If 20 coins are tossed, find the probability that they'll all land on tails.

Answers

Answer 1

Answer: The probability would be split for this problem cause knowing with 20 coins , you can split that even amount ... here's what I would do..

Michelle has 20 coins , but only half of them land on heads and others on tails

As for a half , halves are always equaling to 1/2 or 50 (%) knowing we have 20 , you can evenly split that number of coins in a equal position

So ask your self .. what is half of 20? hmmm .. oh I got it , its 10! so knowing you have 20 but then later you split into half which is 10 ... meaning and knowing you have a 50% probability of getting the 20 coins hitting onto tails

hope this helps!

Answer with explanation:

Ques 1)

$x=(1)/(√(3)-√(2))$

Now we are asked to find the value of:

$√(x)-(1)/(√(x))$

We know that:

$(√(x)-(1)/(√(x)))^2=x+(1)/(x)-2$

Also:

$x=(1)/(√(3)-√(2))$ could be written as:

$x=(1)/(√(3)-√(2))* (√(3)+√(2))/(√(3)+√(2))\n\n\nx=(√(3)+√(2))/((√(3))^2-(√(2))^2)$

since, we know that:

$(a+b)(a-b)=a^2-b^2$

Hence,

$x=(√(3)+√(2))/(3-2)\n\n\nx=√(3)+√(2)$

Also,

$(1)/(x)=√(3)-√(2)$

Hence, we get:

$(√(x)-(1)/(√(x)))^2=√(3)+√(2)+√(3)-√(2)-2\n\n\n(√(x)-(1)/(√(x)))^2=2√(3)-2\n\n\n√(x)-(1)/(√(x))=\sqrt{2√(3)-2}$

Hence,

$√(x)-(1)/(√(x))=\sqrt{2√(3)-2}$

Ques 2)

$x=(√(a+2b)+√(a-2b))/(√(a+2b)-√(a-2b))$

on multiplying and dividing by conjugate of denominator we get:

$x=(√(a+2b)+√(a-2b))/(√(a+2b)-√(a-2b))* (√(a+2b)+√(a-2b))/(√(a+2b)+√(a-2b))\n\n\nx=((√(a+2b)+√(a-2b))^2)/((√(a+2b))^2-(√(a-2b))^2)\n\n\nx=((√(a+2b))^2+(√(a-2b))^2+2√(a+2b)√(a-2b))/(a+2b-a+2b)\n\n\nx=(a+2b+a-2b+2√(a+2b)√(a-2b))/(4b)\n\n\nx=(2a+2√(a^2-4b^2))/(4b)\n\n\nx^2=((2a+2√(a^2-4b^2))/(4b))^2\n\n\nx^2=((2a+2√(a^2-4b^2))^2)/(16b^2)$

Hence, we have:

$x^2=(4a^2+4(a^2-4b^2)+8a√(a^2-4b^2))/(16b^2)\n\n\nx^2=(4a^2+4a^2-16b^2+8a√(a^2-4b^2))/(16b^2)\n\n\n\nx^2=(8a^2-16b^2+8a√(a^2-4b^2))/(16b^2)\n\n\nbx^2=(8a^2-16b^2+8a√(a^2-4b^2))/(16b)\n\n\nbx^2=(8a(a+√(a^2-4b^2))-16b^2)/(16b)\n\n\nbx^2=(8a(a+√(a^2-4b^2)))/(16b)-(16b^2)/(16b)\n\n\nbx^2=(a(a+√(a^2-4b^2)))/(2b)-b\n\n\nbx^2=ax-b\n\n\ni.e.\n\n\nbx^2-ax+b=0$

1. x = 1/ ( $√(3) - √(2))$ = $√(3)+ √(2)$ ;
( $√(x) -1/ √(x) )^(2) = x + 1/x - 2 =$ =

A landscaper is creating a rectangular flower bed such that the width is half of the length.The area of the flower that is 34 ft.² Write an equation to determine the width of the flower bed to the nearest tenth of a foot

Answers

The Area is $A=l * w$ and we know that $w=(l)/(2)$

Therefore: $34=l * (l)/(2) = (l^(2))/(2)$
$68=l^(2)$
$√(68)=l=8.246$
and thus:
$w=(8.246)/(2)$
$w=4.123$
round to nearest tenth:
$w=4.1feet$

area=LW
W=L/2
2W=L
A=34
sub 34 for A and 2W for L
34=(2W)W
34=2W^2
divide 2
17=W^2
square root both sides
4.12=W
round tenth
4.1 ft=W

Which of the following is equivalent to 3a + 4b – (–6a – 3b) ? A. 16ab
B. –3a + b
C. –3a + 7b
D. 9a + b
E. 9a + 7b

Answers

$3a + 4b - (-6a - 3b) =3a + 4b +6a +3b = 9a +7b \n \n \n Answer : \ E. \ \ 9a + 7b$

What’s the answer for this

Answers

To find an area of a square you do Length x width
= 9km

Answer:

A=lw

A=3×3

A=9

Explanation

all sides of a square are the same

What is the surface area of a rectangular prism with a base length of 9 yd, a base width of 7 yd, and a height of 4 yd? A.504 yd2

B.127 yd2

C.254 yd2

D.252 yd2