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2 years ago

5 1/4 divided by 1/3 =

Mathematics

2 answers:

shepuryov [24]2 years ago

5 0

I think it's 1 3/4. I had one similar earlier.

Andrej [43]2 years ago

3 0

I think it's 4 7/12 I use to do stuff like that all the time

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Help needed ASAP will give BRAINLIEST

grin007 [14]

Answer:

Step-by-step explanation:

Recall that the sum of the (3) interior angles of a triangle <em>must</em> equate to 180.

In other words:

$(55)+(20)+(x)=180$

Where x is our unknown angle. So, we need to solve for x.

On the left, add 55 and 20:

$75+x=180$

Subtract 75 from both sides. The left side cancels:

$x=105$

Therefore, the third angle is 105 degrees.

Our answer is C.

7 0

2 years ago

Write 19/41 in lowest forms

Verizon [17]

Answer:

19/41

Step-by-step explanation:

this fraction cannot be simplified since 19 is a prime number.

5 0

2 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

2 years ago

PLEASE HELP ASAP!! ONLY IF YOU KNOW THE ANSWER! THANK YOU SO MUCH!

telo118 [61]

Answer:

f(x)=-x

I hope this helps :)

8 0

2 years ago

Beginning at sunrise, the temperature rose 2.8 degrees per hour.

Dahasolnce [82]

19.18 / 2.8 = 6.85

So answer is D.6.850 h

6 0

2 years ago

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