Answer:
b) The null hypothesis should be rejected.
Explanation:
The null hypothesis is that the mean shear strength of spot welds is at least
3.1 MPa
H0: u ≥3.1 MPa against the claim Ha: u< 3.1 MPa
The alternate hypothesis is that the mean shear strength of spot welds is less than 3.1 MPa.
This is one tailed test
The critical region Z(0.05) < ± 1.645
The Sample mean= x`= 3.07
The number of welds= n= 15
Standard Deviation= s= 0.069
Applying z test
z= x`-u/s/√n
z= 3.07-3.1/0.069/√15
z= -0.03/0.0178
z= -1.68
As the calculated z= -1.68 falls in the critical region Z(0.05) < ± 1.645 the null hypothesis is rejected and the alternate hypothesis is accepted that the mean shear strength of spot welds is less than 3.1 MPa
Answer:
lead dioxide,sulfate and lead acid
Answer:
t = 30.1 sec
Explanation:
If the ant is moving at a constant speed, the velocity vector will have the same magnitude at any point, and can be decomposed in two vectors, along directions perpendicular each other.
If we choose these directions coincident with the long edge of the paper, and the other perpendicular to it, the components of the velocity vector, along these axes, can be calculated as the projections of this vector along these axes.
We are only interested in the component of the velocity across the paper, that can be calculated as follows:
vₓ = v* sin θ, where v is the magnitude of the velocity, and θ the angle that forms v with the long edge.
We know that v= 1.3 cm/s, and θ = 61º, so we can find vₓ as follows:
vₓ = 1.3 cm/s * sin 61º = 1.3 cm/s * 0.875 = 1.14 cm/s
Applying the definition of average velocity, we can solve for t:
t =
= 
⇒ t = 30.1 sec
The answer is b i did the same thing and i got it right
Answer:
See explanation
Explanation:
Solution:-
- Three students measure the volume of a liquid sample which is 6.321 L.
- Each student measured the liquid sample 4 times. The data is provided for each measurement taken by each student as follows:
Students
Trial A B C
1 6.35 6.31 6.38
2 6.32 6.31 6.32
3 6.33 6.32 6.36
4 6.36 6.35 6.36
- We will define the two terms stated in the question " precision " and "accuracy"
- Precision refers to how close the values are to the sample mean. The dense cluster of data is termed to be more precise. We will use the knowledge of statistics and determine the sample standard deviation for each student.
- The mean measurement taken by each student would be as follows:

- The precision can be quantize in terms of variance or standard deviation of data. Therefore, we will calculate the variance of each data:

- We will rank each student sample data in term sof precision by using the values of variance. The smallest spread or variance corresponds to highest precision. So we have:
Var ( A ) < Var ( B ) < Var ( C )
most precise Least precise
- Accuracy refers to how close the sample mean is to the actual data value. Where the actual volume of the liquid specimen was given to be 6.321 L. We will evaluate the percentage difference of sample values obtained by each student .

- Now we will rank the sample means values obtained by each student relative to the actual value of the volume of liquid specimen with the help of percentage difference calculated above. The least percentage difference corresponds to the highest accuracy as follows:
P ( B ) < P ( A ) < P ( C )
most accurate least accurate