Answer: 0.9147
Step-by-step explanation:
Given: 20% of the plants ordered are petunias.
Let p = 20% = 0.2
Sample size: n = 120
If they follow normal distribution, then mean = np = 120x 0.2 = 24
Standard deviation = ![\sqrt{n p(1-p)}=\sqrt{120\times0.2\times.8}](https://tex.z-dn.net/?f=%5Csqrt%7Bn%20p%281-p%29%7D%3D%5Csqrt%7B120%5Ctimes0.2%5Ctimes.8%7D)
![=\sqrt{19.2}\\\\ =4.38](https://tex.z-dn.net/?f=%3D%5Csqrt%7B19.2%7D%5C%5C%5C%5C%20%3D4.38)
The probability that no more than 30 plants are petunias : ![P(\dfrac{X-\mu}{\sigma}\leq \dfrac{30-24}{4.38})](https://tex.z-dn.net/?f=P%28%5Cdfrac%7BX-%5Cmu%7D%7B%5Csigma%7D%5Cleq%20%5Cdfrac%7B30-24%7D%7B4.38%7D%29)
![=P(Z](https://tex.z-dn.net/?f=%3DP%28Z%3C1.37%29%3D%200.9147)
Hence, the probability that no more than 30 plants are petunias = 0.9147
Answer:
5
Step-by-step explanation:
5 100
500
100
600
10000
Answer:
i don't know
Step-by-step explanation:
bro I really don't know for this one
Answer:
Option D. ![y=-4x^{2} -16x-12](https://tex.z-dn.net/?f=y%3D-4x%5E%7B2%7D%20-16x-12)
Step-by-step explanation:
using a graphing tool
Graph and determine the vertex in each case
we know that
If the equation has a maximum value at x=-2, then the x-coordinate of the vertex must be equal to -2 and the parabola open downward
case A) ![y=-x^{2} -20x-16](https://tex.z-dn.net/?f=y%3D-x%5E%7B2%7D%20-20x-16)
The vertex is the point ![(-10,84)](https://tex.z-dn.net/?f=%28-10%2C84%29)
case B) ![y=-x^{2} -16x-12](https://tex.z-dn.net/?f=y%3D-x%5E%7B2%7D%20-16x-12)
The vertex is the point ![(-8,52)](https://tex.z-dn.net/?f=%28-8%2C52%29)
case C) ![y=-4x^{2} -20x-16](https://tex.z-dn.net/?f=y%3D-4x%5E%7B2%7D%20-20x-16)
The vertex is the point ![(-2.5,9)](https://tex.z-dn.net/?f=%28-2.5%2C9%29)
case D) ![y=-4x^{2} -16x-12](https://tex.z-dn.net/?f=y%3D-4x%5E%7B2%7D%20-16x-12)
The vertex is the point
-------> is the answer
see the attached figure
Answer:
The 98% confidence interval to estimate the proportion of watermelon seeds that germinate is (0.4443, 0.6557). This means that we are 98% sure that the true proportion of all watermalong seeds of the company that germinate is between these two values, which means that there is good evidence that the proportion is below the 70% standard.
Step-by-step explanation:
In a sample with a number n of people surveyed with a probability of a success of
, and a confidence level of
, we have the following confidence interval of proportions.
![\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}](https://tex.z-dn.net/?f=%5Cpi%20%5Cpm%20z%5Csqrt%7B%5Cfrac%7B%5Cpi%281-%5Cpi%29%7D%7Bn%7D%7D)
In which
z is the z-score that has a p-value of
.
Once a week for 12 weeks, he purchases a pack of 10 watermelon seeds to act as his sample. He plants the seeds in a greenhouse with good soil to maintain a consistent temperature and watering routine. He finds that the germination rate for the company's watermelon seeds is 55%.
This means that ![n = 12*10 = 120, \pi = 0.55](https://tex.z-dn.net/?f=n%20%3D%2012%2A10%20%3D%20120%2C%20%5Cpi%20%3D%200.55)
98% confidence level
So
, z is the value of Z that has a p-value of
, so
.
The lower limit of this interval is:
![\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.55 - 2.327\sqrt{\frac{0.55*0.45}{120}} = 0.4443](https://tex.z-dn.net/?f=%5Cpi%20-%20z%5Csqrt%7B%5Cfrac%7B%5Cpi%281-%5Cpi%29%7D%7Bn%7D%7D%20%3D%200.55%20-%202.327%5Csqrt%7B%5Cfrac%7B0.55%2A0.45%7D%7B120%7D%7D%20%3D%200.4443)
The upper limit of this interval is:
![\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.55 + 2.327\sqrt{\frac{0.55*0.45}{120}} = 0.6557](https://tex.z-dn.net/?f=%5Cpi%20%2B%20z%5Csqrt%7B%5Cfrac%7B%5Cpi%281-%5Cpi%29%7D%7Bn%7D%7D%20%3D%200.55%20%2B%202.327%5Csqrt%7B%5Cfrac%7B0.55%2A0.45%7D%7B120%7D%7D%20%3D%200.6557)
The 98% confidence interval to estimate the proportion of watermelon seeds that germinate is (0.4443, 0.6557). This means that we are 98% sure that the true proportion of all watermalong seeds of the company that germinate is between these two values, which means that there is good evidence that the proportion is below the 70% standard.