In the figure, if m∠ABD = 120º, then m∠ADC =? º.
2 answers:
The figure that this question is referring to is attached. We must use the Law of Sines to solve this question, which is as follows:
a/sinA = b/sinB = c/sinC
This applies to any triangle. We are told that ∠ABD = 120º. We are asked to solve for ∠ADC. We know that ∠ADC + ∠ADB = 180º. If we assign ∠ADC = x, then ∠ADB = 180<span>º - x. We can now apply the law of sines to this data.
35/sin120 = 30/sin(180-x)
sin(180-x) = (30/35)(sin120)
sin(180-x) = 0.742
sin-1(sin(180-x)) = sin-1(0.742)
180 - x = 48</span><span>º
x = 132</span><span>º
</span>
We have already assigned x = ∠ADC; therefore, ∠ADC = 132<span>º.</span>
a/sinA = b/sinB = c/sinC
This applies to any triangle. We are told that ∠ABD = 120º. We are asked to solve for ∠ADC. We know that ∠ADC + ∠ADB = 180º. If we assign ∠ADC = x, then ∠ADB = 180<span>º - x. We can now apply the law of sines to this data.
35/sin120 = 30/sin(180-x)
sin(180-x) = (30/35)(sin120)
sin(180-x) = 0.742
sin-1(sin(180-x)) = sin-1(0.742)
180 - x = 48</span><span>º
x = 132</span><span>º
</span>
We have already assigned x = ∠ADC; therefore, ∠ADC = 132<span>º.</span>
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Answer:
(a) The probability that the thickness is less than 3.0 mm is 0.119.
(b) The probability that the thickness is more than 7.0 mm is 0.119.
(c) The probability that the thickness is between 3.0 mm and 7.0 mm is 0.762.
Step-by-step explanation:
We are given that thickness measurements of ancient prehistoric Native American pot shards discovered in a Hopi village are approximately normally distributed, with a mean of 5.0 millimeters (mm) and a standard deviation of 1.7 mm.
Let X = <u><em>thickness measurements of ancient prehistoric Native American pot shards discovered in a Hopi village.</em></u>
So, X ~ Normal()
The z-score probability distribution for the normal distribution is given by;
Z = ~ N(0,1)
where, = population mean thickness = 5.0 mm
= standard deviation = 1.7 mm
(a) The probability that the thickness is less than 3.0 mm is given by = P(X < 3.0 mm)
P(X < 3.0 mm) = P( <
) = P(Z < -1.18) = 1 - P(Z
1.18)
= 1 - 0.8810 = <u>0.119</u>
The above probability is calculated by looking at the value of x = 1.18 in the z table which has an area of 0.881.
(b) The probability that the thickness is more than 7.0 mm is given by = P(X > 7.0 mm)
P(X > 7.0 mm) = P( >
) = P(Z > 1.18) = 1 - P(Z
1.18)
= 1 - 0.8810 = <u>0.119</u>
The above probability is calculated by looking at the value of x = 1.18 in the z table which has an area of 0.881.
(c) The probability that the thickness is between 3.0 mm and 7.0 mm is given by = P(3.0 mm < X < 7.0 mm) = P(X < 7.0 mm) - P(X 3.0 mm)
P(X < 7.0 mm) = P( <
) = P(Z < 1.18) = 0.881
P(X 3.0 mm) = P(
) = P(Z
-1.18) = 1 - P(Z < 1.18)
= 1 - 0.8810 = 0.119
The above probability is calculated by looking at the value of x = 1.18 in the z table which has an area of 0.881.
Therefore, P(3.0 mm < X < 7.0 mm) = 0.881 - 0.119 = 0.762.