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nika2105 [10]
3 years ago
15

How can you check that k ≥ 11 is the solution set to the inequality k + 14 ≥ 25?

Mathematics
1 answer:
Andreyy893 years ago
8 0

Answer:

K= 11

Step-by-step explanation:

You solve for K by subtracting 14 by 25.

25-14= 11

When k is pluged into k\geq 11 the statement is true because it is equal to 11.

When plugged into k+14\geq25;

11+14 =25.

That statement is also true because 25 is equel to 25.

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ANSWER FAST (95 POINTS)
lilavasa [31]

Answer:

d) 140

< 2  =   < 3 = 40 \\ so \:  < 3 = 40 \\  < 3 +  < 5 = 180 \\  < 5 = 180 - 40 \\  < 5 = 140 \\ so \: answer \: is \: 140

6 0
3 years ago
Read 2 more answers
For a group of 100 people, compute(a) the expected number of days of the year that are birthdays of exactly 3 people.(b) the exp
Scilla [17]

Answer:

Step-by-step explanation:

Leaving leap years, a year contains 365 days.

For a group of 100 people, each person is independent of the other and probability of any day being his birthday has a chance of

\frac{1}{365}

a) Probability that  exactly 3 people have same birthday = \frac{1}{365^3}

Each day is independent of the other

And hence no of days having exactly 3 persons birthday out of 100 persons is binomial with n = 365 and p = \frac{1}{365^3}

Expected value of days = np = \frac{1}{365^2}

b) Distinct birthdays is binomail with p =1/365 and n = 365

Hence

expected value = np =1

4 0
4 years ago
The following blank indicates a missing operation in the equation.
Natalija [7]
The answer would be 5/2q+9r+6r-20s

Explanation
We have the equation
5(1/2q+3r-4s)
which would equal
5/2q+15r-20s

so with the information
We can subtract 6r from 15r to get 9r
15r-6r=9r
+9r would be the missing piece

So we substitute it in to make the equation true.
5/2q(+9r)+6r-20s
Hope this helps or was correct
3 0
3 years ago
The figure shows the letter Z and four of its transformed images—A, B, C, and D:
tiny-mole [99]

Given:

The figure shows the letter Z and four of its transformed images—A, B, C, and D.

To find:

Which of the following rules will transform the pre-image of Z in quadrant 2 into its image in quadrant 1?

Solution:

From the figure it is clear that the pre-image of Z in quadrant 2 and its image in quadrant 1 (image A) are the mirror image of each other along the y-axis.

It means the pre-image of Z in quadrant 2 reflected across the y-axis to get the image in quadrant 1.

If a figure reflected across the y-axis, then rule of transformation is

(x,y)\to (-x,y)

So, the rule (x,y)\to (-x,y) transform the pre-image of Z in quadrant 2 into its image in quadrant 1.

Therefore, the correct option is c.

3 0
3 years ago
How do you write one million, twelve thousand, sixty widgets in standard form
bekas [8.4K]
One million = 1000000
Twelve thousand = 12000
Sixty = 60

One million, twelve thousand and sixty = 1 012 060

Standard form is given by A × 10ⁿ
Where A is a number  in unit and 'n' is an integer

1 012 060 = 1.012060 × 10⁶

4 0
3 years ago
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