Answer:
y=-2/3x-3
Explanation
Answer:
EY≅ 5.33 cm (nearest to tenth)
Step-by-step explanation:
Two triangles are said to be similar to each other if they have the same shape
(not necessarily the same size) and their corresponding sides are in
proportion
Given that,
AB = 12
DE = 8
BX = 8
ΔABC ∼ ΔDEF means that
- ∠A ∼= ∠D, ∠B ∼=∠E and ∠C ∼= ∠F.
- Further more, their corresponding sides are in proportion.
So,
12/8 = 8/EY
(by doing cross multiplication)
12EY = 8 × 8
12EY = 64
EY = 64 / 12
EY= 5.3333
EY≅ 5.33 cm (nearest to tenth)
Answer:
<h3>#1</h3>
The normal overlaps with the diameter, so it passes through the center.
<u>Let's find the center of the circle:</u>
- x² + y² + 2gx + 2fy + c = 0
- (x + g)² + (y + f)² = c + g² + f²
<u>The center is:</u>
<u>Since the line passes through (-g, -f) the equation of the line becomes:</u>
- p(-g) + p(-f) + r = 0
- r = p(g + f)
This is the required condition
<h3>#2</h3>
Rewrite equations and find centers and radius of both circles.
<u>Circle 1</u>
- x² + y² + 2ax + c² = 0
- (x + a)² + y² = a² - c²
- The center is (-a, 0) and radius is √(a² - c²)
<u>Circle 2</u>
- x² + y² + 2by + c² = 0
- x² + (y + b)² = b² - c²
- The center is (0, -b) and radius is √(b² - c²)
<u>The distance between two centers is same as sum of the radius of them:</u>
<u>Sum of radiuses:</u>
<u>Since they are same we have:</u>
- √(a² + b²) = √(a² - c²) + √(b² - c²)
<u>Square both sides:</u>
- a² + b² = a² - c² + b² - c² + 2√(a² - c²)(b² - c²)
- 2c² = 2√(a² - c²)(b² - c²)
<u>Square both sides:</u>
- c⁴ = (a² - c²)(b² - c²)
- c⁴ = a²b² - a²c² - b²c² + c⁴
- a²c² + b²c² = a²b²
<u>Divide both sides by a²b²c²:</u>
Proved
Answer:
The number of students we expect to have an interval that does not contain the true mean value is, ![255\times [\alpha\%]](https://tex.z-dn.net/?f=255%5Ctimes%20%5B%5Calpha%5C%25%5D)
.
Step-by-step explanation:
A [100(1 - α)%] confidence interval for true parameter implies that if 100 confidence intervals are created then [100(1 - α)] of these 100 confidence intervals will consist the true population parameter value.
Here α is the significance level. It is defined as the probability rejecting the claim that the true parameter value is not included in the 100(1 - α)% confidence interval.
It is provided that 255 students create the same confidence interval, correctly.
Then the number of students we expect to have an interval that does not contain the true mean value is,
For instance, if the students are creating a 95% confidence interval for mean then the number of students we expect to have an interval that does not contain the true mean will be:
The significance level is:
Number of students we expect to have an interval that does not contain the true mean will be: ![255\times [\alpha\%]=255\times 0.05=12.75\approx13](https://tex.z-dn.net/?f=255%5Ctimes%20%5B%5Calpha%5C%25%5D%3D255%5Ctimes%200.05%3D12.75%5Capprox13)
Thus, 13 of the 255 confidence intervals will not consist the true mean value.
