Use the formula of an area of a tringle:
We have b = 11 yd, c = 24 yd and m∠A = 67°
Answer:
The first number is 20, the second one is 11
Step-by-step explanation:
Let the first number be x and the second number be y
Since it says "9 is the difference between two numbers", the first equation will be x-y=9
It also says that "their sum is 31". That means the second equation will be x+y=31
here are the two equations:
x-y=9
x+y=31
You can solve this any way you like, but I'll show elimination
add the two equations
x-y=9
x+y=31
2x=40
divide by 2
x=20
Substitute 20 as x in one of the equations
20+y=31
subtract 20 from both sides
y=11
That means the two numbers are 20 and 11
Hope this helps!
They sold 187 small bags and 75 large bags.
Two variables L and S (large & small)
We know that L + S = 262 -> the amount of bags sold
We also know that 3L + 1S = $412 -> total money made
L = 262 - S
Plug that into the second equation:
3(262 - S) + 1S = 412
786 - 3S + 1S = 412
786 - 2S = 412
786 - 412 = 2S
374 = 2S
S = 187
187 small bass and
So they sold 187 small bags
L = 262 - 187 = 75
They sold 75 large bags.
Answer:
The sampling distribution of the sample proportion of adults who have credit card debts of more than $2000 is approximately normally distributed with mean 
and standard deviation 
Step-by-step explanation:
Central Limit Theorem
The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean 
and standard deviation 
, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean and standard deviation 
.
For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.
For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean 
and standard deviation 
In this question:
Then
By the Central Limit Theorem:
The sampling distribution of the sample proportion of adults who have credit card debts of more than $2000 is approximately normally distributed with mean and standard deviation
