Answer:
After 33 weeks.
Step-by-step explanation:
Let w be number of weeks.
We have been given that Gina opened a bank account with $40. She plans to add $20 per week to the account and not make any withdrawals. So the balance after w weeks will be 20w+40.
To figure out number of weeks Gina will have exactly $700 in her account, we will equate the balance after w weeks to 700.
Let us subtract 40 from both sides of equation.
Upon Dividing both sides of our equation by 20 we will get,
Therefore, after 33 weeks Gina will have exactly $700 in her account, excluding interest.
Answer:
The answer is 2427.5
Step-by-step explanation:
The fraction consists of two numbers and a fraction bar: 4,855/200
The number above the bar is called numerator: 4,855
The number below the bar is called denominator: 200
The fraction bar means that the two numbers are dividing themselves.
To get fraction's value divide the numerator by the denominator:
Value = 4,855 ÷ 200
To calculate the greatest common factor, GCF:
1. Build the prime factorizations of the numerator and denominator.
2. Multiply all the common prime factors, by the lowest exponents.
Factor both the numerator and denominator, break them down to prime factors:
Prime Factorization of a number: finding the prime numbers that multiply together to make that number.
4,855 = 5 × 971;
4,855 is a composite number;
In exponential notation:
200 = 2 × 2 × 2 × 5 × 5 = 23 × 52;
200 is a composite number;
I order to solve this you have to find out how how much root beer there is to the total amount of candy. 12/27. Then you find out what the percentage of root beer there is by dividing 12 by 27. It’ll give you a decimal point. Percentage has a maximum of 100%. And you’ll find out what percent based on this decimal. Factor it out of 1. The percentage you get is .44. By factoring out of 1 you can find out that the percentage is 44%. So the probability of finding a root beer out of all the candy is roughly 44%
We have a sample of 28 data points. The sample mean is 30.0 and the sample standard deviation is 2.40. The confidence level required is 98%. Then, we calculate α by:
The confidence interval for the population mean, given the sample mean μ and the sample standard deviation σ, can be calculated as:
![CI(\mu)=\lbrack x-Z_{1-\frac{\alpha}{2}}\cdot\frac{\sigma}{\sqrt[]{n}},x+Z_{1-\frac{\alpha}{2}}\cdot\frac{\sigma}{\sqrt[]{n}}\rbrack](https://tex.z-dn.net/?f=CI%28%5Cmu%29%3D%5Clbrack%20x-Z_%7B1-%5Cfrac%7B%5Calpha%7D%7B2%7D%7D%5Ccdot%5Cfrac%7B%5Csigma%7D%7B%5Csqrt%5B%5D%7Bn%7D%7D%2Cx%2BZ_%7B1-%5Cfrac%7B%5Calpha%7D%7B2%7D%7D%5Ccdot%5Cfrac%7B%5Csigma%7D%7B%5Csqrt%5B%5D%7Bn%7D%7D%5Crbrack)
Where n is the sample size, and Z is the z-score for 1 - α/2. Using the known values:
![CI(\mu)=\lbrack30.0-Z_{0.99}\cdot\frac{2.40}{\sqrt[]{28}},30.0+Z_{0.99}\cdot\frac{2.40}{\sqrt[]{28}}\rbrack](https://tex.z-dn.net/?f=CI%28%5Cmu%29%3D%5Clbrack30.0-Z_%7B0.99%7D%5Ccdot%5Cfrac%7B2.40%7D%7B%5Csqrt%5B%5D%7B28%7D%7D%2C30.0%2BZ_%7B0.99%7D%5Ccdot%5Cfrac%7B2.40%7D%7B%5Csqrt%5B%5D%7B28%7D%7D%5Crbrack)
Where (from tables):
Finally, the interval at 98% confidence level is:
