MATHEMATICS COLLEGE

Describe the translation of the graph of y = x2 that results in the graph of y = (x - 3)2.up 3 units
down 3 units
Oright 3 units
left 3 units

Answers

Answer 1
Answer:

Answer: right 3

Step-by-step explanation: The equation is in child(y = (x-3)2) to parent(y=x2). So, if we move right 3, the equation can will be y = (x-3)2
Answer 2
Answer:

Answer: it’s to the right 3 units

Step-by-step explanation:

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In a random sample of 500 college students, 23% say that they read or watch the news every day. Develop a 90% confidence interval for the proportion of all students who read or watch the news on a daily basis. Interpret your results. If you wanted to develop a 95% confidence interval with a margin of error of .01, how many students would need to be surveyed?

Answers

Answer:

The 90% confidence interval is  0.199 <  p < 0.261

The sample size to develop a 95% confidence interval is n = 2032  

Step-by-step explanation:

From the question we are told that

   The sample size is n =500

    The sample proportion is  \^ p = 0.23

From the question we are told the confidence level is  90% , hence the level of significance is    

      \alpha = (100 - 90 ) \%

=>   \alpha = 0.10

Generally from the normal distribution table the critical value  of  (\alpha )/(2) is  

   Z_{(\alpha )/(2) } =  1.645

Generally the margin of error is mathematically represented as  

     E =  Z_{(\alpha )/(2) } * \sqrt{(\^ p (1- \^ p))/(n) }

=>   E =  1.645 * \sqrt{(0.23 (1- 0.23))/(500) }

=>   E =  0.03096

Generally 90% confidence interval is mathematically represented as  

      \^ p -E <  p <  \^ p +E

=>    0.23  -0.03096  <  p < 0.23  +  0.03096

=>   0.199 <  p < 0.261

From the question we are told the confidence level is  95% , hence the level of significance is    

      \alpha = (100 - 95 ) \%

=>   \alpha = 0.05

Generally from the normal distribution table the critical value  of  (\alpha )/(2) is  

   Z_{(\alpha )/(2) } =  1.96

The margin of error is given as E = 0.01

Generally the sample size is mathematically represented as  

    n = [\frac{Z_{(\alpha )/(2) }}{E} ]^2 * \^ p (1 - \^ p )

=>    n = [(1.96 )/(0.01) ]^2 *0.23 (1 - 0.23 )      

=>    n = 2032  

What is The greatest common factor of 42 and 96 I’ll give you brainliest

Answers

The greatest common factor of 42 and 96 is 6.

What is greatest common factor?

The greatest common factor (GCF) of a set of numbers is the largest factor that all the numbers share.

42 = 2 x 3 x 7

96 = 2 x 2 x 2 x 2 x 2 x 3

Greatest common factor = 2 x 3 = 6

Learn more about greatest common factor here

brainly.com/question/11221202

#SPJ3

Answer:

6

Step-by-step explanation:

42 and 96

first we will find the factors of 42:

2 21

3 7

7 7

hence, factors of 42 are 2,7 and 3

then, we will find the factors of 96:

2 96

2 48

2 24

2 12

2 6

3 3

1

hence, factors of 96 are 2,2,2,2,2 and 3

Common factors are: 2 and 3

Highest common factor: 6

6.2.Find the triangle bounded by the y-axis the line f(x)=9- 6/7x and the line perpendicular to it that passes through the origin

Answers

First.

We need to find the triangle bounded by the y-axis the line f(x)=9-(6)/(7)x and the line perpendicular to it that passes through the origin.

This triangle is illustrated in the First Figure bellow. The triangle has been dotted with black dots. So, let's identify each part of this graph:


  • The y-axis has been identified by the green line.
  • The linef(x)=9-(6)/(7)x has been identified by the red line.
  • To plot the line perpendicular to the previous one that passes through the origin, we need to know that the product of the slopes of two perpendicular lines is m_(1)m_(2)=-1 \therefore Since \ m_(1)=-(6)/(7) \ then \ m_(2)=(7)/(6)

        Therefore, this line is the blue one and is written as:

         f(x)=(7)/(6)x


Second.

a.

To find y(t) we need to find the slope and y-intercept. So, taking two points:

P_(1)(15,150) \ and \ P_(2)(25,450) \n \nm=(450-150)/(25-15)=30


So:

Using \ P_(1)(15,150) \n\ny=30t+b\n\n150=30(15)+b \therefore b=-300 \n\n\boxed{y(t)=30t-300}


Third.

b.

The y-intercept is b=-300 and has been calculated above. This value represents the profit in thousands of dollars in 1980, that is, there is no any profit or there is a lost of $300.000.


Fourth.

c.

The x-intercept can be found when y=0 (year 1980), that is:

y(t)=30t-300 \n\n0=30t-300 \therefore t=10


This value shows that there is no any profit for the company at the year that represents t=10 (year 1990), because at this point the company starts earning money.


Fifth.

d.

The slope of this function is m=30


This value tells you that the profit increases $30.000 each year.

Hello,

Please, see the attached files.

Thanks.

ABCD is a trapezoid.
What is the area of trapezoid?

Answers

Let h represent the height of the trapezoid, the perpendicular distance between AB and DC. Then the area of the trapezoid is
  Area = (1/2)(AB + DC)·h
We are given a relationship between AB and DC, so we can write
  Area = (1/2)(AB + AB/4)·h = (5/8)AB·h

The given dimensions let us determine the area of ∆BCE to be
  Area ∆BCE = (1/2)(5 cm)(12 cm) = 30 cm²

The total area of the trapezoid is also the sum of the areas ...
  Area = Area ∆BCE + Area ∆ABE + Area ∆DCE
Since AE = 1/3(AD), the perpendicular distance from E to AB will be h/3. The areas of the two smaller triangles can be computed as
  Area ∆ABE = (1/2)(AB)·h/3 = (1/6)AB·h
  Area ∆DCE = (1/2)(DC)·(2/3)h = (1/2)(AB/4)·(2/3)h = (1/12)AB·h

Putting all of the above into the equation for the total area of the trapezoid, we have
  Area = (5/8)AB·h = 30 cm² + (1/6)AB·h + (1/12)AB·h
  (5/8 -1/6 -1/12)AB·h = 30 cm²
  AB·h = (30 cm²)/(3/8) = 80 cm²

Then the area of the trapezoid is
  Area = (5/8)AB·h = (5/8)·80 cm² = 50 cm²

30 + (-87) – 140 + 8A

Answers

Simplify the expression.

−197+8A, Hope this helps, Have a wonderful day!~

1 3/7+2 2/6 the answer

Answers

Answer: 3 16/21

Step-by-step explanation:

10/7 + 2 2/6

10/7 + 2 1/3

= 10/7 + 7/3

= 79/21

= 3 16/21

Hope this helps!

Answer:

Fraction form 79/21 or 21/79

Decimal form 3.76

Mixed fraction form 3 16/21
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