Solve: 3x^3-x^2-38x-24

laiz [17]

$\bf \stackrel{map}{1~cm}\qquad :\qquad \stackrel{actual}{77~km}\qquad \cfrac{map}{actual}=\cfrac{1}{77}~\hfill \cfrac{\stackrel{map}{3}}{\underset{actual}{a}}=\cfrac{1}{77}\implies 231=a$

4 0

3 years ago

Can someone help me 19 to 32

andreev551 [17]

Answer:

simple just divide it by finding the reciprocal for the second number

ex: 2/4=4/2

then multiply

Step-by-step explanation:

8 0

3 years ago

I Need Help Please And This Is Due In 1 Hour

alex41 [277]

So the first one is 1 square meter of ceiling, and it says 1 square meter of cieling is 10.75. So the first one is just that, 10.75

The second one says 10 square meters, so just multiply 10.75, by 10.

In which case you get 107.5. So that's the answer.

Third one, it gives you how many ceiling tiles you have. So you divide the 100 tiles by 10.75, which gives you 9.30232558, which becomes 9.3

I'm sure you can figure out the last one.

7 0

3 years ago

Consider a uniform distribution from aequals4 to bequals29. (a) Find the probability that x lies between 7 and 27. (b) Find th

weeeeeb [17]

Answer:

a) 80% probability that x lies between 7 and 27.

b) 28% probability that x lies between 6 and 13.

c) 44% probability that x lies between 9 and 20.

d) 28% probability that x lies between 11 and 18.

Step-by-step explanation:

An uniform probability is a case of probability in which each outcome is equally as likely.

For this situation, we have a lower limit of the distribution that we call a and an upper limit that we call b.

The probability that we find a value x between c and d, in which d is larger than c, is given by the following formula.

$P(c \leq x \leq d) = \frac{d - c}{b - a}$

Uniform distribution from a = 4 to b = 29

(a) Find the probability that x lies between 7 and 27.

So $c = 7, d = 27$

$P(7 \leq x \leq 27) = \frac{27 - 7}{29 - 4} = 0.8$

80% probability that x lies between 7 and 27.

(b) Find the probability that x lies between 6 and 13. 

So $c = 6, d = 13$

$P(6 \leq x \leq 13) = \frac{13 - 6}{29 - 4} = 0.28$

28% probability that x lies between 6 and 13.

(c) Find the probability that x lies between 9 and 20.

So $c = 9, d = 20$

$P(9 \leq x \leq 20) = \frac{20 - 9}{29 - 4} = 0.44$

44% probability that x lies between 9 and 20.

(d) Find the probability that x lies between 11 and 18.

So $c = 11, d = 18$

$P(11 \leq x \leq 18) = \frac{18 - 11}{29 - 4} = 0.28$

28% probability that x lies between 11 and 18.

3 0

3 years ago

Find the unknown side length

Westkost [7]

For any right triangle, we can use the Pythagorean Theorem. The Pythagorean Theorem states that for any right triangle, the legs when squared and added together will be equal to the hypotenuse squared.

In mathematical notation: $a^2 + b^2 = c^2$

Where the variables a and b are the legs and the variable c is the hypotenuse.

Because we know the two side lengths of the triangle, we can solve for the unknown side.

We know the length of one of the legs and the hypotenuse.

Plug in the values.

$10^2 + b^2 = 24^2 \\ \\ 
100 + b^2 = 576 \\ \\ 
b^2 = 476 \\ \\ 
$

$b = \sqrt{476}$

So, the square root of 476 is the unknown length.

3 0

3 years ago