The possible digits are:
5, 6, 7, 8 and
9. Let's mark the case when the locker code begins with a prime number as
A and the case when <span>the locker code is an odd number as
B. We have
5 different digits in total,
2 of which are prime (
5 and
7).
First propability:
</span>
<span>
By knowing that digits don't repeat we can say that code is an odd number in case it ends with
5, 7 or
9 (three of five digits).
Second probability:
</span>
Answer:
what the h3ll type of question is this ?-
Step-by-step explanation:
your never gonna get one because you dont have any so the probability is 0%.
Answer:
a) It can be used because np and n(1-p) are both greater than 5.
Step-by-step explanation:
Binomial distribution and approximation to the normal:
The binomial distribution has two parameters:
n, which is the number of trials.
p, which is the probability of a success on a single trial.
If np and n(1-p) are both greater than 5, the normal approximation to the binomial can appropriately be used.
In this question:
So, lets verify the conditions:
np = 201*0.45 = 90.45 > 5
n(1-p) = 201*(1-0.45) = 201*0.55 = 110.55 > 5
Since both np and n(1-p) are greater than 5, the approximation can be used.
Answer:
x = -2
Step-by-step explanation:
Solve for x:
(2 (3 x - 4))/5 = -4
Multiply both sides of (2 (3 x - 4))/5 = -4 by 5/2:
(5×2 (3 x - 4))/(2×5) = -4×5/2
5/2×2/5 = (5×2)/(2×5):
(5×2)/(2×5) (3 x - 4) = -4×5/2
5/2 (-4) = (5 (-4))/2:
(5×2 (3 x - 4))/(2×5) = (-4×5)/2
(5×2 (3 x - 4))/(2×5) = (2×5)/(2×5)×(3 x - 4) = 3 x - 4:
3 x - 4 = (-4×5)/2
(-4)/2 = (2 (-2))/2 = -2:
3 x - 4 = 5×-2
5 (-2) = -10:
3 x - 4 = -10
Add 4 to both sides:
3 x + (4 - 4) = 4 - 10
4 - 4 = 0:
3 x = 4 - 10
4 - 10 = -6:
3 x = -6
Divide both sides of 3 x = -6 by 3:
(3 x)/3 = (-6)/3
3/3 = 1:
x = (-6)/3
The gcd of -6 and 3 is 3, so (-6)/3 = (3 (-2))/(3×1) = 3/3×-2 = -2:
Answer: x = -2
