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snow_lady [41]
3 years ago
8

What are three ways to say x>5 in words

Mathematics
1 answer:
SSSSS [86.1K]3 years ago
3 0
1.) x is greater than 5

2.) an unknown number is greater than five

3.) x is greater than five
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A colony of insects doubles every day. If the colony has 5 insects todayHow many will be present in 20 days?
Alex787 [66]

Answer: b is the correct answer.

Step-by-step explanation:

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3 years ago
The length of a text messaging conversation is normally distributed with a mean of 2 minutes and a standard deviation of 0.5 min
zhenek [66]

Answer:

Let x be a random variable representing the length of a text messaging conversation, then

P(x > 3) = 1 - P(x < 3) = 1 - P(z < (3 - 2)/0.5) = 1 - P(z < 2) = 1 - 0.97725 = 0.02275

5 0
3 years ago
Help please solve<br> <img src="https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cfrac%7B6x%5E5%2B11x%5E4-11x-6%7D%7B%282x%5E2-3x%2B1
Shkiper50 [21]

Answer:

\displaystyle  -\frac{1}{2} \leq x < 1

Step-by-step explanation:

<u>Inequalities</u>

They relate one or more variables with comparison operators other than the equality.

We must find the set of values for x that make the expression stand

\displaystyle \frac{6x^5+11x^4-11x-6}{(2x^2-3x+1)^2} \leq 0

The roots of numerator can be found by trial and error. The only real roots are x=1 and x=-1/2.

The roots of the denominator are easy to find since it's a second-degree polynomial: x=1, x=1/2. Hence, the given expression can be factored as

\displaystyle \frac{(x-1)(x+\frac{1}{2})(6x^3+14x^2+10x+12)}{(x-1)^2(x-\frac{1}{2})^2} \leq 0

Simplifying by x-1 and taking x=1 out of the possible solutions:

\displaystyle \frac{(x+\frac{1}{2})(6x^3+14x^2+10x+12)}{(x-1)(x-\frac{1}{2})^2} \leq 0

We need to find the values of x that make the expression less or equal to 0, i.e. negative or zero. The expressions

(6x^3+14x^2+10x+12)

is always positive and doesn't affect the result. It can be neglected. The expression

(x-\frac{1}{2})^2

can be 0 or positive. We exclude the value x=1/2 from the solution and neglect the expression as being always positive. This leads to analyze the remaining expression

\displaystyle \frac{(x+\frac{1}{2})}{(x-1)} \leq 0

For the expression to be negative, both signs must be opposite, that is

(x+\frac{1}{2})\geq 0, (x-1)

Or

(x+\frac{1}{2})\leq 0, (x-1)>0

Note we have excluded x=1 from the solution.

The first inequality gives us the solution

\displaystyle  -\frac{1}{2} \leq x < 1

The second inequality gives no solution because it's impossible to comply with both conditions.

Thus, the solution for the given inequality is

\boxed{\displaystyle  -\frac{1}{2} \leq x < 1 }

7 0
3 years ago
Solve by graphingggggggg
olga_2 [115]

Answer:

Y can't i see the pic

Step-by-step explanation:

6 0
2 years ago
Read 2 more answers
What is 5/8 as a decimal
lara31 [8.8K]
The fraction 5/8 is also the equation of 5 divided by 8, so we just do 5 divided by 8, which is 
0.625 as a decimal.
8 0
3 years ago
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