Answer:
a) 0.00019923%
b) 47.28%
Step-by-step explanation:
a) To determine the likelihood that all sockets in the sample are defective, we can use the following approach:
The first socket is among a group that has 5 defective out of 38, leading to a probability of 5/38.
The second socket is then taken from a group of 4 defective out of 37, following the selection of the first defective socket, resulting in a probability of 4/37.
Extending this logic, the chance of having all 5 defective sockets is computed as: (5/38)*(4/37)*(3/36)*(2/35)*(1/34) = 0.0000019923 = 0.00019923%.
b) Using similar reasoning as in part a, the first socket has a probability of 33/38 of not being defective as it's chosen from a set where 33 sockets are functionally sound. The next socket has a proportion of 32/37, and this continues onward.
The overall probability calculates to (33/38)*(32/37)*(31/36)*(30/35)*(29/34) = 0.4728 = 47.28%.
Response:
Detailed explanation:
The final result is 3 /8/33.
step by step breakdown
Initially, we write:
x
=
3
.
¯¯¯¯
24
After that, we will multiply each side by
100
leading to:
100
x
=
324
.
¯¯¯¯
24
Subsequently, we will subtract the first equation from the second equation:
100
x
−
x
=
324
.
¯¯¯¯
24
−
3
.
¯¯¯¯
24
We can then solve for
x
in the following manner:
100
x
−
1
x
=
(
324
+
0
.
¯¯¯¯
24
)
−
(
3
+
0
.
¯¯¯¯
24
)
(
100
−
1
)
x
=
324
+
0
.
¯¯¯¯
24
−
3
−
0
.
¯¯¯¯
24
99
x
=
(
324
−
3
)
+
(
0
.
¯¯¯¯
24
−
0
.
¯¯¯¯
24
)
99
x
=
321
+
0
99
x
=
321
99
x
99
=
321
99
99
x
99
=
3
×
107
3
×
33
x
=
3
×
107
3
×
33
x
=
107
33
Next, we convert this improper fraction to a mixed numeral:
x
=
107
33
=
99
+
8
33
=
99
33
+
8
33
=
3
+
8
33
=
3
8
33
3
.
¯¯¯¯
=
3
8
33
Robert covered a distance of 292 miles.
He took 4 hours to complete this journey.
Next, let's convert miles to feet.
1 mile equals 5280 feet.
Thus, 292 miles in feet equals 292 * 5280 = 1541760 feet.
Now, converting 4 hours into seconds.
1 hour translates to 3600 seconds.
Therefore, 4 hours equals 3600 * 4 = 14400 seconds.
The question asks for speed in feet per second.
Thus,
is 107.067 feet per second.
The test statistic (Z) is 2.5767, and the p-value of the test is 0.009975. The null hypothesis suggests that the smoking rate among students has not changed, while the alternative indicates otherwise. The z-statistic for the sampled proportion is computed, yielding z ≈ 2.5767. As we investigate whether the smoking percentage has shifted over the preceding five years, the two-tailed p-value is found to be 0.009975. This result is significant at a 99% confidence level, demonstrating substantial evidence that the percentage of smoking students has changed.
