Answer:
0.9938
Step-by-step explanation:
Let mean blood pressure be = x
Pr ( x < 119 ) can be found using ' z value ' = ( x - u ) ÷ ( s / √n ) ; where x = 119 , u = mean = 117 , n = no. of sample items = 144 , s = standard deviation = 9.6
z = (119 - 117) ÷ (9.6 / 144) = -2 ÷ (9.6 / 12) = -2 / 0.8 → z = 2.5
Pr (x < 119) = P (z < 2.5)
P = 0.9938
Answer:
The value of
.
Step-by-step explanation:
Consider the provided information.
The provided formula is ![f(x)=x^3+x+6](https://tex.z-dn.net/?f=f%28x%29%3Dx%5E3%2Bx%2B6)
Substitute
in above equation.
![f(x_1)=(-2)^3+(-2)+6](https://tex.z-dn.net/?f=f%28x_1%29%3D%28-2%29%5E3%2B%28-2%29%2B6)
![f(x_1)=-8-2+6](https://tex.z-dn.net/?f=f%28x_1%29%3D-8-2%2B6)
![f(x_1)=-4](https://tex.z-dn.net/?f=f%28x_1%29%3D-4)
Differentiate the provided function and calculate the value of ![f'(x_1)](https://tex.z-dn.net/?f=f%27%28x_1%29)
![f'(x)=3x^2+1](https://tex.z-dn.net/?f=f%27%28x%29%3D3x%5E2%2B1)
![f'(x)=3(-2)^2+1](https://tex.z-dn.net/?f=f%27%28x%29%3D3%28-2%29%5E2%2B1)
![f'(x)=13](https://tex.z-dn.net/?f=f%27%28x%29%3D13)
The Newton iteration formula: ![x_2=x_1-\frac{f(x_1)}{f'(x_1)}](https://tex.z-dn.net/?f=x_2%3Dx_1-%5Cfrac%7Bf%28x_1%29%7D%7Bf%27%28x_1%29%7D)
Substitute the respective values in the above formula.
![x_2=-2-\frac{(-4)}{13}](https://tex.z-dn.net/?f=x_2%3D-2-%5Cfrac%7B%28-4%29%7D%7B13%7D)
![x_2=-2+0.3077](https://tex.z-dn.net/?f=x_2%3D-2%2B0.3077)
![x_2=-1.6923](https://tex.z-dn.net/?f=x_2%3D-1.6923)
Hence, the value of
.
Answer:
x = 6
Step-by-step explanation:
10 - 4 = 6
The x means the number is unkown, well, we know that 6+4=10, so there you go!
Answer:
Kindly check explanation
Step-by-step explanation:
H0 : μ = 5500
H1 : μ > 5500
The test statistic assume normal distribution :
Test statistic :
(Xbar - μ) ÷ s/sqrt(n)
(5625.1 - 5500) ÷ 226.1/sqrt(15) = 2.1429 = 2.143
Pvalue from test statistic :
The Pvalue obtained using the calculator at df = 15 - 1 = 14 is 0.025083
α = 0.05
Since ;
Pvalue < α
0.025083 < 0.05 ; Reject H0
The confidence interval :
Xbar ± Tcritical * s/sqrt(n)
Tcritical at 95% = 1.761 ;
margin of error = 1.761 * 226.1/sqrt(15) = 102.805
Lower boundary : (5625.1 - 102.805) = 5522.295
(5522.295 ; ∞)
The hypothesized mean does not occur within the constructed confidence boundary. HENCE, there is significant eveidbce to support the claim that the true mean life of a biomedical device is greater than 5500