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>
42 rhhrhjejrjfitkkrjrjrhrt
Answer:
Note: The full question is attached as picture below
a) Hо : p = 0.71
Ha : p ≠ 0.71
<em>p </em>= x / n
<em>p </em>= 91/110
<em>p </em>= 0.83.
1 - Pо = 1 - 0.71 = 0.29.
b) Test statistic = z
= <em>p </em>- Pо / [√Pо * (1 - Pо ) / n]
= 0.83 - 0.71 / [√(0.71 * 0.29) / 110]
= 0.12 / 0.043265
= 2.77360453
Test statistic = 2.77
c) P-value
P(z > 2.77) = 2 * [1 - P(z < 2.77)] = 2 * 0.0028
P-value = 0.0056
∝ = 0.01
P-value < ∝
Reject the null hypothesis. There is sufficient evidence to support the researchers claim at the 1% significance level.
Answer:
x = -1
Step-by-step explanation:
5 + 2x + 6 = x + 10
11 + 2x = x + 10
-x -x
11 + x = 10
-11 -11
x= -1
Answer:
a) 0.172
b) 0.167
c) 0.1404
Step-by-step explanation:
Margin of error, E = 
here,
p = probability of the event
n = sample size
a) n = 30
p = 10 ÷ 30 = 0.333
Therefore,
E = 
= 2 × 0.0861
= 0.172
b) n = 30
p = 21 ÷ 30 = 0.7
Therefore,
E = 
= 2 × 0.0836
= 0.167
c) n = 30
p = 22 ÷ 50 = 0.44
Therefore,
E = 
= 2 × 0.0702
= 0.1404
