Answer:
Step-by-step explanation:
Given that taxi Fares are normally distributed with a mean fare of $22.27 and a standard deviation of $2.20.
For a random single taxi std deviation is 2.20
But for a sample of size 10, std deviation would be
This would be less than the 2.20
Because std devition is less for sample we get a big z score for the sample than the single.
As positive values of z increase we find that probability would decrease since normal curve is bell shaped.
So single taxi fare would have higher probability than sample.
B) Here >24.
By the same argument we have z value less for single taxi hence the probability for more than that would be less than that of sample size 10
Answer: Option C.
Step-by-step explanation:
A) The result of the substitution shown in Option A is obtained by solving for y from the first equation and substituting into the second equation:
Therefore, this is a result of a substitution in the given system.
B) The result of the substitution shown in Option B is obtained by solving for y from the second equation and substituting into the first equation:
Therefore, this is a result of a substitution in the given system.
C) The result of the substitution shown in Option C is not a result of a substitution in the given system, because if you solve for x from the second equation and substitute into the first one, you get:
Yes. If you have very high or very low outliers in your data set, it is generally preferred to use the median - the mid-point when all data points are arranged from least to greatest.
<span>A good example for when to avoid the mean and prefer the median is salary. The mean is less good here as there are a few very high salaries which skew the distribution to the right. This drags the mean higher to the point where it is disproportionately affected by the few higher salaries. In this case, the median would only be slightly affected by the few high salaries and is a better representation of the whole of the data. </span>
<span>In general, if the distribution is not normal, the mean is less appropriate than the median.</span>
Answer:
187.06 units²
Step-by-step explanation:
Area of kite:
½ × d1 × d2
½d1 = 18sin(30)
½d1 = 9
d1 = 18
d2 = 9/tan(60) + 18cos(30)
= 12sqrt(3)
Area = ½ × 18 × 12sqrt(3)
108sqrt(3) units²
187.0614872 units²
To determine the cost to fertilize the trees, you need to figure out how many kilograms of fertilizer are needed. With the information given, you can determine The number of grams needed to fertilize 1200 trees. To do this you would multiply 200 times 1200. This equals 240,000 grams. To convert this to kilograms, divide 240000g by 1000 g. Every group of 1000 g is 1 kg. The answer is 240 kilograms. Multiply 240 kg by the price of $2.75 per kilogram to get $660 as the cost.
