<span>Given:
Matrix = 60 elements</span>
<span>To solve this, we need to
take into account that each row must contain the same number of elements. So, we
need to find which of the options do not divide evenly into 60 (the options are
30, 60, 10, and 18).
So we check each of the choices to see if 1 of them divides evenly with 60.
60 / 60 = 1 (divides evenly)</span>
60 / 30 = 2 (divides
evenly)
<span>60 / 10 = 6 (divides
evenly)
</span>60 / 18 = 3.3.3333333333333333333333333333333 (does not divide evenly)
Therefore, 18 cannot equal the number of rows of the matrix.
You need to add the number for the supplement for us to find this out
We check first the numerical coefficients of both sides of the equation if they match if we perform the operation.
(6)(4) = 24
Then, the variables. For multiplication with the same variables, the exponents are added. In the given above,
n + 2 = 6
The value of n should be 4.
Let
x = pounds of peanuts
y = pounds of cashews
z = pounds of Brazil nuts.
The total pounds is 50, therefore
x + y + z = 50 (1)
The total cost is $6.60 per pound for 50 pounds of mixture.
The total is equal to the sum of the costs of the different nuts.
Because the cost for peanuts, cashews, and Brazil nuts are $3, $10, and $9 respectively, therefore
3x + 10y + 9z = 50*6.8
3x + 10y + 9z = 340 (2)
There are 10 fewer pounds of cashews than peanuts, therefore
x = y + 10 (3)
Substitute (3) into (1) and (2).
y + 10 + y + z = 50
2y + z = 40 (4)
3(y + 10) + 10y + 9z = 340
13y + 9z = 310 (5)
From (4),
z = 40 - 2y (6)
Substitute (6) into (5).
13y + 9(40 - 2y) = 310
-5y = -50
y = 10
z = 40 - 2y = 40 - 20 = 20
x = y + 10 = 20
Answer:
Peanuts: 20 pounds
Cashews: 10 pounds
Brazil nuts: 20 pounds
Answer:
One sample t-test for population mean would be the most appropriate method.
Step-by-step explanation:
Following is the data which botanist collected and can use:
- Sample mean
- Sample Standard Deviation
- Sample size (Which is 10)
- Distribution is normal
We have to find the best approach to construct the confidence interval for one-sample population mean. Two tests are used for constructing the confidence interval for one-sample population mean. These are:
- One-sample z test for population mean
- One-sample t test for population mean
One sample z test is used when the distribution is normal and the population standard deviation is known to us. One sample t test is used when the distribution is normal, population standard deviation is unknown and sample standard deviation is known.
Considering the data botanist collected, One-sample t test would be the most appropriate method as we have all the required data for this test. Using any other test will result in flawed intervals and hence flawed conclusions.
Therefore, One-sample t-test for population mean would be the most appropriate method.