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3 years ago

What would be the nearest ten thoustand for 168356

Mathematics

2 answers:

11111nata11111 [884]3 years ago

7 0

It would be 170000 well that is what i got

marysya [2.9K]3 years ago

4 0

$168,356\approx170,000$

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astraxan [27]

Answer:

5cm

Step-by-step explanation:

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3 years ago

Write an equation in slope-intercept form and then in standard form for each line described. y-intercept -2; x-intercept 5

Xelga [282]

Hello
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3 years ago

Determine whether the random variable is discrete or continuous. a. The number of statistics students now reading a book. b. The

andrey2020 [161]

Answer:

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Step-by-step explanation:

We have to identify whether the random variable is discrete or continuous.

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Discrete random variable since number of students cannot take decimal values.

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Discrete random variable since number of textbooks cannot be expressed in decimals but counted.

c. The exact time it takes to evaluate 27 plus 72.

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d. The number of free dash throw attempts before the first shot is made.

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3 years ago

Natalie just rented an apartment for $900 a month. She was told that the rent will increase 5.5% each year. About how much will

belka [17]

$1246.

This is a guess, though.

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3 years ago

Read 2 more answers

Help please solve <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:

Inequalities

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

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