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iren2701 [21]
2 years ago
8

The waiting time in line at an ice cream shop has a uniform distribution between 3 and 14 minutes. What is the 75th percentile o

f this distribution? (Recall: The 75th percentile divides the distribution into 2 parts so that 75% of area is to the left of 75th percentile) _______ minutes Answer: (Round answer to two decimal places.)
Mathematics
1 answer:
lesya692 [45]2 years ago
3 0

Answer:

The 75th percentile of this distribution is 11 .25 minutes.

Step-by-step explanation:

The random variable <em>X</em> is defined as the waiting time in line at an ice cream shop.

The random  variable <em>X </em>follows a Uniform distribution with parameters <em>a</em> = 3 minutes and <em>b</em> = 14 minutes.

The probability density function of <em>X</em> is:

f_{X}(x)=\frac{1}{b-a};\ a

The <em>p</em>th percentile is a data value such that at least p% of the data-set is less than or equal to this data value and at least (100-p)% of the data-set are more than or equal to this data value.

Then the 75th percentile of this distribution is:

P (X < x) = 0.75

\int\limits^{x}_{3} {\frac{1}{14-3}} \, dx=0.75\\\\ \frac{1}{11}\ \cdot\ \int\limits^{x}_{3} {1} \, dx=0.75\\\\\frac{x-3}{11}=0.75\\\\x-3=8.25\\\\x=11.25

Thus, the 75th percentile of this distribution is 11 .25 minutes.

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REASONING An even integer can be represented by the expression $2n$ , where $n$ is any integer. Find three consecutive even inte
san4es73 [151]

Answer:

First Integer = 16

Second Integer = 18

Third integer = 20

Step-by-step explanation:

An even integer is represented by 2n

Where n is any integer

Let :

First Integer = 2n - 2

Second Integer = 2n

Third integer = 2n + 2

The sum of three even consecutive numbers = 2n - 2 + 2n + 2n + 2

= 2n + 2n + 2n - 2 + 2 = 54

= 6n = 54

n = 54/6

n = 9

First Integer = 2n - 2 = 2(9) - 2

= 16

Second Integer = 2n = 2(9)

= 18

Third integer = 2n + 2 = 2(9) + 2

= 20

8 0
2 years ago
You deposit $300 in a savings account that pays 6% interest compounded semiannually. How much will you have at the middle of the
Makovka662 [10]

Answer:

  • The total amount accrued, principal plus interest,  from compound interest on an original principal of  $ 300.00 at a rate of 6% per year  compounded 2 times per year  over 0.5 years is $ 309.00.

  • The total amount accrued, principal plus interest,  from compound interest on an original principal of  $ 300.00 at a rate of 6% per year  compounded 2 times per year  over 1 year is $ 318.27.

Step-by-step explanation:

a)  How much will you have at the middle of the first year?

Using the formula

A\:=\:P\left(1+\frac{r}{n}\right)^{nt}

where

  • Principle = P
  • Annual rate = r
  • Compound = n
  • Time  = (t in years)
  • A = Total amount

Given:

Principle P = $300

Annual rate r = 6% = 0.06 per year

Compound n = Semi-Annually = 2

Time (t in years) = 0.5 years

To determine:

Total amount = A = ?

Using the formula

A\:=\:P\left(1+\frac{r}{n}\right)^{nt}

substituting the values

A=300\left(1+\frac{0.06}{2}\right)^{\left(2\right)\left(0.5\right)}

A=300\cdot \frac{2.06}{2}

A=\frac{618}{2}

A=309 $

Therefore, the total amount accrued, principal plus interest,  from compound interest on an original principal of  $ 300.00 at a rate of 6% per year  compounded 2 times per year  over 0.5 years is $ 309.00.

Part b) How much at the end of one year?

Using the formula

A\:=\:P\left(1+\frac{r}{n}\right)^{nt}

where

  • Principle = P
  • Annual rate = r
  • Compound = n
  • Time  = (t in years)
  • A = Total amount

Given:

Principle P = $300

Annual rate r = 6% = 0.06 per year

Compound n = Semi-Annually = 2

Time (t in years) = 1 years

To determine:

Total amount = A = ?

so using the formula

A\:=\:P\left(1+\frac{r}{n}\right)^{nt}

so substituting the values

A\:=\:300\left(1+\frac{0.06}{2}\right)^{\left(2\right)\left(1\right)}

A=300\cdot \frac{2.06^2}{2^2}

A=318.27 $

Therefore, the total amount accrued, principal plus interest,  from compound interest on an original principal of  $ 300.00 at a rate of 6% per year  compounded 2 times per year  over 1 year is $ 318.27.

3 0
2 years ago
The coordinates of the midpoint of line GH are M(−132,−6) and the coordinates of one endpoint are G(−4, 1). The coordinates of t
Darina [25.2K]

Answer:

Since, the coordinates of the midpoint of line GH are M(\frac{-13}{2}, -6)(

2

−13

,−6) .

The coordinates of endpoint G are (-4,1)

We have to determine the coordinates of endpoint H.

The midpoint of the line segment joining the points (x_1, y_1)(x

1

,y

1

) and (x_2, y_2)(x

2

,y

2

) is given by the formula (\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})(

2

x

1

+x

2

,

2

y

1

+y

2

) .

Here, The endpoint G is (-4,1) So, x_1 = -4 , y_1=1x

1

=−4,y

1

=1

Let the endpoint H be (x_2,y_2)(x

2

,y

2

)

The midpoint coordinate M is (\frac{-13}{2}, -6)(

2

−13

,−6) .

So, \frac{-13}{2} = \frac{-4+x_2}{2}

2

−13

=

2

−4+x

2

{-13} = {-4+x_2}−13=−4+x

2

{-13}+4 = {x_2}−13+4=x

2

{x_2}=9x

2

=9

Now, -6 = \frac{1+y_2}{2}−6=

2

1+y

2

-12 = {1+y_2}−12=1+y

2

y_2= -13y

2

=−13

So, the other endpoint H is (-9,-13).

7 0
1 year ago
What is the area of △FGH to the nearest tenth of a square meter? The image is of a triangle GHF with base GH length 2m, FG is 2.
Gnesinka [82]
First, we are going to use the law of cosines to find the length of the line segment FH:
FH= \sqrt{2.5^{2}+2^{2}-(2)(2.5)Cos(121)}
FH= \sqrt{2.5^{2}+2^{2}-5Cos(121)}
FH=3.2

Next, we are going to use the semi-perimeter formula: s= \frac{GH+FG+FH}{2}
s= \frac{2+2.5+3.2}{2}
s= \frac{7.7}{2}
s=3.9

Now that we have the semi-perimeter of our triangle, we can find its area using Heron's formula:
A= \sqrt{s(s-GH)(s-FG)(s-FH)}
A= \sqrt{3.9(3.9-2)(3.9-2.5)(3.9-3.2)}
A= \sqrt{3.9(1.9)(1.4)(0.7)}
A=2.7m^{2}

We can conclude that the area of the triangle <span>GHF is 2.7 </span>m^{2}.

3 0
2 years ago
Agnes wants to buy a gallon of iced tea. If a 1-gallon jug costs $3.05 and a
AleksandrR [38]

Answer:

c

Step-by-step explanation:

8 pints in a gal

.45 x 8 = 3.60

3.60 - 3.05 = . 55

6 0
1 year ago
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