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
Note:xy means x times y and x(y) means the same thing
so
first we get rid of square root then
make the equation equal to zero becaues if
xy=0 then x or/and y=0
squareroot(y-1)+3=y
isolate the squareroot
subtrac 3 from boht sides
squareroot(y-1)=y-3
square both sides (since they are equal, you should be able to square both sides and still make it true)
(squareroot(y-1))^2=(y-3)^2
(y-1)=(y-3)(y-3)
y-1=y^2-6y+9
subtrac y from both sides
-1=y^2-7y+9
add 1 to both sides
0=y^2-7y+10
find what two number multiply to make 10 and add to get -7
the answer is -2 and -5
0=(y-5)(y-2)
therfore
y-5=0
and/or
y-2=0
therefor
y=5 or/and 2 might work
let's try out 2
square root(2-1)+3=2
square root(1)+3=2
1+3=2
false
so 2 doesn't work
let's try 5
squareroot(5-1)+3=5
squareroot(4)+3=5
2+3=5
5=5
true
y=5
.3455 is the correct answer.
Answer:s=10.50 and p=27
Step-by-step explanation:
I think the answer's C: $7000/$8000
That's a surprisingly easy question for high school