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aleksklad [387]
3 years ago
5

Choose the equation that could be used to solve this problem.

Mathematics
2 answers:
olganol [36]3 years ago
8 0
23 is the number the answer is c
wolverine [178]3 years ago
6 0
X-45= 22
Add 45 to both sides

X=67
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How to solve this problem 14-6= 10-6=?
Gennadij [26K]
I really don't think you wrote the question properly. Or maybe you did, but this question doesn't make any sense! :/
8 0
3 years ago
The graph of the function f(x)=-(x+6)(x+2) is shown below. What statement about the function is true ?
arlik [135]

Answer:

A

Step-by-step explanation:

in x<-4 the derivative is positive so the function is increasing.

in x>-4 the derivative is negative so the function is decreasing

increasing in (-infinity,-4)

decreasing in (-4,infinity)

7 0
3 years ago
The following are the ages (years) of 5 people in a room:
MA_775_DIABLO [31]

Answer:

The person is 14 years old

Step-by-step explanation:

1. Multiply 6 times 18 which is 108

2. Add all numbers together in list which is 94

3. Subtract 94 from 108 which is 14

4. To check answer add 94 plus 14 and divide by 6 and you will get the median which is 18, now you know this answer is correct.

8 0
3 years ago
Simplify 6 to the 5th power over 7 to the 3rd power times 2
Anon25 [30]
My answer -

<span>1. Use symbols (not words) to express quotient
 

2. Use exponent symbol (^) to denote exponents

3. Just write out question number, question, and choices. No need for extra information (such as points). Also, don't leave blank lines between choices. This extraneous that we don't need just makes your whole question very very long, and means a lot of scrolling on our part.
 

4. You should only post 2 or 3 questions at a time.


1) (6x^3 − 18x^2 − 12x) / (−6x) = −x^2 + 3x + 2 ----> so much simpler to read !

2) (d^7 g^13) / (d^2 g^7) = d^(7−2) g^(13−7) = d^5 g^6 ----> much easier to read !

3) (4x − 6)^2 = 16x^2 − 24x − 24x + 36 = 16x^2 − 48x + 36

4) (x^2 / y^5)^4 = (x^2)^4 / (y^5)^4 = x^8 / y^20

5) (3x + 5y)(4x − 3y) = 12x^2 − 9xy + 20xy − 15y^2 = 12x^2 + 11xy − 15y^2

6) (3x^3y^4z^4)(2x^3y^4z^2) = (3*2) x^(3+3) y^(4+4) z^(4+2) = 6 x^6 y^8 z^6

7) 5x + 3x^4 − 7x^3 ----> Fourth degree trinomial

8) (5x^3 − 5x − 8) + (2x^3 + 4x + 2) = 7x^3 − x − 6

9) (x − 1) + (2x + 5) − (x + 3) = x + 1

10) (−4g^8h^5k^2)0(hk^2)^2 = 0 (anything multiplied by 0 = 0)
or.. (−4g^8h^5k^2)^0(hk^2)^2 = 1 (h^2 (k^2)^2) = h^2 k^4

Last question shows why it is so important to use proper symbols (such as ^ to indicate exponents). Without such symbols, I could not tell if the 0 was an actual number and part of multiplication, of if 0 was an exponent of the expression preceding it.


P.S

Glad to help you have an AWESOME!!! day :)
</span>
6 0
3 years ago
Approximately 5% of calculators coming out of the production lines have a defect. Fifty calculators are randomly selected from t
baherus [9]

Answer:

0.2611 = 26.11% probability that exactly 2 calculators are defective.

Step-by-step explanation:

For each calculator, there are only two possible outcomes. Either it is defective, or it is not. The probability of a calculator being defective is independent of any other calculator, which means that the binomial probability distribution is used to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

5% of calculators coming out of the production lines have a defect.

This means that p = 0.05

Fifty calculators are randomly selected from the production line and tested for defects.

This means that n = 50

What is the probability that exactly 2 calculators are defective?

This is P(X = 2). So

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 2) = C_{50,2}.(0.05)^{2}.(0.95)^{48} = 0.2611

0.2611 = 26.11% probability that exactly 2 calculators are defective.

3 0
2 years ago
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