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ExtremeBDS [4]
3 years ago
6

What are the solutions of this quadratic? Y=X^2 -2X

Mathematics
1 answer:
sergejj [24]3 years ago
3 0

Answer:

this is my answer but it's on you if you think this answer is correct or wrong

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Which is equivalent to 3/8x?
g100num [7]

Answer:

b.

Step-by-step explanation:

hope this helps, could i get brainliest?

5 0
3 years ago
What is the value of f(2) the function f(x)=4x^2-5x+1
snow_tiger [21]

Answer: f(2)=7

Step-by-step explanation:

F(2)= 4x2^2-5x2+1

4x4-10+1

6+1=7

7 0
2 years ago
A group of church members chartered a bus for $832. After making the arrangements, 6 more members decided to join the excursion
g100num [7]
Let x = # of original members
Cost per person (originally) = 832/x
Cost per person after 6 more people = 832/(x+6)
Solve 832/x=832/(x+6) +6  (because the cost was reduced by 6 
Solve for x.
x = 26 or -32 (but x is number of members, so it can't be negative)
# of original members is 26.
Original cost is 832/26 = $32 per person.
3 0
3 years ago
What is the mean of Geometric 9 and 36?
Snowcat [4.5K]

Answer:

Mean is 18

Step-by-step explanation:

We would set up the proportion as X/9=36/X so after we cross multiply we get x^2=324. Then we find the square root of both sides to simplify. ... And the square root of 324 is 18. So the final answer is x=18 or the geometric mean is 18

4 0
3 years ago
A small business owner estimates his mean daily profit as $970 with a standard deviation of $129. His shop is open 102 days a ye
Katena32 [7]

Answer:

The probability that the shopkeeper's annual profit will not exceed $100,000 is 0.2090.

Step-by-step explanation:

According to the Central Limit Theorem if we have a population with mean <em>μ</em> and standard deviation <em>σ</em> and we select appropriately huge random samples (<em>n</em> ≥ 30) from the population with replacement, then the distribution of the sum of values of <em>X</em>, i.e ∑<em>X</em>, will be approximately normally distributed.  

Then, the mean of the distribution of the sum of values of X is given by,  

 \mu_{x}=n\mu

And the standard deviation of the distribution of the sum of values of X is given by,  

 \sigma_{x}=\sqrt{n}\sigma

The information provided is:

<em>μ</em> = $970

<em>σ</em> = $129

<em>n</em> = 102

Since the sample size is quite large, i.e. <em>n</em> = 102 > 30, the Central Limit Theorem can be used to approximate the distribution of the shopkeeper's annual profit.

Then,

\sum X\sim N(\mu_{x}=98940,\ \sigma_{x}=1302.84)

Compute the probability that the shopkeeper's annual profit will not exceed $100,000 as follows:

P (\sum X \leq  100,000) =P(\frac{\sum X-\mu_{x}}{\sigma_{x}}

                              =P(Z

*Use a <em>z</em>-table for the probability.

Thus, the probability that the shopkeeper's annual profit will not exceed $100,000 is 0.2090.

6 0
3 years ago
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