Listing is for finite or "countable" sets -whatever that means to you.
49. list x-value(s) for which the equality holds
50. 0,1,2,3,4 since 5!=120
51. 0,1,2,3,4,5,6,7
52.a. P (10,3) order matters
b. P (6,3)
c. could be answer to a but I am not entirely sure.
Follow for solutions.
Answer:
P(≥ 7 males) = 0.0548
Step-by-step explanation:
This is a binomial probability distribution problem.
We are told that Before 1918;
P(male) = 40% = 0.4
P(female) = 60% = 0.6
n = 10
Thus;probability that 7 or more were male is;
P(≥ 7 males) = P(7) + P(8) + P(9) + P(10)
Now, binomial probability formula is;
P(x) = [n!/((n - x)! × x!)] × p^(x) × q^(n - x)
Now, p = 0.4 and q = 0.6.
Also, n = 10
Thus;
P(7) = [10!/((10 - 7)! × 7!)] × 0.4^(7) × 0.6^(10 - 7)
P(7) = 0.0425
P(8) = [10!/((10 - 8)! × 8!)] × 0.4^(8) × 0.6^(10 - 8)
P(8) = 0.0106
P(9) = [10!/((10 - 9)! × 9!)] × 0.4^(9) × 0.6^(10 - 9)
P(9) = 0.0016
P(10) = [10!/((10 - 10)! × 10!)] × 0.4^(10) × 0.6^(10 - 10)
P(10) = 0.0001
Thus;
P(≥ 7 males) = 0.0425 + 0.0106 + 0.0016 + 0.0001 = 0.0548
Answer:
3X/20 (option a) of the pastries submitted by Rashid and Mikhail were brushed with butter
Step-by-step explanation:
Rashid pastries (R)
Mikhail pastries (M)
Rashid and Mikhail submitted a total of x pastries
R+M=x (I)
Rashid made 2/3 as many pastries as Mikhail
(2/3)*R=M (II)
Using II in I
R+(2/3)*R=x
(5/3)*R = x
R=(3/5)*x (III)
Using III in I
(3/5)*x+M=x
M=x-(3/5)*x
M=(2/5)*x (IV)
Mikhail filo dough (MF)
Mikhail shortcrust dough (MS)
Rashid filo dough (RF)
Rashid shortcrust dough (RS)
Mikhail used filo dough for all of his pastries
MF=M
MS=0
Rashid used shortcrust dough for all of his pastries
RS=R
RF=0
Filo dough (FD)
FD=RF+MF=0+MF=MF=M (V)
5/8 of the filo dough pastries were brushed with olive oil
pastries brushed with olive oil (OI)
(5/8)*FD=OI
Using V
(5/8)*M=OI
Using IV
(5/8)*(2/5)*x=OI
(1/4)*x=OI (VI)
pastries brushed with butter (B)
Pastries made out of filo dough are brushed with either olive oil or butter (but not both)
FD=OI+B
B=FD-OI
Using V and VI
B= M - (1/4)*x
Using IV
B = (2/5)*x - (1/4)*x
B= (3/20)*x
3X/20 (option a) of the pastries submitted by Rashid and Mikhail were brushed with butter
Answer:
00:13 mm:ss
Step-by-step explanation:
There are 60 seconds in a minute. This fact can be used to convert the time period(s) to minutes and seconds either before or after you do the subtraction.
<h3>Difference</h3>
It is often convenient to do arithmetic with all of the numbers having the same units. Here, we are given two values in seconds and asked for their difference.
100 s - 87 s = (100 -87) s = 13 s
The difference between the two time periods is 0 minutes and 13 seconds.
<h3>Conversion</h3>
If you like, the numbers can be converted to minutes and seconds before the subtraction. Since there are 60 seconds in a minute, the number of minutes is found by dividing seconds by 60. The remainder is the number of seconds that will be added to the time in minutes:
87 seconds = ⌊87/60⌋ minutes + (87 mod 60) seconds
= 1 minute 27 seconds
100 seconds = ⌊100/60⌋ minutes + (100 mod 60) seconds
= 1 minute 40 seconds
Then the difference is found in the same way we would find a difference involving different variables. (A unit can be treated as though it were a variable.)
(1 min 40 s) -(1 min 27 s) = (1 -1 min) + (40-27 s) = 0 min 13 s
The difference between the two time periods is 0 minutes and 13 seconds.
3 √10
I think since a^2+b^2=c^2
