Half the cards are even, 70/2 = 35 even cards.
The probability of picking an even card = 35/70, which reduces to 1/2
Answer:
The 95% confidence interval estimate for the mean highway mileage for SUVs is (18.29mpg, 20.91mpg).
Step-by-step explanation:
Our sample size is 96.
The first step to solve this problem is finding our degrees of freedom, that is, the sample size subtracted by 1. So
Then, we need to subtract one by the confidence level and divide by 2. So:
Now, we need our answers from both steps above to find a value T in the t-distribution table. So, with 95 and 0.025 in the t-distribution table, we have .
Now, we find the standard deviation of the sample. This is the division of the standard deviation by the square root of the sample size. So
Now, we multiply T and s
For the lower end of the interval, we subtract the mean by M. So
For the upper end of the interval, we add the mean to M. So
The 95% confidence interval estimate for the mean highway mileage for SUVs is (18.29mpg, 20.91mpg).
So she has 400 and 500 so far.
now if she were to have an average of 2300 for the next 5 years, that means some years she made more, some she made less, but in the 5 years run, she made 2300 * 5 or 11500 total.
Neverminding the lows and highs, if we bundle up the 11500 bucks and divide by 5, that'd 2300, which is what's called the average, the average amount doesn't take into account that some numbers are higher than others, it kinda just flattens them out.
so in total she would have made 11500 in each account.
so in the account with 400, she needs 11100 more, how much will it be 1100 for the next three years on average? 11100/3 or 3700.
so if she earns 3700 each year, for the next 3 years in the account with 400 already, she'd have 11500 bucks in 5 years with an average of 2300.
now let's look at the 500 account, she needs 11000 more, how much will it be for 11000 for the next three years on average? 11000/3 or about 3666.67.
so if she earns 3666.67 each year, for the next 3 years, she'd have 11500 bucks in 5 yeas with an average of 2300.