Amira is writing a coordinate proof to show that the area of a triangle created by joining the midpoints of an isosceles triangl
es is one-fourth the area of the isosceles triangle. She starts by assigning coordinates as given.
Triangle D E F in the coordinate plane so that vertex D is at the origin and is labeled 0 comma 0, vertex E is in the first quadrant and is labeled 2 a comma 2 b, and vertex F is on the positive side of the x-axis and is labeled 4a comma 0. Point Q is between points D and E. Point R is between points E and F. Point P is between points D and F and is labeled 2 a comma 0.
Enter your answers, in simplest form, in the boxes to complete the coordinate proof.
Point Q is the midpoint of DE¯¯¯¯¯ , so the coordinates of point Q are (a, b) .
Point R is the midpoint of FE¯¯¯¯¯ , so the coordinates of point R are (
, b).
In △DEF , the length of the base, DF¯¯¯¯¯ , is
, and the height is 2b, so its area is
.
In △QRP , the length of the base, QR¯¯¯¯¯ , is
, and the height is b, so its area is ab .
Comparing the expressions for the areas proves that the area of the triangle created by joining the midpoints of an isosceles triangle is one-fourth the area of the
Triangle D E F in the coordinate plane so that vertex D is at the origin and is labeled 0 comma 0, vertex E is in the first quadrant and is labeled 2 a comma 2 b, and vertex F is on the positive side of the x-axis and is labeled 4a comma 0. Point Q is between points D and E. Point R is between points E and F. Point P is between points D and F and is labeled 2 a comma 0.
Enter your answers, in simplest form, in the boxes to complete the coordinate proof.
Point Q is the midpoint of DE¯¯¯¯¯ , so the coordinates of point Q are (a, b) .
Point R is the midpoint of FE¯¯¯¯¯ , so the coordinates of point R are (
, b).
In △DEF , the length of the base, DF¯¯¯¯¯ , is
, and the height is 2b, so its area is
.
In △QRP , the length of the base, QR¯¯¯¯¯ , is
, and the height is b, so its area is ab .
Comparing the expressions for the areas proves that the area of the triangle created by joining the midpoints of an isosceles triangle is one-fourth the area of the
2 answers:
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Answer:
P(X ≤ 46 | X~B(821, 0.078)) = 0.00885745584
0.00885... < 0.01
The test statistic of 46 is significant
There is sufficient evidence to reject H₀ and accept H₁
Air bags are more effective as protection than safety belts
Step-by-step explanation:
821 crashes
46 hospitalisations where car has air bags
7.8% or 0.078 probability of hospitalisations in cars with automatic safety belts
α = 0.01 or 1% ← level of significance
One-tailed test
We are testing whether hospitalisations in cars with air bags are less likely than in a car with automatic safety belts;
The likelihood of hospitalisation in a car with automatic safety belts, we are told, is 7.8% or 0.078;
So we are testing if hospitalisations in cars with air bags is less than 0.078;
So, firstly:
Let X be the continuous random variable, the number of hospitalisations from a car crash with equipped air bags
X~B(821, 0.078)
Null hypothesis (H₀): p = 0.078
Alternative hypothesis (H₁): p < 0.078
According to the information, we reject H₀ if:
P(X ≤ 46 | X~B(821, 0.078)) < 0.01
To find P(X ≤ 46) or equally P(X < 47), it could be quite long-winded to do manually for this particular scenario;
If you are interested, the manual process involves using the formula for every value of x up to and including 46, i.e. x = 0, x = 1, x = 2, etc. until x = 46, the formula is:
You can find binomial distribution calculators online, where you input n (i.e. the number of trials or 821 in this case), probability (i.e. 0.078) and the test statistic (i.e. 46), it does it all for you, which gives:
P(X ≤ 46 | X~B(821, 0.078)) = 0.00885745584
Now, we need to consider if the condition for rejecting H₀ is met and recognise that:
0.00885... < 0.01
There is sufficient evidence to reject H₀ and accept H₁.
To explain what this means:
The test statistic of 46 is significant according to the 1% significance level, meaning the likelihood that only 46 hospitalisations are seen in car crashes with air bags in the car as compared to the expected number in car crashes with automatic safety belts is very unlikely, less than 1%, to be simply down to chance;
In other words, there is 99%+ probability that the lower number of hospitalisations in car crashes with air bags is due to some reason, such as air bags being more effective as a protective implement than the safety belts in car crashes.
Answer:
volume of cube=441cm³
l³=441cm³
l=
length=7.6cm