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notka56 [123]
3 years ago
13

D. 2xy + 4y + x + 2 =​

Mathematics
1 answer:
Leona [35]3 years ago
8 0
The answer is 2xy + 4y + x + 2 because you cannot add variables that aren’t the same
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(3.1 · 10² ) (2 · 10³) Write in Scientific Notation.
Nonamiya [84]
Covert into scientific notation?
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3 years ago
Read 2 more answers
the eccentricity of a hyperbola is defined as e=c/a. find an equation with vertices (1,-3) and (-3,-3) and e=5/2
Ksivusya [100]

Answer:

\frac{(x + 1 )^{2} }{4} - \frac{(y + 3 )^{2} }{21} = 1.

Step-by-step explanation:

If (α, β) are the coordinates of the center of the hyperbola, then its equation of the hyperbola is \frac{(x - \alpha )^{2} }{a^{2} } - \frac{(y - \beta )^{2} }{b^{2} } = 1.

Now, the vertices of the hyperbola are given by (α ± a, β) ≡ (1,-3) and (-3,-3)

Hence, β = - 3 and α + a = 1 and α - a = -3

Now, solving those two equations of α and a we get,  

2α = - 2, ⇒ α = -1 and

a = 1 - α = 2.

Now, eccentricity of the hyperbola is given by b^{2} = a^{2}(e^{2}  - 1) = 4[(\frac{5}{2} )^{2} -1] = 21 {Since e = \frac{5}{2} given}

Therefore, the equation of the given hyperbola will be  

\frac{(x + 1 )^{2} }{4} - \frac{(y + 3 )^{2} }{21} = 1. (Answer)

6 0
3 years ago
In March 2007, Business Week reported that at the top 50 business schools, students studied an average of 14.6 hours. You wonder
Rasek [7]

Answer:

The hypotheses used in this situation

H_0:\mu = 14.6

H_a:\mu \neq 14.6

Step-by-step explanation:

We are given that  Business Week reported that at the top 50 business schools, students studied an average of 14.6 hours.

Mean = \mu = 14.6

Claim : The amount UMSL students study is different from this 14.6 hour benchmark.

The hypotheses used in this situation

H_0:\mu = 14.6

H_a:\mu \neq 14.6

5 0
3 years ago
Two weeks in a row, the golf course hosts a group of golfers. The second week had 10 more golfers than the first week. Use the d
Trava [24]

Answer:

Option A.

Step-by-step explanation:

From the first dot plot the given data set for week 1 is

62, 68, 68, 68, 69, 70, 70, 72, 72, 72, 75, 75, 76, 78, 78, 80, 80, 80, 89, 89

Divide the data in 4 equal parts.

(62, 68, 68, 68, 69), (70, 70, 72, 72, 72), (75, 75, 76, 78, 78), (80, 80, 80, 89, 89)

Range=Maximum-Minimum=89-62=27

Median=\frac{72+75}{2}=73.5

Mean=\frac{\sum x}{n}=\frac{1491}{20}=74.55

From the second dot plot the given data set for week 2 is

62, 68, 68, 68, 69, 70, 70, 72, 72, 72, 75, 75, 76, 78, 78, 80, 80, 80, 85, 86, 86, 86, 88, 88, 88, 88, 89, 89, 89, 89

Divide the data in 4 equal parts.

(62, 68, 68, 68, 69, 70, 70), 72, (72, 72, 75, 75, 76, 78, 78), (80, 80, 80, 85, 86, 86, 86), 88, (88, 88, 88, 89, 89, 89, 89)

Range=Maximum-Minimum=89-62=27

Median=\frac{78+80}{2}=79

Mean=\frac{\sum x}{n}=\frac{2364}{30}=78.8

The range for Week 1 is equal to the range for Week 2.

The median for Week 2 is more than the median for Week 1.

The mean for Week 2 is more than the mean for Week 1.

Therefore, the correct option is A.

6 0
3 years ago
Which system of equations has no solutions? y - ​
Alona [7]
The second one on the top right
7 0
3 years ago
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