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Troyanec [42]
2 years ago
13

Edgar has a goal to run 60 miles. He has already completed twelve miles. If he averages three miles per day, how many more days

will it take Edgar to reach his goal?
Mathematics
2 answers:
frosja888 [35]2 years ago
3 0

Answer:

16

Step-by-step explanation:

60-12=48

48÷3=16

Kruka [31]2 years ago
3 0

Answer:

16

Step-by-step explanation:

60m-12m=48

48m÷3m\day

=16 days to reach his goal

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Step-by-step explanation:

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A data set is made up of these values.
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Answer:

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Step-by-step explanation:

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Why is the pope an important figure in Christianity?
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2 years ago
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Suppose that wait times for customers at a grocery store cashier line are uniformly distributed between one minute and twelve mi
olchik [2.2K]

a) Mean: 6.5 min, variance: 10.1 min

b) 0.54 (54%)

c) 0.73 (73%)

d) 0.34

Step-by-step explanation:

a)

Here we can call X the variable indicating the waiting time for the customers:

X = waiting time

We are told that the waiting time is distributed uniformly between 1 and 12; this means that

1\leq X \leq 12

And the probability is equal for each value of X, so:

p(X=1)=p(X=2)=....=p(X=12)

The mean of a uniform distribution is given by:

E[X]=\frac{b+a}{2}

where a and b are the minimum and maximum values of the variable X. In this case,

a = 1

b = 12

So the mean value of X is

E[X]=\frac{12+1}{2}=6.5 (minutes)

The variance of a uniform distribution is given by:

Var[X]=\frac{1}{12}(b-a)^2

And substituting the values of this problem,

Var[X]=\frac{1}{12}(12-1)^2=10.1 (minutes)

b)

Since the distribution is uniform between 1 and 12, we can write the probability density function as

f(x)=\frac{1}{b-a}

The cumulative function gives the probability that the values of X is less than a certain value t:

p(X (1)

In this case, we want to find the probability that the waiting time is less than 7 minutes, so

t = 7

We also have:

a = 1

b = 12

Therefore, calculating (1) and substituting, we find:

p(X

c)

The probability that a customer waits between four and twenty minutes can be rewritten as

p(4

This can be written as:

p(4 (1)

However, the probabilty of X>4 can be written as

p(X>4)=1-p(X

Also, we notice that

p(X because the maximum value of X is 12; therefore, we can rewrite (1) as

p(4

We can calculate p(X by using the same method as in part b:

p(X

So, we find

p(4

d)

In this part, we know that a customer waits for

X = k

minutes in line, and he receives a coupon worth

0.2k^{\frac{1}{4}} dollars.

Here we want to find the mean of the coupon value.

Here therefore we have a new variables defined as

Y=0.2X^{\frac{1}{4}}

Given a variable with standard (between 0 and 1) uniform distribution X, the variable

Y=X^n

follows a beta distribution, with parameters (\frac{1}{n},1), and whose mean value is given by

E[Y]=\frac{1/n}{1+\frac{1}{n}}

In this case,

n=\frac{1}{4}

So the mean value of X^{1/4} is

E[X^{1/4}]=\frac{1/(1/4)}{1+\frac{1}{1/4}}=\frac{4}{1+4}=\frac{4}{5}=0.8

However, our variable is distribution is non-standard, because its values are between 1 and 12, so the range is

Min = 1^{1/4}=1\\Max =12^{1/4}=1.86

So, the actual mean value of X^{1/4} is

E[X^{1/4}]=0.8\cdot (1.86-1)+1=1.69

However, in the  definition of Y we also have a factor 0.2; therefore, the mean value of Y is

E[Y]=0.2E[X^{1/4}]=0.2\cdot 1.69 =0.34

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7 0
3 years ago
Read 2 more answers
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