Answer:
There is no enough evidence that the viscosity is not 3000. The viscosity is not significantly different from 3000.
Step-by-step explanation:
We have to perform an hypothesis test on the mean.
The null and alternative hypothesis are:
The significance level is 0.05.
The mean of the sample is:
The standard deviation of the sample is:
The statistic t can be calculated as:
The degrees of freedom are
The P-value for t=-1.338 and df=4 is P=0.2519. The P-value is greater than the significance level, so it failed to reject the null hypothesis.
There is no enough evidence that the viscosity is not 3000.
Here the unknown number is n . Product of 6 and n is 6n. And 6n is increased by 13 that is 6n +13. And the result is -14 , that is
6n+13=-14
We need to get rid of 13, so we subtract 13 to both sides, that is
6n = -14-13
6n =-27
Now we need to get rid of 6,and for that we divide both sides by 6, that is
n = -27/6
And that's the required value of n .
W(h) = h/3 + 2.5.....or it could also be : w(h) = 1/3h + 2.5
Answer:
15.87% of blue whales eat more than 5,850 pounds of fish
Step-by-step explanation:
Mean = 5000
Standard Deviation = 850
We need to find P(x>5850) = ?
We can find using z-score
z = x - mean/ standard deviation
z = 5850 - 5000/850
z = 850/850
z = 1
Now, P(x>5850) = P(z>1)
Finding value of P(z>1) by looking at the z-score table
P(z>1) = 0.8413
The normal distribution gives area to the left , so subtracting it from 1 to gain the answer
1-0.8413 = 0.1587
Now, to find percentage multiply it with 100
0.1587 * 100 = 15.87 %
15.87% of blue whales eat more than 5,850 pounds of fish