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Cerrena [4.2K]
3 years ago
15

There are 38 pockets on an american roulette wheel: 18 red, 18 black, and 2 green. if you were to randomly spin an american roul

ette wheel, what is the probability that it would land on a green pocket
Mathematics
1 answer:
Pavel [41]3 years ago
8 0
You have about 5.23% chance to land in a green pocket. You can find this by dividing 2/38 or the questioned amount by the total amount of options.
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A candy bar is rectangular in shape and measures 3 inches by 4 inches by .5 inches. How many of the candy bars will fit into the
allsm [11]
It will fit 17 candy bars. Hope this helps!
4 0
3 years ago
Vern drinks three 6-packs of beer each week at a cost of $7
I am Lyosha [343]

Answer:

Vern’s total beer cost for the year is 52×3×$7 = $1092. On an annual basis basis,his beer expense is$1092/$700= 156%of his texbook expense

hope i helped

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5 0
3 years ago
0.5 log3 x=2 x = what
FinnZ [79.3K]
0.5\log_3x=2

Domain:x\in\mathbb{R^+}

0.5\log_3x=2\ \ \ |\cdot2\\\\\log_3x=4

Use\ the\ de finition\ of\ the\ logarithm:\\\\\log_ab=c\iff a^c=b

Therefore\ we\ have:\\\\\log_3x=4\iff3^4=x\to x=81

Answer: C. 81.
8 0
3 years ago
Read 2 more answers
Mike weighs 24 more pounds than Mark. Together, they weigh 250 pounds. How many do they each weigh?
djverab [1.8K]
To get your answer, follow my method :

First you take away 24 from 250=226
Then you half 226=113 which is Mark's weight
As Mike weights 24 pounds more that Mark, you have to add 24 back=137
So your answer is :
Mark weights 113 pounds and Mike weights 137 pounds.
6 0
3 years ago
The average THC content of marijuana sold on the street is 10.5%. Suppose the THC content is normally distributed with standard
MA_775_DIABLO [31]

a. This information is given to you.

b. We want to find

\mathrm{Pr}\{X > 8.9\}

so we first transform X to the standard normal random variable Z with mean 0 and s.d. 1 using

X = \mu + \sigma Z

where \mu,\sigma are the mean/s.d. of X. Now,

\mathrm{Pr}\left\{\dfrac{X - 10.5}2 > \dfrac{8.9 - 10.5}2\right\} = \mathrm{Pr}\{Z > -0.8\} \\\\~~~~~~~~= 1 - \mathrm{Pr}\{Z\le-0.8\} \\\\ ~~~~~~~~ = 1 - \Phi(-0.8) \approx \boxed{0.7881}

where \Phi(z) is the CDF for Z.

c. The 76th percentile is the value of X=x_{76} such that

\mathrm{Pr}\{X \le x_{76}\} = 0.76

Transform X to Z and apply the inverse CDF of Z.

\mathrm{Pr}\left\{Z \le \dfrac{x_{76} - 10.5}2\right\} = 0.76

\dfrac{x_{76} - 10.5}2 = \Phi^{-1}(0.76)

\dfrac{x_{76} - 10.5}2 \approx 0.7063

x_{76} - 10.5 \approx 1.4126

x_{76} \approx \boxed{11.9126}

5 0
1 year ago
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