Answer:
See attachment
Step-by-step explanation:
We graph equations by drawing a number line and placing an open circle on the two values. We fill in the circles if we have
. Since we don't, we leave it open. We then shade between the two.
Answers:
1. The n-intercept is 12. That means after 12 visits the amount of money on the gift car is $0.
2. The A(n)-intercept is 150. Before the visits, the amount of money on the gift car is $150.
Solution:
Amount of money on the gift card after n number of visits: A(n)=$150-$12.50 n
A(n)=150-12.50 n
1. n-intercept
A(n)=0→150-12.50 n =0
Solving for n: Subtracting 150 both sides of the equation:
150-12.50 n-150 = 0-150
-12.50 n = -150
Dividing both sides of the equation by -12.50:
(-12.50 n) / (-12.50) = (-150) / (-12.50)
n=12
The n-intercept is n=12; for n=12→A(12)=0. Point (n, A(n))=(12,0)
2. A(n) intercept
n=0→A(0)=150-12.50 (0)
A(0)=150-0
A(0)=150
The A(n) intercept is 150; for n=0→A(0)=150. Point (n, A(n))=(0,150)
Answer:
The probability that at most 1 car will require major engine repair next year is P=0.3392.
Step-by-step explanation:
This can be modeled as a binomial random variable, with n=22 and p=0.1.
The probability that exavtly k cars will require major engine repair next year is:
![P(x=k) = \dbinom{n}{k} p^{k}(1-p)^{n-k}](https://tex.z-dn.net/?f=P%28x%3Dk%29%20%3D%20%5Cdbinom%7Bn%7D%7Bk%7D%20p%5E%7Bk%7D%281-p%29%5E%7Bn-k%7D)
Then, the probability that at most 1 car will require major engine repair next year is:
![P(x\leq1)=P(x=0)+P(x=1)\\\\\\P(x=0) = \dbinom{22}{0} p^{0}(1-p)^{22}=1*1*0.0985=0.0985\\\\\\P(x=1) = \dbinom{22}{1} p^{1}(1-p)^{21}=22*0.1*0.1094=0.2407\\\\\\P(x\leq1)=0.0985+0.2407\\\\P(x\leq1)=0.3392](https://tex.z-dn.net/?f=P%28x%5Cleq1%29%3DP%28x%3D0%29%2BP%28x%3D1%29%5C%5C%5C%5C%5C%5CP%28x%3D0%29%20%3D%20%5Cdbinom%7B22%7D%7B0%7D%20p%5E%7B0%7D%281-p%29%5E%7B22%7D%3D1%2A1%2A0.0985%3D0.0985%5C%5C%5C%5C%5C%5CP%28x%3D1%29%20%3D%20%5Cdbinom%7B22%7D%7B1%7D%20p%5E%7B1%7D%281-p%29%5E%7B21%7D%3D22%2A0.1%2A0.1094%3D0.2407%5C%5C%5C%5C%5C%5CP%28x%5Cleq1%29%3D0.0985%2B0.2407%5C%5C%5C%5CP%28x%5Cleq1%29%3D0.3392)
Answer:
<h3>H. 150 ft</h3>
Step-by-step explanation:
Given the height of a golf ball in feet modelled by the formula h(t) = -16t² + 100x where t is in seconds, we are to find the height of the golf after 2.5 seconds. To do that we will simply substitute t = 2.5secs into the formula given;
![h(t) = -16t^2 + 100x\\\\h(2.5) = -16(2.5)^2 + 100(2.5)\\\\h(2.5) = -16(6.25) + 250\\\\h(2.5) = -100+250\\\\h(2.5) = 150\\\\](https://tex.z-dn.net/?f=h%28t%29%20%3D%20-16t%5E2%20%2B%20100x%5C%5C%5C%5Ch%282.5%29%20%3D%20-16%282.5%29%5E2%20%2B%20100%282.5%29%5C%5C%5C%5Ch%282.5%29%20%3D%20-16%286.25%29%20%2B%20250%5C%5C%5C%5Ch%282.5%29%20%3D%20-100%2B250%5C%5C%5C%5Ch%282.5%29%20%3D%20150%5C%5C%5C%5C)
Hence the height of the golf after 2.5 seconds is 150ft.
Answer:
Theres no picture........
Step-by-step explanation: