Answer: ![\begin{bmatrix}\mathrm{Solution:}\:&\:x\le \frac{17}{3}\:\\ \:\mathrm{Decimal:}&\:x\le \:5.66666\dots \\ \:\mathrm{Interval\:Notation:}&\:(-\infty \:,\:\frac{17}{3}]\end{bmatrix}](https://tex.z-dn.net/?f=%5Cbegin%7Bbmatrix%7D%5Cmathrm%7BSolution%3A%7D%5C%3A%26%5C%3Ax%5Cle%20%5Cfrac%7B17%7D%7B3%7D%5C%3A%5C%5C%20%5C%3A%5Cmathrm%7BDecimal%3A%7D%26%5C%3Ax%5Cle%20%5C%3A5.66666%5Cdots%20%5C%5C%20%5C%3A%5Cmathrm%7BInterval%5C%3ANotation%3A%7D%26%5C%3A%28-%5Cinfty%20%5C%3A%2C%5C%3A%5Cfrac%7B17%7D%7B3%7D%5D%5Cend%7Bbmatrix%7D)
Step-by-step explanation:
Answer:
The answer is "It would decrease, but not necessarily by 8%".
Step-by-step explanation:
They know that width of the confidence level is proportional to a confidence level. As just a result, reducing the confidence level decreases the width of a normal distribution, but not with the amount of variance in the confidence level. As just a result, when a person teaches a 90% standard deviation rather than a 98 percent normal distribution, the width of the duration narrows.
A₅ = 1/16 and r= 1/4
Let see how to build up this formula that is going to give that term of rank n
1st term =a₁ = To be calculated
1st a₁ = a₁ x r°
2nd a₂ = a₁ x r¹
3rd a₃ = a₁ x r²
4th a₄ = a₁ x r³
5th a₅ = a₁ x r⁴
.......................
.......................
nth : a(n) = a₁ x r⁽ⁿ-¹)
Note when that the subscript of a is the same as the exponent mines 1
We know the ratio r =1/4 & the fifth term, a₅ =1/16 (given). Now let's apply the formula to calculate the unknown a₁.
a(n) = a₁ x r⁽ⁿ-¹) ==>a₅ = a₁ x (1/4)⁽⁵⁺¹⁾ ===> 1/16 = a₁ x (1/4)⁴
1/167 = a₁ (1/256) ==> a₁ =16 & the formula becomes
a₅ = 16(1/4)⁴
344444444;3333333+5655434455432=45666788999765433
Looking at the units, 756 meters divided by 6 m/s will give you the seconds that you need.
So for A) 756/6 = 126
For B, you take the answer from A, and subtract 18 seconds, which would make it 108. Then you take the distance, 756 and divide by 108, that will be the speed.
