The table shows the ages and weights of babies at a hospital.
2 answers:
Answer:
Option (b) is correct.
The correlation between age and weight shown in the table is positive.
Step-by-step explanation:
Given: The table shows the ages and weights of babies at a hospital.
We have to determine the correlation between age and weight as shown in the table.
We first Plot these points to obtain the graph as show in attachment below.
Since, Using correlation calculator , we find the correlation as 0.83113
Which is positive
Thus, The correlation between age and weight shown in the table is positive.
As age increases the weight increase.Hence, positive.
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<u>ANSWER: </u>
170 liters of 10% hydrochloric acid must be mixed with 1700 liters of 65% hydrochloric acid solution to obtain 1870 liters of 60% solution.
<u>SOLUTION: </u>
First, set up a table. Fill in the unknowns with variables x and y. The table is attached below
From this, we can easily set up the two equations.
Sum of values of two acids = Value of mixture
0.1x + 0.65y = 1122
For convenience, we'll multiply the entire equation by 100,
10x + 65y = 112200 ------ (1)
Now, Sum of amounts of each acid = Amount of mixture
x + y = 1870 ---------(2)
multiply (2) with 10 for easy calculation and derive the equation into one variable.
10x + 10y = 18700
Subtracting equation (2) from (1),
( + ) 10x + 65y = 112200
( − ) 10x + 10y = 18700
we get the solution as,
0 + 55y = 93500
Thus, 55y = 93500
y = 1700
Substituting y = 1700 in (2),
x + 1700 = 1870
x = 1870 - 1700
x = 170
So, we have x = 170 and y = 1700
We can conclude that 170 liters of 10% hydrochloric acid must be mixed with 1700 liters of 65% hydrochloric acid solution to obtain 1870 liters of 60% solution.
Answer:
Step-by-step explanation:
Hello!
The variable of interest is X: ounces per cup dispensed by the cola-dispensing machine.
The population mean is known to be μ= 8 ounces and its standard deviation σ= 1.0 ounce. Assuming the variable has a normal distribution.
A sample of 34 cups was taken:
a. You need to calculate the Z-values corresponding to the top 5% of the distribution and the lower 5% of it. This means you have to look for both Z-values that separates two tails of 5% each from the body of the distribution:
The lower value will be:
= -1.648
You reverse the standardization using the formula
X[bar]= 7.72ounces
The lower control point will be 7.72 ounces.
The upper value will be:
X[bar]= 8.28ounces
The upper control point will be 8.82 ounces.
b. Now μ= 7.6, considering the control limits of a.
P(7.72≤X[bar]≤8.28)= P(X[bar]≤8.28)- P(X[bar]≤7.72)
P(Z≤(8.28-7.6)/(1/√34))- P(Z≤7.72-7.6)/(1/√34))
P(Z≤7.11)- P(Z≤0.70)= 1 - 0.758= 0.242
There is a 0.242 probability of the sample means being between the control limits, this means that they will be outside the limits with a probability of 1 - 0.242= 0.758, meaning that the probability of the change of population mean being detected is 0.758.
b. For this item μ= 8.7, the control limits do not change:
P(7.72≤X[bar]≤8.28)= P(X[bar]≤8.28)- P(X[bar]≤7.72)
P(Z≤(8.28-8.7)/(1/√34))- P(Z≤7.72-8.7)/(1/√34))
P(Z≤-2.45)- P(Z≤-5.71)=0.007 - 0= 0.007
There is a 0.007 probability of not detecting the mean change, which means that you can detect it with a probability of 0.993.
I hope it helps!