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3 years ago

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Charlie wants to simulate a random selection from a large population of 40% males and 60% females. How could he use slips of pap

er for the simulation? A) Cut 5 equal-sized slips of paper, 2 for males and 3 for females. Then, randomly select slips of paper from the 5. B) Cut 2 equal-sized slips of paper, 1 for males and 1 for females. Then, randomly select slips of paper from the 2. C) Cut 3 equal-sized slips of paper, 2 for males and 1 for females. Then, randomly select slips of paper from the 3. D) Cut 50 equal-sized slips of paper, 25 for males and 25 for females. Then, randomly select slips of paper from the 50.

2 answers:

Harman [31]3 years ago

6 0

Answer: a

Step-by-step explanation:

Georgia [21]3 years ago

4 0

Answer:a

Step-by-step explanation:

Cause boys is 2/5 and girls are 3/5

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I don’t know I’m lost use a calculator

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Learning Task 3

ASHA 777 [7]

Answer:

1. The sides are: 38.5, 39.5, 40.5 and 41.5

2. Presently, Inzo is 15 years old and Miz is 10 years old

3. $Distance = 3682km$

4. Possible pairs are: (12, 14), (16, 18) and (20, 24)

5. $L < 38m$

Step-by-step explanation:

The question has lots of missing details. See the comment section for complete question

Solving (a):

Given

Shape: Quadrilateral

Sides: Consecutive

$Perimeter = 160$

Let one of the sides be x. The other sides will be x + 1, x + 2 and x + 3.

So, the perimeter is:

$Perimeter = x + x + 1 + x + 2 + x + 3$

This gives:

$160 = x + x + 1 + x + 2 + x + 3$

Collect Like Terms

$160 -1-2-3= x+x+x+x$

$154= 4x$

Divide both sides by 4

$38.5 = x$

$x =38.5$

<em>So, the sides are: 38.5, 39.5, 40.5 and 41.5</em>

<em></em>

Solving (b):

Given

Presently: $Inzo = 10 + Miz$

Five years ago: $Miz - 5= \frac{1}{3} * (Inzo-5)$ --- (Missing from the question)

Required: Determine their present ages

Substitute $10 + Miz$ for Inzo in the second equation

$Miz - 5 = \frac{1}{3} * (10 + Miz - 5)$

$Miz - 5 = \frac{1}{3} * (Miz + 5)$

Multiply both sides by 3

$3Miz - 15 = Miz + 5$

Collect Like Terms

$3Miz -Miz = 15 + 5$

$2Miz= 20$

Divide both sides by 2

$Miz = 10$

Substitute 10 for Miz in $Inzo = 10 + Miz$

$Inzo = 10 + 5$

$Inzo = 15$

<em>So presently, Inzo is 15 years old and Miz is 10 years old</em>

Solving (3):

Given

Express Train:

$Speed = 100kph$

$Time = t$

Local Train:

$Speed = 45kph$

$Time = t + 45$

Required: Determine the distance between A and B

Distance is calculated as:

$Distance = Speed * Time$

For the express train:

$Distance = 100 * t$

For the local train

$Distance = 45 * (45 + t)$

Distance between both stations are equal; So,

$100 * t = 45 * (45 + t)$

$100 t = 2025 + 45t$

Collect Like Terms

$100 t -45t= 2025$

$55t= 2025$

Divide by 55

$t = 36.82$

Substitute 36.82 for t in $Distance = 100 * t$

$Distance = 100 * 36.82$

$Distance = 3682km$

Solving (4):

Let the even numbers be: n and n + 2.

Sum greater than 24: $n + n + 2 > 24$

Sum less than 60: $n + n + 2 < 60$

So, we have:

$n + n + 2 > 24$

$2n + 2 > 24$

Collect Like Terms

$2n > 24-2$

$2n > 22$

$n > 11$

$n + n + 2 < 60$

$2n + 2< 60$

Collect Like Terms

$2n < 60-2$

$2n < 58$

Divide by 2

$n$

So, we have:

$n > 11$ and $n < 29$

So, possible pairs are: (12, 14), (16, 18) and (20, 24)

Solving (5):

Given

$Perimeter < 152m$

Required: Find the possible lengths

Let the length be L.

Perimeter is calculated as:

$Perimeter =4 * L$

$Perimeter =4 L$

So, we have:

$4L < 152m$

Divide through by 4

$L < 38m$

Hence, length is less than 38m (e.g. 37m)

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2 years ago

Let T be the plane-2x-2y+z =-13. Find the shortest distance d from the point Po=(-5,-5,-3) to T, and the point Q in T that is cl

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Answer:

d=10u

Q(5/3,5/3,-19/3)

Step-by-step explanation:

The shortest distance between the plane and Po is also the distance between Po and Q. To find that distance and the point Q you need the perpendicular line x to the plane that intersects Po, this line will have the direction of the normal of the plane $n=(-2,-2,1)$ , then r will have the next parametric equations:

$x=-5-2\lambda\\y=-5-2\lambda\\z=-3+\lambda$

To find Q, the intersection between r and the plane T, substitute the parametric equations of r in T

$-2x-2y+z =-13\\-2(-5-2\lambda)-2(-5-2\lambda)+(-3+\lambda) =-13\\10+4\lambda+10+4\lambda-3+\lambda=-13\\9\lambda+17=-13\\9\lambda=-13-17\\\lambda=-30/9=-10/3$

Substitute the value of $\lambda$ in the parametric equations:

$x=-5-2(-10/3)=-5+20/3=5/3\\y=-5-2(-10/3)=5/3\\z=-3+(-10/3)=-19/3\\$

Those values are the coordinates of Q

Q(5/3,5/3,-19/3)

The distance from Po to the plane

$d=\left| {\to} \atop {PoQ}} \right|=\sqrt{(\frac{5}{3}-(-5))^2+(\frac{5}{3}-(-5))^2+(\frac{-19}{3}-(-3))^2} \\d=\sqrt{(\frac{5}{3}+5))^2+(\frac{5}{3}+5)^2+(\frac{-19}{3}+3)^2} \\d=\sqrt{(\frac{20}{3})^2+(\frac{20}{3})^2+(\frac{-10}{3})^2}\\d=\sqrt{\frac{400}{9}+\frac{400}{9}+\frac{100}{9}}\\d=\sqrt{\frac{900}{9}}=\sqrt{100}\\d=10u$

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Step-by-step explanation:

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