Y = 2x² + 30x + 200
y = 2(x² + 15x + 100)

x² = x*x
100 ⇒ 1 x 100 ; 2 x 50 ; 4 x 25 ; 5 x 20 ; 10 x 10

200 = 2x² + 30x + 200
0 = 2x² + 30x + 200 - 200
0 = 2x² + 30x + 0
0 = 2(x² + 15x + ?)

−5 ≤ x ≤ 20

7 0

3 years ago

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Consider the reduction of the rectangle. A large rectangle has a length of 16.8 feet and width of 2.3 feet. A smaller rectangle

vodomira [7]

Answer:

0.6

Step-by-step explanation:

7 0

3 years ago

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Log 3243 a)5 b)7 c)4 d)6

kramer

What’s the question

3 0

3 years ago

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Blood pressure values are often reported to the nearest 5 mmHg (100, 105, 110, etc.). The actual blood pressure values for nine

Paladinen [302]

Answer:

a) 117.4

b) 117.9

c) Option A) When there is rounding or grouping, the median can be highly sensitive to small change

Step-by-step explanation:

We are given the following data set in the question:

108.6, 117.4, 128.4, 120.0, 103.7, 112.0, 98.3, 121.5, 123.2

$Median:\\\text{If n is odd, then}\\\\Median = \displaystyle\frac{n+1}{2}th ~term \\\\\text{If n is even, then}\\\\Median = \displaystyle\frac{\frac{n}{2}th~term + (\frac{n}{2}+1)th~term}{2}$

n = 9

a) Median of the reported blood pressure values

Sorted Values: 98.3, 103.7, 108.6, 112.0, 117.4, 120.0, 121.5, 123.2, 128.4

Median =

$\dfrac{9 + 1}{2}^{th}\text{ term} = 5^{th}\text{ term} = 117.4$

b) New median of the reported values

Data: 108.6, 117.9, 128.4, 120.0, 103.7, 112.0, 98.3, 121.5, 123.2

Sorted Values: 98.3, 103.7, 108.6, 112.0, 117.9, 120.0, 121.5, 123.2, 128.4

New Median =

$\dfrac{9 + 1}{2}^{th}\text{ term} = 5^{th}\text{ term} = 117.9$

c) Since median is a position based descriptive statistics, a small change in values can bring a change in the median value as the order of the data may change.

Option A) When there is rounding or grouping, the median can be highly sensitive to small change

7 0

3 years ago