Answer:
Total distance mouse traveled in 3 hours = 
of a mile
The mouse traveled the same distance in each hour. So in order to find the distance covered in 1 hour we have to divide the distance covered in 3 hours by 3. This will give us the distance that the mouse traveled in one hour.
So, the distance traveled in one hour will be = 
of a mile
The error which Matt made was that he divided only the denominator of the expression by 3, this probably was a calculation error.
Correct conclusion will be: Mouse travel 1/24 of a mile each hour
Answer:
The store paid 6.67 times the profit made on the jeans
Step-by-step explanation:
Let the amount the clothing store pay for Jean be X
Let the amount the clothing store sells Jean be Y = X ×1.15
The profit (P) made is the difference between amount the clothing store sells Jean and the amount paid for Jean = Y - X = 1.15X - X
Profit (P) = 0.15X
X = P/0.15 = 6.67P
Therefore, the store paid 6.67 times the profit made on the jeans
Each child will get 1/4 of the sandwich.
(1/2)/2 = 1/4
Options
A. UV = 14 ft and m∠TUV = 45°
B. TU = 26 ft
C. m∠STU = 37° and m∠VTU = 37°
D. ST = 20 ft, UV = 14 ft, and m∠UST = 98°
E. m∠UST = 98° and m ∠TUV = 45°
Answer:
A. UV = 14 ft and m∠TUV = 45°
D. ST = 20 ft, UV = 14 ft, and m∠UST = 98°
Step-by-step explanation:
Given
See attachment for triangle
Required
What proves that: ΔSTU ≅ ΔVTU using SAS
To prove their similarity, we must check the corresponding sides and angles of both triangles
First:
must equal 
So:
Next:
UV must equal US.
So:
Also:
ST must equal VT
So:
Lastly
must equal 
So:
Hence: Options A and D are correct
(a) 0.059582148 probability of exactly 3 defective out of 20
(b) 0.98598125 probability that at least 5 need to be tested to find 2 defective.
(a) For exactly 3 defective computers, we need to find the calculate the probability of 3 defective computers with 17 good computers, and then multiply by the number of ways we could arrange those computers. So
0.05^3 * (1 - 0.05)^(20-3) * 20! / (3!(20-3)!)
= 0.05^3 * 0.95^17 * 20! / (3!17!)
= 0.05^3 * 0.95^17 * 20*19*18*17! / (3!17!)
= 0.05^3 * 0.95^17 * 20*19*18 / (1*2*3)
= 0.05^3 * 0.95^17 * 20*19*(2*3*3) / (2*3)
= 0.05^3 * 0.95^17 * 20*19*3
= 0.000125* 0.418120335 * 1140
= 0.059582148
(b) For this problem, let's recast the problem into "What's the probability of having only 0 or 1 defective computers out of 4?" After all, if at most 1 defective computers have been found, then a fifth computer would need to be tested in order to attempt to find another defective computer. So the probability of getting 0 defective computers out of 4 is (1-0.05)^4 = 0.95^4 = 0.81450625.
The probability of getting exactly 1 defective computer out of 4 is 0.05*(1-0.05)^3*4!/(1!(4-1)!)
= 0.05*0.95^3*24/(1!3!)
= 0.05*0.857375*24/6
= 0.171475
So the probability of getting only 0 or 1 defective computers out of the 1st 4 is 0.81450625 + 0.171475 = 0.98598125 which is also the probability that at least 5 computers need to be tested.
