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REY [17]
3 years ago
11

Is 481216 a rational or irrational

Mathematics
2 answers:
Nastasia [14]3 years ago
7 0

<u><em>4821216 is rational</em></u>. the definition of rational is a number that ends. this number ends in the ones place which means it is rational

DerKrebs [107]3 years ago
4 0

481216 is rational because rational numbers are numbers that are fractions. If you do 481216/1, you get a fraction.

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The sum of two consecutive integers is -95. Find the integers. State the unknov-
kvasek [131]

Let the unknown consecutive integers be x, x+1

x+x+1=-95

2x+1=-95

2x=-95-1

2x=-96

x=-96/2

x=-48

x=-48

x=-48x+1=-48+1=-47

-48 and -47 are the two consecutive integers

Please mark as Brainliest

8 0
2 years ago
The sum of the measures of the angles of a triangle is 180. The sum of the measures of the second and third angles is four times
prohojiy [21]

Answer:

{36, 66, 78}

Step-by-step explanation:

Let the measures of the three angles be f, s and t, for first, second and third.  Then s + t = 4f, t = 12 + s, and f + s + t = 180.

Subbing 12 + s for t in the first equation, we get s + 12 + s = 4f.

Subbing the same in the third equation, we get f + s + 12 + s = 180

This results in two equations in two unknowns (f and s):

2s - 4f = -12

2s  + f  = 168.

Let's eliminate s by subtracting the first of these two equations from the second.  Doing so yields 5f = 180.  Then the first angle, f, or x, is 36.

Then, from 2s + f = 168, we get 2s + 36 = 168, or 2s = 132.  Thus, the second angle is s = y = 66.

The sum of the three angles must be 180.  Thus, 36 + 66 + t = 180, or

102 + t = 180, or t = z = 78.

The three angles are

{36, 66, 78}.

3 0
3 years ago
a school baseball team raised $810 for new uniforms. each player on the team sold one book of tickets. There were 10 tickets in
artcher [175]
the answer is 27 because all you have to do is divide 810 by 30 which equals 10 times three dollars at the equals 30 soda by 830 and 30 parts and equals 27 soap and try to answer or problem only calculator at home on your phone to add on your tablet on your computer and a half thank you
6 0
3 years ago
Read 2 more answers
Can someone tell me if my answer to this problem is correct, please? And if it isn't how do you solve it? ​
Mrac [35]

Answer:

Height of tree= 13.8m

Step-by-step explanation:

When finding the value of a in your working, I suggest to leave the answer to 5 significant figures so that your final answer is more accurate. The value of a should be 26.048m to 5 s.f.

Then multiplying 26.048 with sin 32° would give you 13.8m too when rounded off to 3 significant figures :)

I have attached my working too.

3 0
3 years ago
Suppose that the length of a side of a cube X is uniformly distributed in the interval 9
Nastasia [14]

Answer:

f(v) = \left \{ {{\frac{1}{3}v^{-\frac{2}{3}}\ 9^3 \le v \le 10^3} \atop {0, elsewhere}} \right.

Step-by-step explanation:

Given

9 < x < 10 --- interval

Required

The probability density of the volume of the cube

The volume of a cube is:

v = x^3

For a uniform distribution, we have:

x \to U(a,b)

and

f(x) = \left \{ {{\frac{1}{b-a}\ a \le x \le b} \atop {0\ elsewhere}} \right.

9 < x < 10 implies that:

(a,b) = (9,10)

So, we have:

f(x) = \left \{ {{\frac{1}{10-9}\ 9 \le x \le 10} \atop {0\ elsewhere}} \right.

Solve

f(x) = \left \{ {{\frac{1}{1}\ 9 \le x \le 10} \atop {0\ elsewhere}} \right.

f(x) = \left \{ {{1\ 9 \le x \le 10} \atop {0\ elsewhere}} \right.

Recall that:

v = x^3

Make x the subject

x = v^\frac{1}{3}

So, the cumulative density is:

F(x) = P(x < v^\frac{1}{3})

f(x) = \left \{ {{1\ 9 \le x \le 10} \atop {0\ elsewhere}} \right. becomes

f(x) = \left \{ {{1\ 9 \le x \le v^\frac{1}{3} - 9} \atop {0\ elsewhere}} \right.

The CDF is:

F(x) = \int\limits^{v^\frac{1}{3}}_9 1\  dx

Integrate

F(x) = [v]\limits^{v^\frac{1}{3}}_9

Expand

F(x) = v^\frac{1}{3} - 9

The density function of the volume F(v) is:

F(v) = F'(x)

Differentiate F(x) to give:

F(x) = v^\frac{1}{3} - 9

F'(x) = \frac{1}{3}v^{\frac{1}{3}-1}

F'(x) = \frac{1}{3}v^{-\frac{2}{3}}

F(v) = \frac{1}{3}v^{-\frac{2}{3}}

So:

f(v) = \left \{ {{\frac{1}{3}v^{-\frac{2}{3}}\ 9^3 \le v \le 10^3} \atop {0, elsewhere}} \right.

8 0
3 years ago
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