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Serggg [28]
2 years ago
7

X^2 +4X+3 Complete the square using the method

Mathematics
1 answer:
gayaneshka [121]2 years ago
7 0

Answer:

(x+3)(x+1)

Step-by-step explanation:

Think of using FOIL

First Outer Inner Last

What can you multiply to get to 3 and add up to 4

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Dr. Wong has a table of data in her veterinary office that shows the average lifespans and weights of several popular dog breeds
SIZIF [17.4K]

Answer:

Step-by-step explanat

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3 years ago
A model of a skyscraper is 1.6 in long, 2.8 in wide, and 11.2 in high. The scale factor is 8 in : 250 ft. What are the actual di
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<span>At 8in to 250 feet the skyscraper converts to the following. 1.6 in/8 in x 250ft = 50 ft long 2.8in/8in x 250ft = 87.5 ft wide 11.2in/8in x 250ft = 350 ft high You get this because every 8 in is 250 feet so you need to calculate the ratio of scale to that 8in per 250 feet.</span>
3 0
3 years ago
If f(x)=3x-2 and g(x)=2x+1,find (f-g)(x) <br><br> A. x - 3<br> B. 5x - 3<br> C. 3 - x<br> D. 5x - 1
Nady [450]

Answer:

A

Step-by-step explanation:

(f - g)(x)

= f(x) - g(x)

= 3x - 2 - (2x + 1)

= 3x - 2 - 2x - 1 ← collect like terms

= x - 3 → A

6 0
3 years ago
Read 2 more answers
The stem-and-leaf plot shows the number of digs for the top 15 players at a volleyball tournament.
NeX [460]

Answer:

Step-by-step explanation:

Hello!

(Data and full text attached)

The stem and leaf plot is a way to present quantitative data.  Considering two-digit numbers, for example 50, the tens digits are arranged in the stem and the units determine the leafs.

So for the stem and leaf showing the digs of the top players of the tournament, the observed data is:

41, 41, 43, 43, 45, 50, 52, 53, 54, 62, 63, 63, 67, 75, 97

n= 15

Note that in the stem it shows the number 8, but with no leaf in that row, that means that there were no "eighties" observed.

a) 6 Players had more than 60 digs.

b)

To calculate the mean you have to use the following formula:

X[bar]= ∑x/n= (41 + 41 + 43 + 43 + 45 + 50 + 42 + 53 + 54 + 62 + 63 + 63 + 67 + 75 + 97)/15= 849/15= 56.6 digs

To calculate the median you have to calculate its position and then identify its value out of the observed data arranged from least to greatest:

PosMe= (n+1)/2= (15+1)/2= 8 ⇒ The median is in the eight place:

41, 41, 43, 43, 45, 50, 42, 53, 54, 62, 63, 63, 67, 75, 97

The median is Me= 53

53 is the value that separates the data in exact halves.

The mode is the most observed value (with more absolute frequency).

Consider the values that were recorded more than once

41, 41

43, 43

63, 63

41, 43 and 63 are the values with most absolute frequency, which means that this distribution is multimodal and has three modes:

Md₁: 41

Md₂: 43

Md₃: 63

The Range is the difference between the maximum value and the minimum value of the data set:

R= max- min= 97 - 41= 56

c)

The distribution is asymmetrical, right skewed and tri-modal.

Md₁: 41 < Md₂: 43 < Me= 53 < X[bar]= 56.6 < Md₃: 63

Outlier: 97

d)

An outlier is an observation that is significantly distant from the rest of the data set. They usually represent experimental errors (such as a measurement) or atypical observations. Some statistical measurements, such as the sample mean, are severely affected by this type of values and their presence tends to cause misleading results on a statistical analysis.  

Considering the 1st quartile (Q₁), the 3rd quartile (Q₃) and the interquartile range IQR, any value X is considered an outlier if:

X < Q₁ - 1.5 IQR

X > Q₃ + 1.5 IQR

PosQ₁= 16/4= 4

Q₁= 43

PosQ₃= 16*3/4= 12

Q₃= 63

IQR= 63 - 43= 20

Q₁ - 1.5 IQR = 43 - 1.5*20= 13 ⇒ There are no values 13 and below, there are no lower outliers.

Q₃ + 1.5 IQR = 63 + 1.5*20= 93 ⇒ There is one value registered above the calculated limit, the last observation 97 is the only outlier of the sample.

The mean is highly affected by outliers, its value is always modified by the magnitude of the outliers and "moves" its position towards the direction of them.

Calculated mean with the outlier: X[bar]= 849/15= 56.6 digs

Calculated mean without the outlier: X[bar]= 752/14= 53.71 digs

I hope this helps.

7 0
3 years ago
Question 1 of 3 NE A scale drawing of a fishing boat has a scale of 1 cmn to 2 meters. If the length of the fishing boat is 6 cm
bagirrra123 [75]

Answer:

12 meters

Step-by-step explanation:

Given

Scale= 1cm:2m

Required

Determine the measurement when the scale reads 6cm

Represent this scale with x;

Such that:

Scale = 6cm:x

Equate

Scale= 1cm:2m and Scale = 6cm:x

1cm:2m = 6cm:x

Express as fraction

\frac{1cm}{2m} = \frac{6cm}{x}

Cross Multiply

x * 1cm = 6cm * 2m

x * 1 = 6 * 2m

x = 6 * 2m

x = 12m

3 0
3 years ago
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