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hodyreva [135]
2 years ago
5

Six years ago annabelle was twice as old as jason was. now, annabelle is four years older than jason. how old is jason now?

Mathematics
1 answer:
baherus [9]2 years ago
4 0
Now: a=j+4
6 years ago: Anabelle: a-6; Jason: j-6

a-6=2(j-6) \\ a=2j-12+6 \\ a=2j-6 \\  \\ j+4=2j-6 \\ 4+6=2j-j \\ j=10 \\  \\ a=2(10)-6 \\ a=14
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The Big River Casino is advertising a new digital lottery-style game called Instant Lotto. The player can win the following mone
zysi [14]

Answer:

(a) The expected value of the prize for one play of Instant Lotto is $3.50.

(b) The probability that the visitor wins some prize at least twice in the 20 free plays is 0.2641.

(c) The probability that a randomly selected day has at least 1000 people play Instant Lotto is 0.2579.

Step-by-step explanation:

(a)

The probability distribution of the monetary prizes that can be won at the game called Instant Lotto is:

<em>X</em>         P (<em>X</em> = <em>x</em>)

$10        0.05

$15        0.04

$30       0.03

$50       0.01

$1000   0.001

$0         0.869

___________

Total =   1.000

Compute the expected value of the prize for one play of Instant Lotto as follows:

E(X)=\sum x\cdot P (X=x)

         =(10\times 0.05)+(15\times 0.04)+(30\times 0.03) \\+ (50\times 0.01)+(1000\times 0.001)+(0\times 0.869)\\=0.5+0.6+0.9+0.5+1+0\\=3.5          

Thus, the expected value of the prize for one play of Instant Lotto is $3.50.

(b)

Let <em>X</em> = number of times a visitor wins some prize.

A visitor to the casino is given <em>n</em> = 20 free plays of Instant Lotto.

The probability that a visitor wins at any of the 20 free plays is, <em>p</em> = 1/20 = 0.05.

The event of a visitor winning at a random free play is independent of the others.

The random variable <em>X</em> follows Binomial distribution with parameters <em>n</em> = 20 and <em>p</em> = 0.05.

Compute the probability that the visitor wins some prize at least twice in the 20 free plays as follows:

P (X ≥ 2) = 1 - P (X < 2)

              = 1 - P (X = 0) - P (X = 1)

              =1-[{20\choose 0}0.05^{0}(1-0.05)^{20-0}]-[{20\choose 1}0.05^{1}(1-0.05)^{20-1}]\\=1-0.3585-0.3774\\=0.2641

Thus, the probability that the visitor wins some prize at least twice in the 20 free plays is 0.2641.

(c)

Let <em>X</em> = number of people who play Instant Lotto each day.

The random variable <em>X</em> is normally distributed with a mean, <em>μ</em> = 800 people and a standard deviation, <em>μ</em> = 310 people.

Compute the probability that a randomly selected day has at least 1000 people play Instant Lotto as follows:

Apply continuity correction:

P (X ≥ 1000) = P (X > 1000 + 0.50)

                    = P (X > 1000.50)

                    =P(\frac{X-\mu}{\sigma}>\frac{1000.50-800}{310})

                    =P(Z>0.65)\\=1-P(Z

Thus, the probability that a randomly selected day has at least 1000 people play Instant Lotto is 0.2579.

5 0
2 years ago
What value of x makes the equation -4x – (-3 − 5x) = -3(2x − 8) true?
mezya [45]

Answer:hrhdhf end b

Step-by-step explanation:

Rndjdbajrcb

5 0
2 years ago
Eva wanted to fill bags with ¾ pounds of trail mix. She started with 11 ⅜ pounds but ate ⅛ pound before she started filling the
Len [333]
She fill 12 bags because if you take 3/4*11 3/8- 1/8=12

3 0
2 years ago
A ferry will safely accommodate 68 tons of passenger cars. Assume that the mean weight of a passenger car is 1.8 tons with stand
nika2105 [10]

Answer:

P (Z

Step-by-step explanation:

We want the probability that the 35 cars are loaded onto the ferry. Therefore:

Since  

μ=1.8  and  σ=0.5  we have:

P(X<35 )=P ( X−μ<35−1.8 )=P((X−μ)/σ<(35−1.8)/0.5)

Since  

(x−μ)/σ=Z and  

(35-1.8)/0.5=66.4

we have:

P (X

Use the standard normal table to conclude that:

P (Z

7 0
2 years ago
Read 2 more answers
Suppose that a jewelry store tracked the amount of emeralds they sold each week to more accurately estimate how many emeralds to
RUDIKE [14]

Answer:

The confidence interval for the mean is given by the following formula:

\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}   (1)

And for this case the 95% confidence interval is given by (2.13; 2.37)

We have a point of estimate for the sample mean with this formula:

\bar X = \frac{Upper+ Lower}{2}= \frac{3.37+2.13}{2}= 2.75

And for the margin of error we have the following estimation:

ME= \frac{Upper -Lower}{2}= \frac{3.37-2.13}{2}= 0.62

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

\bar X represent the sample mean

\mu population mean (variable of interest)

s represent the sample standard deviation

n represent the sample size  

Solution to the problem

The confidence interval for the mean is given by the following formula:

\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}   (1)

And for this case the 95% confidence interval is given by (2.13; 2.37)

We have a point of estimate for the sample mean with this formula:

\bar X = \frac{Upper+ Lower}{2}= \frac{3.37+2.13}{2}= 2.75

And for the margin of error we have the following estimation:

ME= \frac{Upper -Lower}{2}= \frac{3.37-2.13}{2}= 0.62

5 0
2 years ago
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