]

Y = -2x + 3
Hope that helps.

8 0

3 years ago

Read 2 more answers

Lengths of full-term babies in the US are Normally distributed with a mean length of 20.5 inches and a standard deviation of 0.9

mash [69]

Answer:

66.48% of full-term babies are between 19 and 21 inches long at birth

Step-by-step explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Mean length of 20.5 inches and a standard deviation of 0.90 inches.

This means that $\mu = 20.5, \sigma = 0.9$