Answer:
2.5 bottles or 3 bottles
Step-by-step explanation:
Given that :
Dimension of lot :
22.5 yds by 20.4 yds
Area of lot = 22.5 * 20.4 = 459 yd²
Volume per bottle of herbicide = 12 fl oz
1.1 fl oz combined with water treats an area of 155 ft²
Number of 1.1 fl oz from 12 fl oz that can be obtained : 12 / 1.1 = 10.90
1 yard² = 9 feet²
459 yard² = (459 * 9) = 4131 feet²
Number of 151 feet² that can be obtained from 4131 feet²
4131 feet² / 155 ft²
= 27.3576 times
= 27.3576 / 10.909
= 2.507
This means he'll have to buy a minimum of 3 bottles if a fraction of a bottle can't be purchased.
Bacteria growth occurs exponentially; bacteria divides into 2 every t minutes (similar to the penny doubling every day story). If one gets filled, and the contents divides once, there will be enough for 2 bottles; when these two are ready to divide, there will be enough for 4. This growth process begins very slowly - 1, 2, 4, 8, 16, 32; but it soon speeds up greatly, 64, 128, 256, 512, 1024, 2048.
A manufacturer<span> of </span>medical supplies produces 1 liter bags<span> of </span>saline solution. thecost<span> to </span>produce x thousand bags<span> of </span>saline<span> is </span>given<span> by the </span>cost<span> func. ... thousand </span>bags<span> of </span>saline<span> is</span>given<span> by the </span>cost function<span> C(x)=1.7x^2-5.3x+3.2, ...</span>
Answer:
Step-by-step explanation
Hello!
Be X: SAT scores of students attending college.
The population mean is μ= 1150 and the standard deviation σ= 150
The teacher takes a sample of 25 students of his class, the resulting sample mean is 1200.
If the professor wants to test if the average SAT score is, as reported, 1150, the statistic hypotheses are:
H₀: μ = 1150
H₁: μ ≠ 1150
α: 0.05
![Z= \frac{X[bar]-Mu}{\frac{Sigma}{\sqrt{n} } } ~~N(0;1)](https://tex.z-dn.net/?f=Z%3D%20%5Cfrac%7BX%5Bbar%5D-Mu%7D%7B%5Cfrac%7BSigma%7D%7B%5Csqrt%7Bn%7D%20%7D%20%7D%20~~N%280%3B1%29)

The p-value for this test is 0.0949
Since the p-value is greater than the level of significance, the decision is to reject the null hypothesis. Then using a significance level of 5%, there is enough evidence to reject the null hypothesis, then the average SAT score of the college students is not 1150.
I hope it helps!