Answer:
<u>OAmalOHopeO</u>
can someone please help me out with this problem i been trying to figure out what the answer is .. ? or at least give me some ti
ps so i can remember for the other problems ??
1 answer:
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Answer:
Given: BD is an altitude of △ABC .
Prove: sinA/a=sinC/c
Triangle ABC with an altitude BD where D is on side AC. Side AC is also labeled as small b. Side AB is also labeled as small c. Side BC is also labeled as small a. Altitude BD is labeled as small h.
Statement Reason
BD is an altitude of △ABC .
Given △ABD and △CBD are right triangles. (Definition of right triangle)
sinA=h/c and sinC=h/a
Cross multiplying, we have
csinA=h and asinC=h
(If a=b and a=c, then b=c)
csinA=asinC
csinA/ac=asinC/ac (Division Property of Equality)
sinA/a=sinC/c
This rule is known as the Sine Rule.
Answer:
Step-by-step explanation:
Hello!
For me, the first step to any statistics exercise is to determine what is the variable of interest and it's distribution.
In this example the variable is:
X: height of a college student. (cm)
There is no information about the variable distribution. To estimate the population mean you need a variable with at least a normal distribution since the mean is a parameter of it.
The option you have is to apply the Central Limit Theorem.
The central limit theorem states that if you have a population with probability function f(X;μ,δ²) from which a random sample of size n is selected. Then the distribution of the sample mean tends to the normal distribution with mean μ and variance δ²/n when the sample size tends to infinity.
As a rule, a sample of size greater than or equal to 30 is considered sufficient to apply the theorem and use the approximation.
The sample size in this exercise is n=50 so we can apply the theorem and approximate the distribution of the sample mean to normal:
X[bar]~~N(μ;σ2/n)
Thanks to this approximation you can use an approximation of the standard normal to calculate the confidence interval:
98% CI
1 - α: 0.98
⇒α: 0.02
α/2: 0.01
X[bar] ±
174.5 ±
[172.22; 176.78]
With a confidence level of 98%, you'd expect that the true average height of college students will be contained in the interval [172.22; 176.78].
I hope it helps!