To solve this problem, it is easiest to set up a system of equations. Let's let the variable s represent the cost of a shirt and the variable j represent the cost of jackets. According to the given information, we can set up the following equations (because cost multiplied by quantity yields price):
3s + 4j = 360
1s + 3j = 220
Next, we can manipulate the second equation so that it equals s in terms of j. We do this by subtracting 3j from both sides of the equation, as shown below:
s = 220 - 3j
After that, we should substitute in this value for the variable s in the first equation.
3(220-3j) + 4j = 360
Next, we should use the distributive property to simplify the left side of the equation.
660 - 9j + 4j = 360
Then, we should simplify the left side of the equation by combining like terms.
660 - 5j = 360
After, we can subtract 660 from both sides of the equation to get the variable term alone.
-5j = -300
Finally, we should divide both sides of the equation by -5 in order to get the variable j alone.
j = 60
Now that we know the value of the variable j, we should substitute this value into one of the original equations and solve using division and subtraction to isolate the variable.
3s + 4j = 360
3s + 4(60) = 360
3s + 240 = 360
3s = 120
s = 40
Therefore, the cost of one shirt is $40.
Hope this helps!
The answer will be 31 hope it right
Answer: The average person doesn't live 1 million hours. 1 million hour is 114 year. That is much more than 78 years. 1 million ours is too much. You would need 683280 to make it 78 years.
Answer:
Im not quite sure but probably 30-40% at the best?
Step-by-step explanation:
Answer:
The 80% confidence interval for the population proportion of disks which are defective is (0.059, 0.079).
Step-by-step explanation:
In a sample with a number n of people surveyed with a probability of a success of , and a confidence level of , we have the following confidence interval of proportions.
In which
z is the zscore that has a pvalue of .
Suppose a sample of 1067 floppy disks is drawn. Of these disks, 74 were defective.
This means that
80% confidence level
So , z is the value of Z that has a pvalue of , so .
The lower limit of this interval is:
The upper limit of this interval is:
The 80% confidence interval for the population proportion of disks which are defective is (0.059, 0.079).