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astraxan [27]
2 years ago
12

How many solutions are there if m < 0 and p > k ? p + m| x + n| = k

Mathematics
1 answer:
neonofarm [45]2 years ago
4 0

9514 1404 393

Answer:

  2

Step-by-step explanation:

The solutions are ...

  p +m|x +n| = k . . . . . given

  m|x +n| = k -p . . . . . . subtract p

  |x +n| = (k -p)/m = (p -k)/-m . . . . . . must have (p-k)/-m ≥ 0

  x = -n ± (p -k)/-m

For the given conditions, m < 0, p > k, we are guaranteed that (p-k)/-m > 0, so there are 2 solutions.

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serg [7]
I know B is true, but i’m unsure for the other choices.
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3 years ago
Which statement is true about the equation fraction 3 over 4z − fraction 1 over 4z + 1 = fraction 2 over 4z + 1?
nata0808 [166]

(A- No solution)

Ignore the folloqing;

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3 0
3 years ago
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The ages of two brothers are in the ratio two to three, but in eight years, the ratio of their ages will be three to four. What
Darina [25.2K]
X = older brother
y = younger brother 

Equation 1: x/y = 2/3 ⇒ x = 2y/3
Equation 2: x + 8/y + 8 = 3/4

( \frac{2y}{3} + 8) / y +8 =  \frac{3}{4}
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2y + 24 = 3y + 24 =  \frac{3}{4}
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The age of the older brother is 24.

x =  \frac{2y}{3} = \frac{2*24}{3}
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The age of the youngest brother is 16.
5 0
3 years ago
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Emmett gathered some data to compare the monthly cost of renting a one-bedroom apartment in Dallas and Austin. The data collecte
Goryan [66]

Answer:

The correct options are;

a) The city's data has the higher median monthly apartment cost of the two cities by $155

b) The city's data had the greater variability among monthly apartment costs because the standard deviation was $49.13 more than that of the other set of data

Step-by-step explanation:

The given data for the cost of renting a one-bedroom apartment in Dallas is presented as follows;

$994, $1,322, $1,075, $1,189, $1,172, $1,465, $1,215, $930, $1,090, $1,288

Which can be arranged in increasing order and analyzed to find the interquartile range, mean and standard deviation using Microsoft Excel as follows;

$930, $994, $1,075, $1,090, $1,172, $1,189, $1,215, $1,288, $1,322, and $ 1,465

The first quartile, Q₁ = $1,054.75

The third quartile, Q₃ = $1,296.5

The interquartile range = Q₃ - Q₁ = $241.75

The median = $1,180.5

The average monthly cost = $1,174

The standard deviation of the sample = $159.9597

The given data for the cost of renting a one-bedroom apartment in Austin is presented in increasing order using Microsoft Excel as follows;

$900, $950, $1,100, $1,250, $1,296, $1,375, $1,389, $1,400, $1,450, $1,495

The interquartile range, mean and standard deviation are found using Microsoft Excel as follows;

The first quartile, Q₁ = $1,062.5

The third quartile, Q₃ = $1,412.5

The interquartile range = Q₃ - Q₁ = $1,412.5 - $1,062.5 = $350

The median = $1,335.5

The average monthly cost = $1,260.5

The standard deviation of the sample = $209.0945

The difference in the median cost of the two cities is $1,335.5 - $1,180.5 = $155

Therefore, the Austin's city data has the higher median monthly apartment cost of the two cities by $155

b) The difference in the sample standard deviation of the two cities is $209.0945 - $159.9597 = $49.13476

Therefore, the Austin city data had the greater variability among monthly apartment costs because the standard deviation was $49.13 more than that of the other set of data.

7 0
3 years ago
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Can someone help me
FromTheMoon [43]
BC is 8.7 
CA is 20
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6 0
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