Four runners are training for long races. Noah Ran 5.123 miles. Andre ran 6.34 miles. Jada ran 7.1 miles. Diego ran 8 miles.

NEED ANSWERS ASAP!! Help Pls I’d really appreciate it :)

alexandr1967 [171]

Question 23:

The area of the shaded portion of the circle = area of the arc - area of the triangle
Area of arc = $\frac{72}{360} * \pi r^2 = \frac{72}{360}* \pi * 5^2 = \frac{72*25}{360} \pi = 5 \pi$
Area of triangle = 0.5 * base * height = 0.5 (2.9 + 2.9) * 4 = 11.6
Area of the shaded portion = (5π - 11.6) ft²
So, the correct answer is option A
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Question 24:
As shown in the problem the two congruent shaded figures have the following properties:
1. Each figure has four sides.
2. For each figure, each two opposite sides are parallel.

So, The correct statements are D and E
D. The figures are parallelograms because their opposite sides are parallel.
E. The figures are quadrilaterals because each has four sides.
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Question 25:
The length of the line segment GH will be calculated as following:
Δ CEG is aright triangle at E
So the hypotenuse is CG = √(CE² + EG²)
∴ CG = √(9² + 12²) = √225 = 15 →→ (1)
For the circle C
CH = CE = radius of the circle = 9 →→(2)
From (1) and (2)
∴ GH = CG - CH = 15 - 9 = 6

So, the correct answer is option D. 6 units
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Question 26:
For the right cylinder
let the height of the cylinder = h
and let the radius of the base = r
so, the volume = π r² h

Given that: the height of the cylinder is 3 times the radius of the base∴ h = 3 r
Given the volume = 24π
∴ volume = π r² h = π r² * ( 3r ) = 3π r³
3π r³ = 24 π ⇒⇒⇒ divide both sides over 3π
∴ r³ = (24π)/(3π) = 8
∴ r = ∛8 = 2
∴ h = 3r = 3 * 2 = 6

So, the correct answer is option C. 6 units

7 0

3 years ago

Read 2 more answers

Please help Find P2 if possible.

stich3 [128]

Answer:

It's Impossible.

Matrix can only be raised to a power if it has same number of rows and columns

Step-by-step explanation:

$P= \begin{pmatrix}1&3&2\\ 3&2&1\end{pmatrix}\\\\P^2 = \begin{pmatrix}1&3&2\\ 3&2&1\end{pmatrix}^2\\\\\mathrm{Dimensions\:of\:matrix}\:\\\begin{pmatrix}1&3&2\\ 3&2&1\end{pmatrix}:\:2x3\\\mathrm{Matrix\:can\:only\:be\:raised\:to\:a\:power,\\\:if\:it\:has\:same\:number\:of\:rows\:and\:columns}$

7 0

2 years ago

Naval intelligence reports that 4 enemy vessels in a fleet of 17 are carrying nuclear weapons. If 9 vessels are randomly targete

icang [17]

Answer:

0.7588 = 75.88% probability that more than 1 vessel transporting nuclear weapons was destroyed

Step-by-step explanation:

The vessels are destroyed without replacement, which means that the hypergeometric distribution is used to solve this question.

Hypergeometric distribution:

The probability of x successes is given by the following formula:

$P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}$

In which:

x is the number of successes.

N is the size of the population.

n is the size of the sample.

k is the total number of desired outcomes.

Combinations formula:

$C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

In this question:

Fleet of 17 means that $N = 17$

4 are carrying nucleas weapons, which means that $k = 4$

9 are destroyed, which means that $n = 9$

What is the probability that more than 1 vessel transporting nuclear weapons was destroyed?

This is:

$P(X > 1) = 1 - P(X \leq 1)$

In which

$P(X \leq 1) = P(X = 0) + P(X = 1)$

So

$P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}$

$P(X = 0) = h(0,17,9,4) = \frac{C_{4,0}*C_{13,9}}{C_{17,9}} = 0.0294$

$P(X = 1) = h(1,17,9,4) = \frac{C_{4,1}*C_{13,8}}{C_{17,9}} = 0.2118$

Then

$P(X \leq 1) = P(X = 0) + P(X = 1) = 0.0294 + 0.2118 = 0.2412$

$P(X > 1) = 1 - P(X \leq 1) = 1 - 0.2412 = 0.7588$

0.7588 = 75.88% probability that more than 1 vessel transporting nuclear weapons was destroyed

8 0

3 years ago

Are the ratios of 2 apples to 5 oranges and 6 apples to 14

kramer

Answer:

No

Step-by-step explanation:

Multiplying the 2 and the 5 by 3 will get you 6 and 14. 14 is not a factor of 5 making it incorrect.

3 0

3 years ago

Read 2 more answers

Given: AB = 4 AC = 6 What point is in the interior of both circles? H. A. B.

AleksAgata [21]

Answer:

Point A is in the interior of both circles

Step-by-step explanation:

<u><em>Verify each point</em></u>

we know that

Point H is in the interior of greater circle but is outside smaller circle

Point A is the center both circles, then is in the interior of greater circle and is in the interior of smaller circle

Point B is on the smaller circle and is in the interior of greater circle

therefore

Point A is in the interior of both circles

5 0

4 years ago