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3 years ago

Plzz Answer first to answer correctly will get brainliest

Mathematics

2 answers:

sdas [7]3 years ago

6 0

Answer:

1/8

Step-by-step explanation:

1/4*1/2=1/8

Since you are multiplying, you flip the second fraction. So 2/1 becomes 1/2.

RoseWind [281]3 years ago

3 0

Answer: 1/8

Step-by-step explanation:

Multiply by the reciprocal which would be 1/2, so 1/4x1/2 would be 1/8

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Five kites were flying in the sky. Each kite was flying at a different level. The red kite was flying higher than the green one.

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3 years ago

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A group of freshmen at a local university consider joining the equestrian team. Thirty‑five percent of students choose Western r

lesantik [10]

Answer:

a) Option D) 0.75

b) Option D) 0.3

Step-by-step explanation:

We are given the following in the question:

Percentage of students who choose Western riding = 35%

$P(w) = 0.35$

Percentage of students who choose dressage= 45%

$P(d) = 0.45$

Percentage of students who choose jumping = 50%

$P(j) = 0.50$

Percentage of students who choose both dressage and jumping = 20%

$P(d \cap j) = 0.20$

Percentage of students who choose Western and dressage = 10%

$P(w \cap d) = 0.10$

Percentage of students who choose Western and jumping = 0%

$P(w \cap j) = 0$

Thus, we can say

$P(w \cap j \cap d) = 0$

Formula:

$P(A \cup B) = P(A) + P(B) - P(A \cap B)$

a) P(student chooses dressage or jumping)

$P(d \cup j) = P(d) + P(j) -p(d \cap j)\\= 0.45 + 0.50-0.20 = 0.75$

b) P(student chooses neither dressage nor Western riding)

$= 1 - P(d \cup w)\\= 1 - (P(d) + P(w) - P(d \cap w))\\= 1 - (0.45 + 0.35 - 0.10)= 0.3$

5 0

3 years ago

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3 years ago

Quadrilateral OPQR is inscribed in circle N, as shown below. What is the measure of ∠PQR? (5 points)

dimulka [17.4K]

Answer:

is there a picture?

Step-by-step explanation:

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4 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