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AleksandrR [38]
2 years ago
12

A computer gererates 80 integers from 1 to 5 at random. The results are recrded in the table. What is the experimental probabili

ty of the computer generating a 2?​
Mathematics
1 answer:
iVinArrow [24]2 years ago
6 0
1/5 as a percentage can be found by setting it equal to x/100. 1/5=x/100
By cross multiplying, you get 5x=100.
Simplify and x is equal to 20 or 20%.
However, we still need to find how many 2s are in the 80 integers.
This can be found by multiplying the percent by the number 80.
So, 0.20(80)=16
Therefore the probability of getting 2 is 16/80 or 20%.
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Work out the value of R. Give your answer in standard form to an appropriate degree of accuracy.
Mumz [18]

Answer:

R = 8,228,571.43 (Approx)

Step-by-step explanation:

Given:

R = x²/y

x = 4.8 x 10⁵

y = 2.8 x 10⁴

Find:

Value of R

Computation:

R = x²/y

R = (4.8 x 10⁵)²/ (2.8 x 10⁴)

R = 23.04 x 10¹⁰ / (2.8 x 10⁴)

R = 8.22857.. x 10⁶

R = 8,228,571.43 (Approx)

4 0
3 years ago
Evaluate each expression for the given values: 
IrinaK [193]

B)-18

  1. 7(-3)+(-2(-3))+(-3)
  2. (-21)+(-2(-3))+(-3)
  3. (-21)+6+(-3)
  4. (-24)+6
  5. -18
8 0
3 years ago
In a certain Algebra 2 class of 26 students, 8 of them play basketball and 11 of them play baseball. There are 6 students who pl
sasho [114]

Answer:19/26

Step-by-step explanation:

Total number of students=26

Number of students that play basket ball=8

Number of students that play

baseball=11

Probability of playing both=6

Probability of playing either of the 2 games=8/26+11/26=19/26

8 0
3 years ago
Find the additive inverse of each linear expression<br> 1. 3x - 6<br> 2. -9x + 3<br> 3. -4x -8
lubasha [3.4K]

Step-by-step explanation:

G(x) = 2/3 x - 6     switch x and y around and solve for y.

y = 2/3 x - 6

x = 2/3 y - 6

x + 6 = 2/3 y

3/2  x + 9 = y

4 0
3 years ago
Find thhe remainder when 7^203 is divided by 4
Aleksandr [31]
Using the square-and-multiply approach, we have

7^{203}=7\times(7^{101})^2
7^{101}=7\times(7^{50})^2
7^{50}=(7^{25})^2
7^{25}=7\times(7^{12})^2
7^{12}=(7^6)^2
7^6=(7^3)^2
7^3=7\times7^2

and so, using the property that, if a_1\equiv b_1\mod n and a_2\equiv b_2\mod n, then a_1a_2\equiv b_1b_2\mod n, we get

7\equiv3\mod4
7^2\equiv9\equiv1\mod4
7^3\equiv7\times1\equiv7\equiv3\mod4
7^6\equiv9\equiv1\mod4
7^{12}\equiv1\mod4
7^{25}\equiv7\times1\equiv7\equiv3\mod4
7^{50}\equiv9\equiv1\mod4
7^{101}\equiv7\times1\equiv7\equiv3\mod4
7^{203}\equiv7\times9\equiv3\times1\equiv3\mod4
4 0
3 years ago
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