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kumpel [21]
3 years ago
13

Please help find the translations

Mathematics
1 answer:
Irina-Kira [14]3 years ago
7 0

Answer:

1. translation (x, y) --> (5, 1)

2. translation (x, y) --> (-5, -5)

Step-by-step explanation:

Pick any point and count how many units it moves in the x direction. That will be your value for x. Then count how many units it moves in the y direction. That will be your value for y. If the x or y direction the figure moves in is negative, then we will get a negative number. For translation one, I will pick point A. Point A moves 5 units to the right, the positive x direction, and one unit up, the positive y direction. Thus giving us the answer (5, 1). For the second translation, point A moves 5 units left, the negative x direction, and 5 units down, the negative y direction. Thus giving us the answer (-5, -5).

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What is the LCM of 539, and 15?
aev [14]
<u>LCM on 539 and 15 :</u>

539 x 15
= 8 085

Answer : 8 085

8 0
4 years ago
Read 2 more answers
A batch of 200 calculators contains 3 defective units. What the probability that a sample of three calculators will have/be
Anika [276]

Answer:

1. 0.048

2. 0.952

3. 0.6465

Step-by-step explanation:

Requirement 1

the probability of no defective calculator is 0.048

Given,

Total calculator = 200

Number of defective = 3

Probability of defective,  

p = 3/200

= 0.015

And the probability of non-defective,

q = 1 - p

=1-0.015

=0.985

The Probability distribution of defective follows the normal distribution.

P [X=0] = 〖200〗_(c_0 ) 〖(.015)〗^0 〖(.985)〗^(200-0)

P [X=0] = 0.048

Requirement 2

the probability of no defective is 0.048.

here, the probability of at least one defective means minimum 1 calculator can be defective. So the probability of at least one will be the probability of less than or equal 1.

P [X =less than or equal 1] = 1- P [X=0]

= 1 - 0.048  

= 0.952

So the probability of at least one defective is 0.952.

Requirement 3

c) the probability of all defective calculators is 0.6465.

P [X=3] = P[X=0]+ P[X=1]+ P[X=2]+ P[X=3]

Here,

P [X=0] = 0.048

P [X=1] = 〖200〗_(c_1 ) 〖(.015)〗^1 〖(.985)〗^(200-1)

= 0.148228

P [X=2] = 〖200〗_(c_2 ) 〖(.015)〗^2 〖(.985)〗^(200-2)

= 0.2245997

P [X=3] = 〖200〗_(c_3 ) 〖(.015)〗^3 〖(.985)〗^(200-3)

= 0.2257398

So,  P [X=3] = 0.048+0.148228+0.2245997+0.2257398  

= 0.6465675

So, the probability of all defective calculators is 0.6465.

6 0
3 years ago
What are the values of x and y in the matrix equation below?
makvit [3.9K]
-8x-12+y^2
This is my aswer haha
3 0
4 years ago
Emily and jimmy each improved their yards by planting rose bushes and shrubs. they bought their supplies from the same store. em
Ratling [72]

Answer:

rose$7  shrub $11

Step-by-step explanation:

5r+3s=68

1r+14s=161

I will multiply the second by -5 to eliminate r variable

5r+  3s= 68

-5r- 70s = -805

        -67s  = -737

            Divide both by -67

          s= 11

14*11=154

161-154 =7

      5*7 + 3*11 = 68

    35+33 = 68

       7+ 14*11 = 161

7+154 = 161

4 0
3 years ago
A manufacturer of band saws wants to estimate the average repair cost per month for the saws he has sold to certain industries.
vlada-n [284]

Answer:

The average repair cost per saw for the past month is 20.

The bound of error of estimation is ±10.

Step-by-step explanation:

a) Data and Calculations:

Industry        No.of saws   Total repair cost   Average

                                           for past month

1                            3                        50           16.67

2                           7                        110           15.71

3                          11                       230          20.91

4                          9                        140           15.56

5                          2                         60           30

6                         12                       280           23.33

7                         14                       240           17.14

8                          3                         45           15

9                          5                        60           12

10                         9                      230           25.56

11                          8                       140           17.50

12                         6                       130           21.67

13                         3                        70           23.33

14                         2                        50           25

15                         1                          10           10

16                        4                         60           15

17                       12                       280           23.33

18                        6                        150           25

19                        5                         110           22

20                       8                        120           15

Total               130                    2,565           19.73

Average = Sum of the total repair cost for past month divided by the number of saws repaired

= 19.73 (2,565/130)

= 20

The bound on error of estimation = the difference between the upper bound of the interval and the calculated mean

= 30 - 20

= 10

The lower bound = 10

The bound of error is also = (30 -10)/2 = 10

8 0
3 years ago
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