In triangle ΔABC, ∠C is a right angle and CD is the height to AB . Find the angles in ΔCBD and ΔCAD if: Chapter Reference a m∠A
= 20° Answer: m∠CDB = m∠CBD = m∠BCD = m∠CDA = m∠ CAD= m∠ACD =
1 answer:
Answer:
Given A triangle ABC in which
∠C =90°,∠A=20° and CD ⊥ AB.
In Δ ABC
⇒∠A + ∠B +∠C=180° [ Angle sum property of triangle]
⇒20° + ∠B + 90°=180°
⇒∠B+110° =180°
∠B =180° -110°
∠B = 70°
In Δ B DC
∠BDC =90°,∠B =70°,∠BC D=?
∠BDC +,∠B+∠BC D=180°[ angle sum property of triangle]
90° + 70°+∠BC D =180°
∠BC D=180°- 160°
∠BC D = 20°
In Δ AC D
∠A=20°, ∠ADC=90°,∠AC D=?
∠A + ∠ADC +∠AC D=180° [angle sum property of triangle]
20°+90°+∠AC D=180°
110° +∠AC D=180°
∠AC D=180°-110°
∠AC D=70°
So solution are, ∠AC D=70°,∠ BC D=20°,∠DB C=70°
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Answer:
We have something in the form log(x/y) where x = q^2*sqrt(m) and y = n^3. The log is base 2.
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Explanation:
It seems strange how the first two logs you wrote are base 2, but the third one is not. I'll assume that you meant to say it's also base 2. Because base 2 is fundamental to computing, logs of this nature are often referred to as binary logarithms.
I'm going to use these three log rules, which apply to any base.
- log(A) + log(B) = log(A*B)
- log(A) - log(B) = log(A/B)
- B*log(A) = log(A^B)
From there, we can then say the following:
Answer:
The null hypothesis:
The alternative hypothesis:
Test statistics : z = −2.30
P-value: = 0.02144
Decision Rule: Since the p-value is lesser than the level of significance; then we reject the null hypothesis.
Conclusion: We accept the alternative hypothesis and conclude that under the same circumstances the proportion of offspring peas will be yellow is not equal to 0.27
Step-by-step explanation:
From the given information:
Let's state the null and the alternative hypothesis;
Since The claim is that 27% of the offspring peas will be yellow.
The null hypothesis state that the proportion of offspring peas will be yellow is equal to 0.27.
i.e
The alternative hypothesis state that the proportion of offspring peas will be yellow is not equal to 0.27
<u>The test statistics:</u>
we are given 437 green peas and 129 yellow apples;
Hence;
where ;
= sample proportion
x = number of success
n = total number of the sample size
Now; the test statistics can be computed as :
z = −2.30
C. P-value
P-value = P(Z < z)
P-value = P(Z< -2.30)
By using the P-value method and the normal distribution as an approximation to the binomial distribution.
from the table of standard normal distribution
move left until the first column is reached. Note the value as –2.0
move upward until the top row is reached. Note the value as 0.30
find the probability value as 0.010724 by the intersection of the row and column values gives the area to the left of
z = -2.30
P- value = 2P(z ≤ -2.30)
P-value = 2 × 0.01072
P - value = 0.02144
Decision Rule: Since the p-value is lesser than the level of significance; then we reject the null hypothesis.
Conclusion: We accept the alternative hypothesis and conclude that under the same circumstances the proportion of offspring peas will be yellow is not equal to 0.27