The equation would be 40+15p=70+5p. You would subtract 5p from each side (15-5) and (5-5) and you'll have 40+10p=70. You would then subtract 40 from each side (70-40) and (40-40) to have 10p=30. You would divide by both of the sides by 10, you would end up with p=3. P is price training. The reason you divide and subtract on both sides is because you want the months to be on one side and the price training on the other side.
Answer: 2
Step-by-step explanation:
This should be right because 5 * 2 = 10
5x mean that 5 * x = 10
Hope this makes sense
Answer:
1c

1d

Step-by-step explanation:
From the question we are told that
The probability of telesales representative making a sale on a customer call is 
The mean is 
Generally the distribution of sales call made by a telesales representative follows a binomial distribution
i.e
and the probability distribution function for binomial distribution is
Here C stands for combination hence we are going to be making use of the combination function in our calculators
Generally the mean is mathematically represented as

=> 
=> 
Generally the least number of calls that need to be made by a representative for the probability of at least 1 sale to exceed 0.95 is mathematically represented as

=> 
=> ![P( X \ge 1) = 1 - [ ^{n}C_0 * (0.15 )^0 * (1- 0.15)^{n-0}] > 0.95](https://tex.z-dn.net/?f=P%28%20X%20%5Cge%201%29%20%3D%201%20-%20%5B%20%5E%7Bn%7DC_0%20%2A%20%20%280.15%20%29%5E0%20%2A%20%20%281-%200.15%29%5E%7Bn-0%7D%5D%20%3E%200.95)
=> ![1 - [1 * 1* (0.85)^{n}] > 0.95](https://tex.z-dn.net/?f=%201%20-%20%5B1%20%20%2A%20%201%2A%20%20%280.85%29%5E%7Bn%7D%5D%20%3E%200.95)
=> ![[(0.85)^{n}] > 0.05](https://tex.z-dn.net/?f=%20%20%5B%280.85%29%5E%7Bn%7D%5D%20%3E%200.05)
taking natural log of both sides

=> 
To solve this problem, you have to know these two special factorizations:

Knowing these tells us that if we want to rationalize the numerator. we want to use the top equation to our advantage. Let:
![\sqrt[3]{x+h}=x\\ \sqrt[3]{x}=y](https://tex.z-dn.net/?f=%20%5Csqrt%5B3%5D%7Bx%2Bh%7D%3Dx%5C%5C%20%5Csqrt%5B3%5D%7Bx%7D%3Dy%20)
That tells us that we have:

So, since we have one part of the special factorization, we need to multiply the top and the bottom by the other part, so:

So, we have:
![\frac{x+h-h}{h(\sqrt[3]{(x+h)^2}+\sqrt[3]{(x+h)(x)}+\sqrt[3]{x^2})}=\\ \frac{x}{\sqrt[3]{(x+h)^2}+\sqrt[3]{(x+h)(x)}+\sqrt[3]{x^2}}](https://tex.z-dn.net/?f=%20%5Cfrac%7Bx%2Bh-h%7D%7Bh%28%5Csqrt%5B3%5D%7B%28x%2Bh%29%5E2%7D%2B%5Csqrt%5B3%5D%7B%28x%2Bh%29%28x%29%7D%2B%5Csqrt%5B3%5D%7Bx%5E2%7D%29%7D%3D%5C%5C%20%5Cfrac%7Bx%7D%7B%5Csqrt%5B3%5D%7B%28x%2Bh%29%5E2%7D%2B%5Csqrt%5B3%5D%7B%28x%2Bh%29%28x%29%7D%2B%5Csqrt%5B3%5D%7Bx%5E2%7D%7D%20)
That is our rational expression with a rationalized numerator.
Also, you could just mutiply by:
![\frac{1}{\sqrt[3]{x_h}-\sqrt[3]{x}} \text{ to get}\\ \frac{1}{h\sqrt[3]{x+h}-h\sqrt[3]{h}}](https://tex.z-dn.net/?f=%20%5Cfrac%7B1%7D%7B%5Csqrt%5B3%5D%7Bx_h%7D-%5Csqrt%5B3%5D%7Bx%7D%7D%20%5Ctext%7B%20to%20get%7D%5C%5C%20%5Cfrac%7B1%7D%7Bh%5Csqrt%5B3%5D%7Bx%2Bh%7D-h%5Csqrt%5B3%5D%7Bh%7D%7D%20)
Either way, our expression is rationalized.
Answer:
-30√(t/a) cos(√(at)) + 30/a sin(√(at)) + C
Step-by-step explanation:
∫ 15 sin(√(at)) dt
Use substitution:
If x = √(at), then:
dx = ½ (at)^-½ (a dt)
dx = a / (2√(at)) dt
dx = a/(2x) dt
dt = (2/a) x dx
Plugging in:
∫ 15 sin x (2/a) x dx
30/a ∫ x sin x dx
Integrate by parts:
If u = x, then du = dx.
If dv = sin x dx, then v = -cos x.
∫ u dv = uv − ∫ v du
= 30/a (-x cos x − ∫ -cos x dx)
= 30/a (-x cos x + ∫ cos x dx)
= 30/a (-x cos x + sin x + C)
Substitute back:
30/a (-√(at) cos(√(at)) + sin(√(at)) + C)
-30√(t/a) cos(√(at)) + 30/a sin(√(at)) + C