Answer:
k
Step-by-step explanation
k can be 11. If u plug 11: 3-21-12 is bigger than 9
Answer:
80 oz
Step-by-step explanation:
5 pints is 10 cups. 8 oz in a cup. Multiply 10(8) and you get 80 oz
Problem 3: Let x = price of bag of pretzels Let y = price of box of granola bars
We have Lesley's purchase: 4x+2y=13.50
And Landon's: 1x+5y=17.55
We can use the elimination method. Let's negate Landon's purchase by multiplying by -1. -1x-5y=-17.55
We add this four times to Lesley's purchase to eliminate the x variable.
2y-20y=13.50-70.2
-18y=-56.7
y = $3.15 = Price of box of granola bars
Plug back into Landon's purchase to solve for pretzels.
x+5*3.15=17.55
x+15.75=17.55
x = $1.80 = price of bag of pretzels
Problem 4.
Let w = number of wood bats sold
Let m = number of metal bats sold
From sales information we have: w + m = 23
24w+30m=606
Substitution works well here. Solve for w in the first equation, w = 23 - m, and plug this into the second.
24*(23-m)+30m=606
552-24m+30m=606
6m=54
m=9 = number of metal bats sold
Therefore since w = 23-m, w = 23-9 = 14. 14 wooden bats were sold.
Answer:
she is paying back 9112.5 R.O
Interest paid back is 2,403.95
Step-by-step explanation:
To find the amount, we use the compound interest formula.
This is given as;
A = I( 1 + r/n)^nt
where A is the amount we are trying to calculate
I is money borrowed = 6709
r is the interest rate = 12 1/3% = 37/3 = 12.33% which is same as 12.33/100 = 0.1233
n is the number of times interest is compounded. We have 15 2 months in 2 and a half years
t is the number of years = 2.5
Plugging these values, we have;
A = 6709(1 + 0.1233/15)^(15)(2.5)
A = 6709(1.0082)^(37.5)
A = 9112.95 R.O
Interest is amount - principal( money borrowed)
9112.95 - 6709 = 2403.95
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).
