<span>Hey there, Lets solve this together.
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<span>The variable (R) will be representing the rate as a decimal.
When you solve for (R) multiply by 100 to get the percentage. A percentage is </span><span>a rate, number, or amount in each hundred.
</span><span>"0.97" will be replaced by "(97/100)"
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<span>Rearrange the equation by subtracting what is to the right of the equal sign from both sides of the given equation
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Rewrite the whole as a fraction using 10 <span> as the denominator :
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The equation now takes the shape :
t • (10f-291)<span> = 0
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<span><span> f = 291/10 = 29.100
</span> t = 0</span>
<span>Positive = Growth
Negative = Decay
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Answer:
(-1, -1/2)
Step-by-step explanation:
To find the x coordinate of the midpoint, add the x coordinates and divide by 2
( -11+9)/2 = -2/2 = -1
To find the y coordinate of the midpoint, add the y coordinates and divide by 2
(0+-1)/2 = -1/2
(-1, -1/2)
Answer:
2. x = 3
3. x = 10 and y =2
Step-by-step explanation:
2.
From inspection, we can see that EF = 10 and KL = 5
Therefore the ratio KL : EF = 5 : 10 = 5/10 = 1/2
This means that the side measures of ΔJKL are 1/2 the corresponding side measures of ΔGEF
Therefore, if GF = 22, then JL = 22 ÷ 2 = 11
So: 2x + 5 = 11
⇒ 2x = 11 - 5 = 6
⇒ x = 6 ÷ 2 = 3
x = 3
3.
From inspection, we can see that DF = 12 and HJ = 9
Therefore the ratio DF : HJ = 12 : 9 = 12/9 = 4/3
This means that the side measures of ΔDEF are 4/3 times the corresponding side measures of ΔHGJ
Therefore, if HG = 6, then DE = 6 x 4/3 = 8
So: 2y + 4 = 8
⇒ 2y = 8 - 4 = 4
⇒ y = 4 ÷ 2 = 2
y = 2
Similarly, if GJ = 24, then EF = 24 x 4/3 = 32
So: 4x - 8 = 32
⇒ 4x = 32 + 8 = 40
⇒ x = 40 ÷ 4 = 10
x = 10
Answer:
The students should request an examination with 5 examiners.
Step-by-step explanation:
Let <em>X</em> denote the event that the student has an “on” day, and let <em>Y</em> denote the
denote the event that he passes the examination. Then,
The events (
) follows a Binomial distribution with probability of success 0.80 and the events (
) follows a Binomial distribution with probability of success 0.40.
It is provided that the student believes that he is twice as likely to have an off day as he is to have an on day. Then,
Then,
⇒
Then,
Compute the probability that the students passes if request an examination with 3 examiners as follows:
![=[\sum\limits^{3}_{x=2}{{3\choose x}(0.80)^{x}(1-0.80)^{3-x}}]\times\frac{2}{3}+[\sum\limits^{3}_{x=2}{{3\choose x}(0.40)^{3}(1-0.40)^{3-x}}]\times\frac{1}{3}](https://tex.z-dn.net/?f=%3D%5B%5Csum%5Climits%5E%7B3%7D_%7Bx%3D2%7D%7B%7B3%5Cchoose%20x%7D%280.80%29%5E%7Bx%7D%281-0.80%29%5E%7B3-x%7D%7D%5D%5Ctimes%5Cfrac%7B2%7D%7B3%7D%2B%5B%5Csum%5Climits%5E%7B3%7D_%7Bx%3D2%7D%7B%7B3%5Cchoose%20x%7D%280.40%29%5E%7B3%7D%281-0.40%29%5E%7B3-x%7D%7D%5D%5Ctimes%5Cfrac%7B1%7D%7B3%7D)
The probability that the students passes if request an examination with 3 examiners is 0.715.
Compute the probability that the students passes if request an examination with 5 examiners as follows:
![=[\sum\limits^{5}_{x=3}{{5\choose x}(0.80)^{x}(1-0.80)^{5-x}}]\times\frac{2}{3}+[\sum\limits^{5}_{x=3}{{5\choose x}(0.40)^{x}(1-0.40)^{5-x}}]\times\frac{1}{3}](https://tex.z-dn.net/?f=%3D%5B%5Csum%5Climits%5E%7B5%7D_%7Bx%3D3%7D%7B%7B5%5Cchoose%20x%7D%280.80%29%5E%7Bx%7D%281-0.80%29%5E%7B5-x%7D%7D%5D%5Ctimes%5Cfrac%7B2%7D%7B3%7D%2B%5B%5Csum%5Climits%5E%7B5%7D_%7Bx%3D3%7D%7B%7B5%5Cchoose%20x%7D%280.40%29%5E%7Bx%7D%281-0.40%29%5E%7B5-x%7D%7D%5D%5Ctimes%5Cfrac%7B1%7D%7B3%7D)
The probability that the students passes if request an examination with 5 examiners is 0.734.
As the probability of passing is more in case of 5 examiners, the students should request an examination with 5 examiners.
In probability, problems involving arrangements are called combinations or permutations. The difference between both is the order or repetition. If you want to arrange the letters regardless of the order and that there must be no repetition, that is combination. Otherwise, it is permutation. Therefore, the problem of arrange A, B, C, D, and E is a combination problem.
In combination, the number of ways of arranging 'r' items out of 'n' items is determined using n!/r!(n-r)!. In this case, you want to arrange all 5 letters. So, r=n=5. Therefore, 5!/5!(505)! = 5!/0!=5!/1. It is simply equal to 5! or 120 ways.
