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liberstina [14]
3 years ago
5

The number of traffic citations given daily by two police departments over a two-week period is shown in the box-and-whisker plo

ts. Which statement is NOT true?
a.The East department gave the greatest number of citations in one day.
b.The East department gave the least number of citations in one day.
c.The East department has a greater IQR than the West department.
d.The East department has the greater median of citations in one day.

Mathematics
2 answers:
m_a_m_a [10]3 years ago
5 0
D iiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiii
Anna [14]3 years ago
3 0

Answer:

  • b. The East department gave the least number of citations in one day.

Step-by-step explanation:

a. The East department gave the greatest number of citations in one day.

  • True, 210 > 190 (approx.)

b. The East department gave the least number of citations in one day.

  • False, 110 > 90 (approx.)

c. The East department has a greater IQR than the West department.

  • True, 4 out of 5 numbers are greater

d. The East department has the greater median of citations in one day.

  • True, 143 > 135 (approx.)
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Let m the tens digit and n the ones
the Original number is 10m+n
7(m+n) =10m+n
7m+7n=10m+n
6n=3m

Reversing the number 10n+m
10n+m=18+n+m
9n=18
n=2

6n=3m
6(2)=3m
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the number is 10m+n=10(4)+2=42
4 0
3 years ago
Find the inverse f(x)=x+5/3x-1<br><br> f^-1(x)=
Kipish [7]

Answer:

  • f^-1(x) = (3/8)(x +1) . . . . as written
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Step-by-step explanation:

The inverse function can be found by solving for y:

  x = f(y)

  x = y + 5/3y -1 . . . . . . . . . . y +(5/3)y -1 . . . per order of operations

  x+1 = 8/3y . . . . . . . . . . add 1

  (3/8)(x +1) = y . . . . . . . . multiply by 3/8

  f^-1(x) = (3/8)(x +1) . . . . . inverse of the function as written

_____

Perhaps you intend f(x) = (x+5)/(3x-1). The inverse is found the same way.

  x = (y +5)/(3y -1)

  x(3y -1) = y +5

  3xy -x = y +5 . . . . . eliminate parentheses

  3xy -y = x + 5 . . . . . add x-y

  y(3x -1) = x +5 . . . . . factor out y

  y = (x +5)/(3x -1) . . . divide by the coefficient of y

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7 0
3 years ago
Consider the region bounded by 4y=x^2 and 2y=x.
gayaneshka [121]

Answer:

a) ⅓ units²

b) 4/15 pi units³

c) 2/3 pi units³

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2y = x

4y = (2y)²

4y = 4y²

4y² - 4y = 0

y(y-1) = 0

y = 0, 1

x = 0, 2

Area

Integrate: x²/4 - x/2

From 0 to 2

(x³/12 - x²/4)

(8/12 - 4/4) - 0

= -⅓

Area = ⅓

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Squares and then integrate

Integrate: [x²/4]² - [x/2]²

Integrate: x⁴/16 - x²/4

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Volume = 4/15 pi

About the x-axis

x² = 4y

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Integrate the difference

Integrate: 4y² - 4y

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7 0
3 years ago
Question 8
Lelechka [254]

Answer:

$695.88

Step-by-step explanation:

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611 + 48.88 = 659.88

Stay safe! <3

6 0
3 years ago
The International Air Transport Association surveys business travelers to develop quality ratings for transatlantic gateway airp
motikmotik

Answer:

The 95% confidence interval for the population mean rating is (5.73, 6.95).

Step-by-step explanation:

We start by calculating the mean and standard deviation of the sample:

M=\dfrac{1}{n}\sum_{i=1}^n\,x_i\\\\\\M=\dfrac{1}{50}(6+4+6+. . .+6)\\\\\\M=\dfrac{317}{50}\\\\\\M=6.34\\\\\\s=\sqrt{\dfrac{1}{n-1}\sum_{i=1}^n\,(x_i-M)^2}\\\\\\s=\sqrt{\dfrac{1}{49}((6-6.34)^2+(4-6.34)^2+(6-6.34)^2+. . . +(6-6.34)^2)}\\\\\\s=\sqrt{\dfrac{229.22}{49}}\\\\\\s=\sqrt{4.68}=2.16\\\\\\

We have to calculate a 95% confidence interval for the mean.

The population standard deviation is not known, so we have to estimate it from the sample standard deviation and use a t-students distribution to calculate the critical value.

The sample mean is M=6.34.

The sample size is N=50.

When σ is not known, s divided by the square root of N is used as an estimate of σM:

s_M=\dfrac{s}{\sqrt{N}}=\dfrac{2.16}{\sqrt{50}}=\dfrac{2.16}{7.071}=0.305

The degrees of freedom for this sample size are:

df=n-1=50-1=49

The t-value for a 95% confidence interval and 49 degrees of freedom is t=2.01.

The margin of error (MOE) can be calculated as:

MOE=t\cdot s_M=2.01 \cdot 0.305=0.61

Then, the lower and upper bounds of the confidence interval are:

LL=M-t \cdot s_M = 6.34-0.61=5.73\\\\UL=M+t \cdot s_M = 6.34+0.61=6.95

The 95% confidence interval for the mean is (5.73, 6.95).

4 0
3 years ago
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