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Lana71
19 days ago
13

Carmen is going to roll an 8-sided die 200 times. She predicts that she will roll a multiple of 4 twenty-five times. Based on th

e theoretical probability, which best describes Carmen’s prediction?
Carmen’s prediction is exact because 200 times 1/8 is 25.

Carmen’s prediction is low because 200 times 1/4 is 50.

Carmen’s prediction is low because 200 divided by 2 is 100.

Carmen’s prediction is high because200 divided by 25 is 8.
Mathematics
2 answers:
AnnZ [9K]19 days ago
4 0

Answer:

Carmen's estimate is too low, since rolling 200 times \frac{1}{4} results in a total of 50.

Step-by-step explanation:

Initially, we will define the sample space for this scenario.

The sample space is Ω = {1,2,3,4,5,6,7,8}

For the event A: "Rolling an 8-sided die to get a multiple of 4"

The probability for event A is P(A)=\frac{2}{8}=\frac{1}{4}

This is because there are two multiples of 4 (4 and 8) out of a total of eight numbers (1 through 8).

Next, considering the random variable X: "Total count of multiples of 4 if she rolls an 8-sided die 200 times"

X can be described as following a Binomial distribution.

X ~ Bi (n,p)

X ~ Bi (200,\frac{1}{4})

Where n is the number of rolls and p is the probability of success, defined as rolling a multiple of 4.

The mean for this variable is

E(X)=np=200.\frac{1}{4}=50

Thus, we conclude that Carmen's prediction is low, as rolling 200 times \frac{1}{4} yields 50.

Zina [9.1K]19 days ago
3 0

Carmen’s estimate is low since 200 times 1/4 equals 50.

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12-5x-4kx=y solve for x
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12-5x-4kx=y:x 

-5x-4kx=y-12 

-x(5+4k)=y-12 

-x= \dfrac{y-12}{5+4k} 

x=- \dfrac{y-12}{5+4k}
7 0
9 days ago
Question: Sara started her homework at 5:30 p.M. And finished it at 9:15 p.M. How much time did she take to finish her homework?
Svet_ta [9500]

Explicación detallada:

530 p.m. = 17:50 en 24 horas/tipo militar

915 p.m. = 21:15 en 24 horas/tipo militar

por lo tanto..

24 horas/tipo militar. 21:15 - 17:50 = 3.65 horas

en nuestro tiempo. 3 horas 45 minutos

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14 days ago
Let X denote the data transfer time (ms) in a grid computing system (the time required for data transfer) between a "worker" com
Inessa [9000]

Answer:

a. Alpha equals 3.014 while beta equals 12.442

b. The likelihood that the data transfer duration surpasses 50ms is 0.238

c. The chance that data transfer time falls between 50 and 75 ms is 0.176

Step-by-step explanation:

a. Given the data, the mean and standard deviation for the random variable X are 37.5 ms and 21.6, respectively.

Thus, E(X)=37.5 and V(X)=(21.6)∧2  

To find alpha, we need to apply the formula:

alpha=E(X)∧2/V(X)

alpha=(37.5)∧2/21.6∧2

alpha=1,406.25 /466.56

​alpha=3.014

To determine beta, the following formula is employed:

β=  V(X) ∧2/E(X)

β=(21.6)  ∧2/37.5

β=466.56 /37.5

β=12.442

b. With E(X)=37.5 and V(X)=(21.6)∧2,  

Hence, P(X>50)=1−P(X≤50)

To find the probability of data transfer time exceeding 50ms, we use the formula:

P(X>50)=1−P(X≤50)

=1−0.762

=0.238

The chance of data transfer time exceeding 50ms is 0.238

c. With E(X)=37.5 and V(X)=(21.6)∧2,  

Thus, P(50<X<75)=P(X<75)−P(X<50)  

To find the probability that data transfer time is between 50 and 75 ms, we apply the formula:

P(50<X<75)=P(X<75)−P(X<50)

=0.938−0.762

=0.176

​

The probability that data transfer time falls between 50 and 75 ms is 0.176

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12 days ago
A baseball is thrown up in the air. The table shows the heights y (in feet) of the baseball after x seconds.
lawyer [9240]

Answer:

Alright, we can express the baseball's motion with an equation like:

h(x) = a*x^2 + b*x + c

Here, x denotes time, while h(x) indicates height.

Let’s construct this:

The acceleration is:

a(t) = a

For velocity, integrating over time results in:

v(x) = a*x + v0

Where v0 signifies the initial vertical velocity.

Subsequently, we can determine position or height by integrating once more:

h(x) = a*x^2 + v0*x + h0

Here, h0 is the initial height.

<pthus our="" equation="" is:="">

h(x) = a*x^2 + v0*x + h0.

<pexamining the="" table:="">

When x = 0s, h(0s) = 6ft

<pthus:>

h(0s) = a*0s^2 + v0*0s + h0 = 6ft

            h0 = 6ft.

It’s also noted that:

h(2s) = h(4s)

<pthe symmetry="" of="" the="" quadratic="" function="" implies="" that="" axis="" lies="" between="" and="" located="" at="" x="3s.&lt;/p"><pin a="" standard="" quadratic="" function:="">

a*x^2 + b*x + c

The symmetry line is given by:

x = -b/2a

<pin this="" instance:="">

b = v0

a = a

<ptherefore we="" derive:="">

3s  = -v0/(2*a)

v0 = -3s*(2a)

<phaving gathered="" all="" necessary="" data="" for="" our="" equation="" we="" can="" express="" it="" as:="">

h(x) = a*x^2 - 6s*a*x + 6ft

<pnext focusing="" on="" just="" one="" variable="" we="" know="" that="" at="" x="2s," h=""><pso:>

h(2s) = 22ft = a*(2s)^2 - 6s*a*2s + 6ft

<pthus our="" resulting="" equation="" reads:="">

h(x) = (-2ft/s^2)*x^2 + (12ft/s)*x + 6ft

b) The height after 5 seconds is expressed as:

h(5s) =  (-2ft/s^2)*(5s)^2 + (12ft/s)*5s + 6ft = 16ft

</pthus></p></pso:></pnext></phaving></ptherefore></pin></pin></pthe></pthus:></pexamining></pthus>
7 0
1 month ago
Igor recently immigrated to the UK. He was an experienced surgeon in his country but had to re-do his medical training in order
AnnZ [9099]

Answer:

a) The outlier is the point located at the bottom right of the graph

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a) The problem presents a scenario where Igor, who has recently moved, is experienced but needs to retrain medically to practice in the UK

Thus, he corresponds to the outlier situated nearest to the graph's bottom right

b) According to the scatter graph, there's a direct relationship showing that as a doctor's age increases, their annual salary tends to climb as well

Referencing the graph:

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38        £42000

42        £30000

42        £46000

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The data points follow a line demonstrating the proportional increase of salary with age.

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