Answer:
The amount invested at 4% is $20,500 and the amount invested at 10% is $500
Step-by-step explanation:
Let
x -----> the amount invested at 4%
(21,000 -x) ----> the amount invested at 10%
we know that
The interest of the amount invested at 4% plus the interest of the amount invested at 10% must be equal to $870
![4\%=4/100=0.04](https://tex.z-dn.net/?f=4%5C%25%3D4%2F100%3D0.04)
![10\%=10/100=0.10](https://tex.z-dn.net/?f=10%5C%25%3D10%2F100%3D0.10)
so
![0.04(x)+0.10(21,000-x)=870](https://tex.z-dn.net/?f=0.04%28x%29%2B0.10%2821%2C000-x%29%3D870)
Solve for x
![0.04x+2,100-0.10x=870](https://tex.z-dn.net/?f=0.04x%2B2%2C100-0.10x%3D870)
![0.10x-0.04x=2,100-870](https://tex.z-dn.net/?f=0.10x-0.04x%3D2%2C100-870)
![0.06x=1,230](https://tex.z-dn.net/?f=0.06x%3D1%2C230)
![x=\$20,500](https://tex.z-dn.net/?f=x%3D%5C%2420%2C500)
![21,000-x=\$500](https://tex.z-dn.net/?f=21%2C000-x%3D%5C%24500)
therefore
The amount invested at 4% is $20,500 and the amount invested at 10% is $500
Step-by-step explanation:
cos x = 59/95
x = 0.9007
round this and you have your answer
(Daddy Mamma Sissy Bubba Rover=Divide Multiply Subtract Borrow Repeat/remainder)
Answer:
$15600
Step-by-step explanation:
Minimum number of samples can be calculated using the formula:
N≥
where
- N is the sample size
- z is the corresponding z-score for 95% confidence level (1.96)
- s is the standard deviation of the calorie consumption (450 cal.)
- ME is the margin of error that the researcher is willing to accept. (50 cal)
N≥
=311.2
Therefore, minimum required sample size for 50 cal standard error in 95% confidence is 312
Since the survey costs $50 for each person then the survey costs minimum
50*312=$15600
Answer:
2x-1 is an equivalent answer.