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miskamm [114]
2 years ago
8

Morgan is walking her dog on an 8-meter-long leash. She is currently 500 meters from her house, so the maximum and minimum dista

nces that the dog may be from the house can be found using the equation |x – 500| = 8. What are the minimum and maximum distances that Morgan’s dog may be from the house?
Mathematics
1 answer:
MAVERICK [17]2 years ago
4 0

Answer:

Max: 500

Min: 492

Step-by-step explanation:

Her dog is on an 8 meter leash but it doesn't necessarily take up the whole leash. The most is could be closer to her house than Morgan is 8 meters, the least is 0.

Most: 500-0=500

Min: 500-8=492

Please rate, thank, and mark branliest (not necessarily mine) Hope this helps :)

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3. Jacob and Sarah are saving money to go on a trip. They need at least $1975 in order to go. Jacob mows lawns and Sarah walks d
kvv77 [185]
1. The total must be at least $1975:    25x+15y \geq \$1975

2. Dog walks is no more than 4 times lawns: y \leq 4x

3. At least 50 dog walks:  y \geq 50
5 0
2 years ago
1. The following are the number of hours that 10 police officers have spent being trained in how to handle encounters with peopl
dusya [7]

Answer:

Range = 16

Inter\ Quartile\ Range = 6.75

Variance = 20.44

Standard\ Deviation = 4.52

Step-by-step explanation:

Given

4, 17, 12, 9, 6, 10, 1, 5, 9, 3

Calculating the range;

Range = Highest - Lowest

From the given data;

Highest = 17 and Lowest = 1

Hence;

Range = 17 - 1

Range = 16

Calculating the Inter-quartile Range

Inter quartile range (IQR) is calculates as thus

IQR = Q_3 - Q_1

Where

Q3 = Upper Quartile and Q1 = Lower Quartile

<em />

<em>Start by arranging the data in ascending order</em>

1, 3, 4, 5, 6, 9, 9, 10, 12, 17

N = Number of data; N = 10

---------------------------------------------------------------------------------

Calculating Q3

Q_3 = \frac{3}{4}(N+1) th\ item

<em>Substitute 10 for N</em>

Q_3 = \frac{3}{4}(10+1) th\ item

Q_3 = \frac{3}{4}(11) th\ item

Q_3 = \frac{33}{4} th\ item

Q_3 = 8.25 th\ item

Express 8.25 as 8 + 0.25

Q_3 = (8 + 0.25) th\ item

Q_3 = 8th\ item + 0.25 th\ item

Express 0.25 as fraction

Q_3 = 8th\ item +\frac{1}{4} th\ item

Q_3 = 8th\ item +\frac{1}{4} (9th\ item - 8th\ item)

From the arranged data;

8th\ item = 10 and 9th\ item = 12

Q_3 = 8th\ item +\frac{1}{4} (9th\ item - 8th\ item)

Q_3 = 10 +\frac{1}{4} (12 - 10)

Q_3 = 10 +\frac{1}{4} (2)

Q_3 = 10 +0.5

Q_3 = 10.5

Calculating Q1

Q_1 = \frac{1}{4}(N+1) th\ item

<em>Substitute 10 for N</em>

Q_1 = \frac{1}{4}(10+1) th\ item

Q_1 = \frac{1}{4}(11) th\ item

Q_1 = \frac{11}{4} th\ item

Q_1 = 2.75 th\ item

Express 2.75 as 2 + 0.75

Q_1 = (2 + 0.75) th\ item

Q_1 = 2nd\ item + 0.75 th\ item

Express 0.75 as fraction

Q_1 = 2nd\ item +\frac{3}{4} th\ item

Q_1 = 2nd\ item +\frac{3}{4} (3rd\ item - 2nd\ item)

From the arranged data;

2nd\ item = 3 and 3rd\ item = 4

Q_1 = 3 +\frac{3}{4} (4 - 3)

Q_1 = 3 +\frac{3}{4} (1)

Q_1 = 3 +0.75

Q_1 = 3 .75

---------------------------------------------------------------------------------

Recall that

IQR = Q_3 - Q_1

IQR = 10.5 - 3.75

IQR = 6.75

Calculating Variance

Start by calculating the mean

Mean = \frac{1+3+4+5+6+9+9+10+12+17}{10}

Mean = \frac{76}{10}

Mean = 7.6

Subtract the mean from each data, then square the result

(1 - 7.6)^2 = (-6.6)^2 = 43.56

(3 - 7.6)^2 = (-4.6)^2 = 21.16

(4 - 7.6)^2 = (-3.6)^2 = 12.96

(5 - 7.6)^2 = (-2.6)^2 = 6.76

(6 - 7.6)^2 = (-1.6)^2 = 2.56

(9 - 7.6)^2 = (1.4)^2 = 1.96

(9 - 7.6)^2 = (1.4)^2 = 1.96

(10 - 7.6)^2 = (2.4)^2 = 5.76

(12 - 7.6)^2 = (4.4)^2 = 19.36

(17 - 7.6)^2 = (9.4)^2 = 88.36

Sum the result

43.56 + 21.16 + 12.96 + 6.76 + 2.56 + 1.96 + 1.96 + 5.76 + 19.36 + 88.36 = 204.4

Divide by number of observation;

Variance = \frac{204.4}{10}

Variance = 20.44

Calculating Standard Deviation (SD)

SD = \sqrt{Variance}

SD = \sqrt{20.44}

SD = 4.52 <em>(Approximated)</em>

4 0
2 years ago
A semicircular flower bed has a diameter of 3 metres. What is the area of the flower bed?
Yuri [45]

The area of circle is

A_{\text{circle}}=\pi r^2.

Then the area of semicircular region is

A_{\text{semicircular region}}=\dfrac{1}{2} \pi r^2.

If r=3 m, you have

A_{\text{semicircular region}}=\dfrac{1}{2} \pi \cdot 3^2=\dfrac{9\pi}{2}=4.5\pi \approx 14.13 sq. m.

Answer: A_{\text{semicircular region}}=4.5\pi \approx 14.13 sq. m.

3 0
2 years ago
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A sequence is defined recursively by the formula f(n+1)=-2f(n). The first term of the sequence is -1.5. What is the next term in
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Given f(n+1) =-2f(n)
here f(n) =f(1) = -1.5
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