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MrRissso [65]
2 years ago
6

What is m(m+7)=0? Please get right!

Mathematics
2 answers:
Neko [114]2 years ago
8 0

Answer:

m^2+7m=0

Step-by-step explanation:

m(m+7)=0

m^2+7m=0

Deffense [45]2 years ago
5 0

hey your answrr of this question

{{x}^{2}\:+\:30 \:= \:1000 }

{{x}^{2}\:+\:30 \:-\:1000\:= \:0 }

{{x}^{2}\:+\:50x\:-\:20x \:-\:1000\:= \:0}

{(x\:+\:50)\:-\:20(x \:+\:50)\:= \:0 }

{(x\:-\:20)(x \:+\:50)\:= \:0 }

x - 20 = 0. or x + 50 = 0

x = 20. or x = - 50

20 , -50 are the roots of the equation

most easy way

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A nurse collected data about the average birth weight of babies in the hospital that month. Her data is shown using the dot plot
mixer [17]

Answer:

The values of the box-plot are: B = {8.2, 8.3, 8.4, 8.5 and 8.9}.

Step-by-step explanation:

The data provided is as follows:

X     Frequency

8        0

8.1        0

8.2        2

8.3        2

8.4        2

8.5        3

8.6        0

8.7         1

8.8        0

8.9         1

 9           0

So, the actual data is:

S = {8.2 , 8.2 , 8.3 , 8.3 , 8.4 , 8.4 , 8.5 , 8.5 , 8.5 , 8.7 , 8.9 }

A boxplot, also known as a box and whisker plot is a method to demonstrate the distribution of a data-set based on the following 5 number summary,

  1. Minimum (shown at the bottom of the chart)
  2. First Quartile (shown by the bottom line of the box)
  3. Median (or the second quartile) (shown as a line in the center of the box)
  4. Third Quartile (shown by the top line of the box)
  5. Maximum (shown at the top of the chart).

The data set is arranged in ascending order.

The minimum value is, Min. = 8.2

The lower quartile is,

[\frac{n+1}{4}]^{th}\ obs.=[\frac{11+1}{4}]^{th}\ obs.=3^{rd }\ obs. =8.3,

Q₁ = 8.3.

The median value is,

[\frac{n+1}{2}]^{th}\ obs.=[\frac{11+1}{2}]^{th}\ obs.=6^{th}\ obs.=8.4

Median = 8.4

The upper quartile is,

[\frac{3(n+1)}{4}]^{th}\ obs.=[\frac{3(11+1)}{4}]^{th}\ obs.=9^{th }\ obs. =8.5,

Q₃ = 8.5.

The maximum value is, Max. = 8.9.

So, the values of the box-plot are: B = {8.2, 8.3, 8.4, 8.5 and 8.9}.

5 0
3 years ago
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