Answer:
17.5hrs
Step-by-step explanation:
Set they worked for x hours,
10x+35=210
10x=175
x=17.5
I have the same problem here with a slight change in the given values:
radius is 2 & height of 6 indicates the bounding line is y = 3 x---> x = y / 3....
<span>thus the [ π radius ² thickness ] yields π (y² / 9 ) <span>dy ,</span> y in [ 0 , 6 ] for the volume... </span>
a Riemann sum is then : y_i = 0 + i [ 6 / n ] = 6 i / n , i = 1,2,3...n and do a right side sum
<span>π Σ { i = 1,2,3..n } [ 36 i² / 9 n² ] [ 6 / n ]
</span>
I hope my guide has come to your help. God bless and have a nice day ahead!
Answer:
Step-by-step explanation:
Percentage of A
34/40 x 100 = 85%
Percentage of B
54/40 x 100 = 135%
Ans: A) 85% B) 135%
The value of the given trigonometry function is 2√15/15
<h3>Half angles</h3>
Half angles are trigonometric identities used to express sine, cosine and tangent of half angles.
For instance the value of cos theta is expressed as shown below;
cosФ = cos(Ф/2+Ф/2)
cosФ = cos²Ф/2-sin²Ф/2
cosФ = cos²Ф/2-(1-cos²Ф/2)
cosФ = 1 - 2cos²Ф/2
Given the following parameters
cosФ = -7/15
Substitute
cosФ = 1 - 2cos²Ф/2
-7/15 = 1 - 2cos²Ф/2
-2cos²Ф/2 = -7/15 - 1
-2cos²Ф/2 = -8/15
cos²Ф/2 = 4/15
cosФ/2 = 2/√15
Rationalize
2/√15 * √15/√15
2√15/15
Hence the value of the given trigonometry function is 2√15/15
Learn more on trigonometry function here: brainly.com/question/2254074
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Answer:
The 95% CI for the difference of means is:

Step-by-step explanation:
<em>The question is incomplete:</em>
<em>"Find a 95% confidence interval on the difference of the towels mean absorbency produced by the two processes. Assumed that the standard deviations are estimated from the data. Round to two decimals places."</em>
Process 1:
- Sample size: 10
- Mean: 200
- S.D.: 15
Process 2:
- Sample size: 4
- Mean: 300
- S.D.: 50
The difference of the sample means is:

The standard deviation can be estimated as:

The degrees of freedom are:

The t-value for a 95% confidence interval and 12 degrees of freedom is t=±2.179.
Then, the confidence interval can be written as:
