Answer:
f = 2
g = 8
h = -9
k = 40
m = 1
Step-by-step explanation:
Equation 1:
23f - 17 = 29
Add 17 to both sides. This undoes the -17.
23f = 29 + 17
Add 17 to 29 to get 46.
23f = 46
Divide both sides by 23. This undoes the multiplication by 23.
f = 46/23
Divide 46 by 23 to get 2.
f = 2
Equation 2:
2(3g + 4) = 56
Divide both sides by 2. This undoes the multiplication by 2.
3g + 4 = 56/2
Divide 56 by 2 to get 28.
3g + 4 = 28
Subtract 4 from both sides. This undoes the +4.
3g = 28 - 4
Subtract 4 from 28 to get 24.
3g = 24
Divide both sides by 3. This undoes the multiplication by 3.
g = 24/3
Divide 24 by 3 to get 8.
g = 8
Equation 3:
h + 9 = 0
Subtract 9 from both sides. This undoes the +9.
h = 0 - 9
Any number subtracted from 0 gives its negation.
h = -9
Equation 4
3(k - 8) = 96
Divide both sides by 3. This undoes the multiplication by 3.
k - 8 = 96/3
Divide 96 by 3 to get 32.
k - 8 = 32
Add 8 to both sides. This undoes the -8.
k = 32 + 8
Add 8 to 32 to get 40.
k = 40
Equation 5:
5m - 5 = 0
Add 5 to both sides. This undoes the -5
5m = 0 + 5
Anything plus 0 gives itself.
5m = 5
Divide both sides by 5. This undoes the multiplication by 5
m = 5/5
Anything divided by itself gives you 1.
m = 1
Answer:
19
Step-by-step explanation:
PEMDAS
no parenthesis
no exponents
4-6+2 x 7
4-6+ 17
no division
4-6=2
2+17
19 is the answer
Answer:
see the attachment
Step-by-step explanation:
We assume that the question is interested in the probability that a randomly chosen class is a Friday class with a lab experiment (2/15). That is somewhat different from the probability that a lab experiment is conducted on a Friday (2/3).
Based on our assumption, we want to create a simulation that includes a 1/5 chance of the day being a Friday, along with a 2/3 chance that the class has a lab experiment on whatever day it is.
That simulation can consist of choosing 1 of 5 differently-colored marbles, and rolling a 6-sided die with 2/3 of the numbers being designated as representing a lab-experiment day. (The marble must be replaced and the marbles stirred for the next trial.) For our purpose, we can designate the yellow marble as "Friday", and numbers greater than 2 as "lab-experiment".
The simulation of 70 different choices of a random class is shown in the attachment.
_____
<em>Comment on the question</em>
IMO, the use of <em>70 trials</em> is coincidentally the same number as the first <em>70 days</em> of school. The calendar is deterministic, so there will be exactly 14 Fridays in that period. If, in 70 draws, you get 16 yellow marbles, you cannot say, "the probability of a Friday is 16/70." You need to be very careful to properly state the question you're trying to answer.
Answer:
a) 1/800 or 0.00125
b) i) 0.0013
ii) 0.001
c) 60%
Step-by-step explanation:
T = [tan(2×30)+1][2cos(30)-1] ÷ (y²-x²)
T = (tan60 + 1)(2cos30 - 1) ÷ (41² - 9²)
T = (sqrt(3) + 1)(sqrt(3) - 1) ÷ 1600
T = (3-1)/1600
T = 2/1600
T = 1/800
T = 0.00125
Error: 0.002 - 0.00125
0.00075
%error
0.00075/0.00125 × 100
60%
