Answer:
b
Step-by-step explanation:
The median of a triangle is a line that from the vertex touches the middle lets eliminate options
a=ab starts at a vertex but doensn't touch a the middle of a line X
b=cd starts at a vertex and touches the middle of a line Ye
C= ce stars at a vertex but doesn't end at the middle of a line X
d=mb starts in the middle of the triangle and and ends a vertex X
<u>SO THE CORRECT ANSWER IS B</u>
<em>-hope this helps :)</em>
Remmeber you can do anything to an equaiton as long as you do it to btoh sides
5-6(a+2)=7+a
distribute
5-6a-12=7+a
add like terms
-6a-7=7+a
add 6a both sides
-7=7+7a
minus 7 both sides
-14=7a
divide by 7
-2=a
a=-2
1. 1/5 = x - 9 / x
2. Cross multiply x = 5 (x - 9)
3. Distribute x = 5x - 45
4. Move the variable and change the sign x - 5x = -45
5. Add like terms -4x = -45
6. Divide x = 45 / 4
So the solution is x = 45 / 4 or 11.25
Answer:
5.9× 10^4
Step-by-step explanation:
I'm assuming that by "104" you actually meant "10^4," which reads "fourth power of 10." Please be sure to use the symbol " ^ " to express exponentiation.
The sum of 2.5 × 10^4 and 3.4 × 10^4 is (2.5 + 3.4)×10^4
or
5.9× 10^4
Answer:

Step-by-step explanation:
Previous concepts
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".
Solution to the problem
Let X the random variable who represents the variable of interest. We know from the problem that the distribution for the random variable X is given by:
We select a sample of size n=64. That represent the sample size.
From the central limit theorem we know that the distribution for the sample mean
is given by:
The mean for the sample distirbution would be given by:

And the deviation given by:

And then the distribution for the sample mean is:
