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Goryan [66]
3 years ago
13

**IMPORTANT**

Mathematics
1 answer:
koban [17]3 years ago
8 0

Answer:

11/18

5/9

Step-by-step explanation:

we will make punnette square

         dice1

    +

dice2   1       2        3       4        5       6

1            2      3        4       5        6       7

2           3      4         5       6       7        8

3           4      5         6       7       8        9

4           5      6         7       8       9        10

5           6      7         8       9      10       11

6          7       8         9       10     11        12

1)

The probability that the sum of the numbers rolled is either even or a multiple of 5

even number = 18

{(1,1), (1, 3), (1, 5), (2, 2), (2, 4), (2, 6), (3, 1), (3, 3), (3, 5), (4,2), (4, 4), (4, 6), (5, 1), (5, 3), (5, 5), (6, 2), (6, 4), (6, 6)},

and

multiple of 5 = 4

{(1,4), (2, 3), (3, 2), (4, 1), (4, 6), (5, 5), (6, 4)}

<em>(we will skip the repeating)</em>

<em />

18 + 4 = 22

probability = 22/36

                  = 11/18

               

2)

The probability that the sum of the numbers rolled is either a multiple of 3 or 4

multiple of 3 = 12

{(1, 2), (1, 5), (2, 1), (2, 4), (3, 3), (3, 6), (4, 2), (4, 5), (5, 1), (5, 4), (6, 3), (6, 6)}

and

multiple of 4 = 8

D = {(1, 3), (2, 2), (2, 6), (3, 1), (3, 5), (4, 4), (5, 3), (6, 2), (6, 6)}

<em>(we will skip the repeating)</em>

<em />

12 + 8 = 20

probability = 20/36

                  = 5/9

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<h3>Meaning of Trigonometry</h3>

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6 0
2 years ago
You can upgrade lighting at your factory to LED bulbs that cost $6.95 each and last an average of 5 years. It costs S3 in labor
marshall27 [118]

Answer:

11.95 dollars

Step-by-step explanation:

Cost of one LED bulb = 6.95 dollars

Cost of changing one bulb = 3 dollars

Life of each bulb = 5 years.

In 10 years, only one time after 5 years the bulbs would have been changed

Hence cost of 100 lamps = 695 and

cost of changing once = 500

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Divide this by 10 years to get average cost per year

Cost of 100 lamps per year = 1195/100 = 11.95 dollars


8 0
3 years ago
Okay so, “Lynn multiplied a number by 8, and then subtracted 35 from the product. The result was 21. Write an equation to find t
Serhud [2]

............................................................................

7 0
3 years ago
What are the approximate values of the minimum and maximum points of f(x) = x5 − 10x3 + 9x on [-3,3]?
nika2105 [10]

Answer:

Minimum : -37 at x=2.4 and

Maximum = 37 at x=-2.4.

Step-by-step explanation:

Given:

f(x)=x^5-10x^3+9x; [-3,3]

Explanation:

In order to find minimum/maximum of a function, we need to find the first derivative of the function and then set it equal to 0 to get critical points.

Therefore,

f'(x)=5x^4-30x^2+9

Setting derivative equal to 0, we get

5x^4-30x^2+9=0

On applying quadratic formula, we get

x=2.4, -2.4, -0.7, 0.7.

So, those are critical points of the given function.

Plugging the values x=2.4, -2.4, -0.7, 0.7, -3 and 3 in above function, we get

f(2.4)=(2.4)^5-10(2.4)^3+9(2.4)= -37.01376   : Minimum.

f(-2.4)=(-2.4)^5-10(-2.4)^3+9(-2.4)= 37.01376 : Maximum.

f(0.7)=(0.7)^5-10(0.7)^3+9(0.7) = 3.03807

f(-0.7)=(-0.7)^5-10(-0.7)^3+9(-0.7) = -3.03807

f(-3)=(-3)^5-10(-3)^3+9(-3) =0

f(3)=(3)^5-10(3)^3+9(3) =0

Therefore the approximate values of the minimum and maximum points of f(x) = x^5- 10x^3+ 9x on [-3,3] are:

Minimum : -37 at x=2.4 and

Maximum = 37 at x=-2.4.


7 0
3 years ago
If your percentile rank in a group of 500 students is 84, how many persons rank lower than you?
julsineya [31]
If you do 84 x 5 then you have 420 people lower than you 
5 0
3 years ago
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