Which expression is equivalent to -6(3x-2/3)?

Mathematics

1 answer:

Evgesh-ka [11]3 years ago

6 0

Hello!

-6(3x - 2/3)

Distribute the -6

-18x + 4

The answer is D) -18x + 4

Hope this helps!

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Runner A crosses the starting line of a marathon and runs at an average pace of 5.6 miles per hour. Half an hour later, Runner B

marishachu [46]

Time it took runner A to complete the marathon = 26.2 / 5.6 = 4 hrs 41 mins
Time it took runner B to complete the marathon = 26.2 / 6.4 = 4 hrs 6 mins

Time it took runner B to complete the marathon relative to when runner A started = 30 mins + 4 hrs 6 mins = 4 hrs 36 mins

Therefore, runner B will finnish ahead of runner A.

Let x be the time the two runners are at the same point, then
5.6x + 5.6(0.5) = 6.4x
6.4x - 5.6x = 2.8
0.8x = 2.8
x = 2.8/0.8 = 3.5
Therefore, runner B will catch up with runner A 3.5 hours after runner A starts the race.
<span>Runner B; Runner B will catch up to Runner A 3.5 hours after Runner A crosses the starting line.</span>

5 0

3 years ago

Show that if X is a geometric random variable with parameter p, then

Lubov Fominskaja [6]

Answer:

$\sum_{k=1}^{\infty} \frac{p(1-p)^{k-1}}{k}=-\frac{p ln p}{1-p}$

Step-by-step explanation:

The geometric distribution represents "the number of failures before you get a success in a series of Bernoulli trials. This discrete probability distribution is represented by the probability density function:"

$P(X=x)=(1-p)^{x-1} p$

Let X the random variable that measures the number os trials until the first success, we know that X follows this distribution:

$X\sim Geo (1-p)$

In order to find the expected value E(1/X) we need to find this sum:

$E(X)=\sum_{k=1}^{\infty} \frac{p(1-p)^{k-1}}{k}$

Lets consider the following series:

$\sum_{k=1}^{\infty} b^{k-1}$

And let's assume that this series is a power series with b a number between (0,1). If we apply integration of this series we have this:

$\int_{0}^b \sum_{k=1}^{\infty} r^{k-1}=\sum_{k=1}^{\infty} \int_{0}^b r^{k-1} dt=\sum_{k=1}^{\infty} \frac{b^k}{k}$ (a)

On the last step we assume that $0\leq r\leq b$ and $\sum_{k=1}^{\infty} r^{k-1}=\frac{1}{1-r}$ , then the integral on the left part of equation (a) would be 1. And we have: