Answer:
1
Step-by-step explanation:
Given :
Mean, μ = 4
Standard deviation, s = 0.8
Sample size, n = 30
The distribution is independent.
Z = (x - μ) / s /sqrt(n)
Probability that downtime period is between 1 and 5
P(1≤ x ≤ 5) :
[(x - μ) / (s /sqrt(n))] ≤ Z ≤ [(x - μ) / (s /sqrt(n))]
[(1 - 4) / (0.8 /sqrt(30))] ≤ Z ≤ [(5 - 4) / (0.8 /sqrt(30)]
[-3 / 0.1460593] ≤ Z ≤] 1 / 0.1460593]
P(-20.539602 ≤ Z ≤ 6.8465342)
P(Z ≤ 6.8465342) - P(Z ≤ - 20.5396)
P(Z ≤ 6.8465342) = 1 (Z probability calculator)
P(Z ≤ - 20.5396) = 0 (Z probability calculator)
1 - 0 = 1
Answer:
50%
Step-by-step explanation:
The distance between each dot is 25%
From 68 to 70 is 25% and 70 to 74is 25%
Above 68 is 25+25 or 50 %
Answer:
96 cm²
Step-by-step explanation:
The area of a rhombus is usually found as half the product of the diagonals. Here, the lengths of the diagonals are 16 cm and 12 cm, so the area is ...
A = (1/2)(d1)(d2) = (1/2)(16 cm)(12 cm) = 96 cm²
_____
This is one of the area formulas it is useful to keep handy.
_____
Here, the area can also be found as 4 times the area of the marked triangle. That triangle's area is ...
A = (1/2)bh = (1/2)(8 cm)(6 cm) = 24 cm²
Four times that area is the are of the rhombus:
rhombus area = 4 × triangle area = 4 × 24 cm² = 96 cm²
Answer:
a) The volume of the water is approximately 678.6 cm³.
b) The volume of the sphere is approximately 113.1 cm³.
c) The new height of the water is approximately 7 cm, one cm higher than before.
Step-by-step explanation:
To solve this problem we first need to calculate the volume of the water, which is given by the cylinder volume, since it's stored in a cylindrical container.
![\text{water volume} = \pi*r^2*h\\\text{water volume} = \pi*(6)^2*6\\\text{water volume} = \pi*36*6\\\text{water volume} = 678.584 \text{ }cm^3\\](https://tex.z-dn.net/?f=%5Ctext%7Bwater%20volume%7D%20%3D%20%5Cpi%2Ar%5E2%2Ah%5C%5C%5Ctext%7Bwater%20volume%7D%20%3D%20%5Cpi%2A%286%29%5E2%2A6%5C%5C%5Ctext%7Bwater%20volume%7D%20%3D%20%5Cpi%2A36%2A6%5C%5C%5Ctext%7Bwater%20volume%7D%20%3D%20678.584%20%5Ctext%7B%20%7Dcm%5E3%5C%5C)
We now need to calculate the volume of the sphere, by using the appropriate formula:
![\text{volume sphere} = \frac{4}{3}\pi*r^3\\\text{volume sphere} = \frac{4}{3}\pi*(3^3)\\\text{volume sphere} = \frac{4}{3}\pi*27\\\text{volume sphere} = 113.09 \text{ } cm^3](https://tex.z-dn.net/?f=%5Ctext%7Bvolume%20sphere%7D%20%3D%20%5Cfrac%7B4%7D%7B3%7D%5Cpi%2Ar%5E3%5C%5C%5Ctext%7Bvolume%20sphere%7D%20%3D%20%5Cfrac%7B4%7D%7B3%7D%5Cpi%2A%283%5E3%29%5C%5C%5Ctext%7Bvolume%20sphere%7D%20%3D%20%5Cfrac%7B4%7D%7B3%7D%5Cpi%2A27%5C%5C%5Ctext%7Bvolume%20sphere%7D%20%3D%20113.09%20%5Ctext%7B%20%7D%20cm%5E3)
When the sphere is inserted into the cylinder the volume of the things that are inside of the container are added up, so the volume would be the volume of the water plus the volume of the sphere, we can use this information to calculate the height of the water as shown below:
![\text{total volume} = \text{water volume} + \text{sphere volume}\\\text{total volume} = 678.6 + 113.1 = 791.7\\\text{total volume} = \pi*r^2*h_{new}\\791.7 = \pi*(6)^2*h_{new}\\h_{new} = \frac{791.7}{\pi*36} = 7.00016\text{ } cm](https://tex.z-dn.net/?f=%5Ctext%7Btotal%20volume%7D%20%3D%20%5Ctext%7Bwater%20volume%7D%20%2B%20%5Ctext%7Bsphere%20volume%7D%5C%5C%5Ctext%7Btotal%20volume%7D%20%3D%20678.6%20%2B%20113.1%20%3D%20791.7%5C%5C%5Ctext%7Btotal%20volume%7D%20%3D%20%5Cpi%2Ar%5E2%2Ah_%7Bnew%7D%5C%5C791.7%20%3D%20%5Cpi%2A%286%29%5E2%2Ah_%7Bnew%7D%5C%5Ch_%7Bnew%7D%20%3D%20%5Cfrac%7B791.7%7D%7B%5Cpi%2A36%7D%20%3D%207.00016%5Ctext%7B%20%7D%20cm)
a) The volume of the water is approximately 678.6 cm³.
b) The volume of the sphere is approximately 113.1 cm³.
c) The new height of the water is approximately 7 cm, one cm higher than before.
Answer:
the third expression
Step-by-step explanation:
x=10.8
(x2 / 18) + 33.52
10.8 x 10.8=116.64
116.64 divide by 18=6.48
6.48 + 33.52=40