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aev [14]
4 years ago
8

PLEASE HELP Simplify (3x2 − 2) + (2x2 − 6x + 3).

Mathematics
2 answers:
leva [86]4 years ago
6 0
The whole equation simplified to one answer is 5
VARVARA [1.3K]4 years ago
6 0
3x^2-2+2x^2-6x+3=5x^2-6x+1

If you want to factor this:
5x^2-5x-x+1=5x(x-1)-1(x-1)=5x-1(x-1)
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A city park designer is designing a new park. The park will be shaped like a right triangle and there will be two pathways for p
Aloiza [94]

Answer:

VW=36

VT=27

Step-by-step explanation:

VW is the midsegment of the triangle

so

VW=72/2

=36

VXT is a right angle triangle

so

a²+b²=c²

a=√c²-b²

a=√45²-36²

a=√(45-36)(45+36)

a=√9×81

a=√9×√81

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therefore

VT=27

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3 years ago
Use the data below to construct a stem and leaf display on your own paper. Determine if the determination form the book, Last Co
AveGali [126]
Given that a sample of 32 cowboys gave the following years of longevity: 58 52 68 86 72 66 97 89 84 91 91 92 66 68 87 86 73 61 70 75 72 73 85 84 90 57 77 76 84 93 58 47

The stem and leaf display for the data is given as follows:
4   |   7
5   |   2, 7, 8, 8
6   |   1, 6, 6, 8, 8
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7 0
4 years ago
If 35 of the 140 cars delivered to a dealer were blue, what percent of the car is blue
maw [93]
If we take 140 to be the 100%, what is 35 off of it in percentage anyway?

\bf \begin{array}{ccll}
amount&\%\\
\text{\textemdash\textemdash\textemdash}&\text{\textemdash\textemdash\textemdash}\\
140&100\\
35&p
\end{array}\implies \cfrac{140}{35}=\cfrac{100}{p}\implies p=\cfrac{35\cdot 100}{140}
3 0
3 years ago
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Independent random samples from two regions in the same area gave the following chemical measurements (ppm). Assume the populati
Basile [38]

Answer:

d. The interval contains only negative numbers. We cannot say at the required confidence level that one region is more interesting than the other.

Step-by-step explanation:

Hello!

You have the data of the chemical measurements in two independent regions. The chemical concentration in both regions has a Gaussian distribution.

Be X₁: Chemical measurement in region 1 (ppm)

Sample 1

n= 12

981 726 686 496 657 627 815 504 950 605 570 520

μ₁= 678

σ₁= 164

Sample mean X[bar]₁= 678.08

X₂: Chemical measurement in region 2 (ppm)

Sample 2

n₂= 16

1024 830 526 502 539 373 888 685 868 1093 1132 792 1081 722 1092 844

μ₂= 812

σ₂= 239

Sample mean X[bar]₂= 811.94

Using the information of both samples you have to determina a 90% CI for μ₁ - μ₂.

Since both populations are normal and the population variances are known, you can use a pooled standard normal to estimate the difference between the two population means.

[(X[bar]₁-X[bar]₂)±Z_{1-\alpha /2}* \sqrt{\frac{Sigma^2_1}{n1}+\frac{Sigma^2_2}{n_2}  }]

Z_{1-\alpha /2}= Z_{0.95}= 1.648

[(678.08-811.94)±1.648*\sqrt{\frac{164^2}{12}+\frac{239^2}{16}  }]

[-259.49;-8.23]ppm

Both bonds of the interval are negative, this means that with a 90% confidence level the difference between the population means of the chemical measurements of region 1 and region 2 may be included in the calculated interval.

You cannot be sure without doing a hypothesis test but it may seem that the chemical measurements in region 1 are lower than the chemical measurements in region 2.

I hope it helps!

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Circle, A cone is made of circles that get smaller


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4 years ago
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