∠1 and ∠8 are alternate exterior angles while ∠3 and ∠6 are alternate interior angles.
step-by-step explanation:
When transversal l intersects lines m and n, the angles formed outside each line on the opposite side of the transversal constitute alternate angles (like ∠1 and ∠8). In contrast, alternate interior angles appear on opposite sides but are located between the two lines (like ∠3 and ∠6).
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Keywords: Transversal, angles
18.95(0.3 + 0.1) is the formula to use.
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
Hello
the dollar amount of each monthly payment that is interest
11,798.81÷120
=98.32
the percentage of the total payments that is total interest
(11,798.81÷41,798.81)×100
=28.23%
Hope this information is helpful.
The question is missing some information. It should be phrased as follows:
<span><span>A container has 50 electronic components, with 10 identified as defective. If 6 components are randomly selected from the container, what is the probability that at least 4 of them are not defective? Additionally, if 8 components are drawn at random from the container, what is the probability that exactly 3 are defective?
</span>Answers
<span>Part 1. 0.02
Part 2. </span></span>0.0375<span><span>
</span>Explanation
Probability denotes the likelihood of an event occurring. It is computed as:
probability = (Number of favorable outcomes)/(Number of total outcomes)
Part 1
When 6 components are chosen, if 4 are confirmed functioning, then 2 must be defective.
P(at least 4 functional) = 4/40</span>× 2/10
= 1/10 × 1/5
= 1/50
= 0.02
Part 2
Choosing 8 components, if 3 are defective, then 5 are functioning.
P(3 defective) = 3/40 × 5/10
= 15/400
= 3/80
= 0.0375
. Here, mi indicates the expected frequency based on the hypothesis, which is 36, n = 180, and xi corresponds to the actual observations. By substituting the known values, we find that x^2 = 9. The level of association is expressed as 
. This results in approximately 0.10, which surpasses our threshold.