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Katarina [22]
3 years ago
9

Find the range of the parent function below y=x^2

Mathematics
1 answer:
marusya05 [52]3 years ago
6 0

you don't show the graph, but graphing y=x^2 it is a U shaped line that starts at (0,0) and the lines go upwards on both sides of the Y axis,

 This means all real numbers above 0 and 0 will solve it

 so the range is y≥0

You might be interested in
Amanda ate 1/3 of her mom's pie. she gave the remaining 2/3 to 4 friends, What fraction did each friend get.
Semenov [28]
So she divided the 2/3 between 4 friends.

So each friend got 2/3 divided by 4, which gives:

(2/3)*(1/4) = 2/12 = 1/6
7 0
3 years ago
Plz help don’t understand
saul85 [17]

Answer:

Step-by-step explanation:

every time x goes up 1,  y goes up 4  

x = 3  minus x = 2  is a delta (a change of 1)

x = 4  minus x = 3  is a delta (a change of 1)

x = 5  minus x = 4  is a delta (a change of 1)

       the delta x is ALWAYS 1 for this table you can always substract x's to find the delta     delta is symbolized by Δ      Δx  for the delta x

y = 7  minus y = 3  is a delta (a change of 4)

y = 11  minus y = 7  is a delta (a change of 4)

y = 15  minus y = 11  is a delta (a change of 4)

    the delta y is ALWAYS 4 for this table you can always substract y's to find the delta     delta is symbolized by Δ     Δy  for the delta y

the slope m = Δy / Δx      m = 4 / 1     m = 4

     y = mx + b             when x = 0    then  y = b   (called the y intercept)

so how do we find the y intercept?    

look at the pattern in the TABLE  Δx  and Δy

   when x = 1    y = (3 - 4)  = -1

             x = 0    y =( -1 - 4)   = -5

so       y = 4x - 5     is a line equation for the data in this table

                                I believe there are other forms for linear equations

                               I chose the y-intercept form

use the point slope equation

       (y - y1)  =  m(x - x1)            where (x1, y1)  are the coordinates of a point on  

                                                the line

       when you know the slope of the line and a point on the line

       as before

       m = (y2 - y1)  / (x2 - x1)          from the table

           = (7 - 3) / (3 - 2)                 still equals

           = 4/1  =  4        

        pick a point on the line    say x= 4 and y = 11    (4, 11)

       (y - y1)  =  m(x - x1)    

        y - 11   =   4(x - 4)              point slope form

         solving point slope form  for  y

        y  - 11  =   4x - 16               add 11 to both sides    

           y  =  4x - 5                    back to the y-intercept form

7 0
3 years ago
Write the ratio as a fraction in simplest form. 27 to 72
Mademuasel [1]

Answer: The ration for 27:71, would be 3/8. Hope this helps!

4 0
3 years ago
The local branch of the Internal Revenue Service spent an average of 21 minutes helping each of 10 people prepare their tax retu
AURORKA [14]

Answer:

The confidence Interval is [- 0.7053  10.4521]

a: The  hypotheses  are

H0: μ1=μ2    against the claim Ha :μ1≠μ2

b. The critical value for t∝/2 for 17 d.f  t > 2.508 and  t < -2.111

c. t= -2.8422

d. The calculated value of t= -2.8422 is less than t < -2.11 the critical value therefore we reject H0 and conclude there is a difference between the two means.

Step-by-step explanation:

When the standard deviations are not the same then the confidence intervals for mean differences are calculated as

(x1`-x2`)- t∝/2 √s1²/n1 + s2²/n2 < u1-u2 < (x1`-x2`)+ t∝/2 √s1²/n1 + s2²/n2

x1`= 21        x2`= 27

n1=  10       n2= 14

s1= 5.6       s2= 4.3

The degrees of freedom is calculated using

υ = [s₁²/n1 + s₂²/n2]²/ (s₁²/n1 )²/ n1-1 + (s₂²/n2)²/n2-1

= 17

The t∝/2 for 17 d.f = 2.11

Putting the values

(x1`-x2`)- t∝/2 √s1²/n1 + s2²/n2 < u1-u2 < (x1`-x2`)+ t∝/2 √s1²/n1 + s2²/n2

(21-27) - 2.11√5.6²/10+ 4.3²/14 < u1-u2 <(21-27)  +2.11√5.6²/10+4.3²/14

6- 2.11*2.111 < u1-u2 <  ( 6 )  +2.11*2.111

6- 4.4521 < u1-u2 <  ( 6 )  +5.294

- 1.5479 < u1-u2 <  10.4521

The confidence Interval is [- 0.7053  10.4521]

a: The  hypotheses  are

H0: μ1=μ2    against the claim Ha :μ1≠μ2

The claim is that there is a difference in the average time spent by the two services

b. The critical value for t∝/2 for 17 d.f  t > 2.508 and  t < -2.111

The degrees of freedom is calculated using

υ = [s₁²/n1 + s₂²/n2]²/ (s₁²/n1 )²/ n1-1 + (s₂²/n2)²/n2-1

= 17

c. The test statistic is

t= (x1`-x2`)  /√s1²/n1 + s2²/n2

t= (21-27)  /√5.6²/10+ 4.3²/14

t= -6/2.111

t= -2.8422

d. The calculated value of t= -2.8422 is less than t < -2.11 the critical value therefore we reject H0 and conclude there is a difference between the two means.

5 0
3 years ago
Anyone help? don’t understand this question?
Anit [1.1K]

Answer:

Step-by-step explanation:

Let the fraction is "x"

hence putt in box we get,

1-x=3/7

x=1-3/7

taking LCM

x=7-3/7

x=4/7 Ans....

7 0
3 years ago
Read 2 more answers
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