How do you do the work for 4-4 study guide and intervention parallel and perpendicular lines

A veterinary study of horses looked water sources for horses and the investigators found that horses received water from either

Irina18 [472]

Answer:

d. Water Source Well City Water Stream Expected Count 45.33 45.33 45.33

Step-by-step explanation:

Hello!

The objective of this study is to test whether the horses are equally likely to get water from one of three sources (Well, City or Stream), i.e. that the water sources have the same probability of being used.

The study variable is categorical:

X: Origin of the drinking water of the horses, categorized: Well "W", City "C" and Stream "S"

And one study population:

"Horses"

If the study hypothesis is that the three water sources are equally to be selected, you can symbolize it as P(W)=P(C)=P(S)=1/3

You want to test if the observed distribution follows an expected distribution, the test to use is a Chi-Square Goodness of Fit Test.

The hypothesis is:

H₀: P(W)=P(C)=P(S)=1/3

Under the null hypothesis, the formula to calculate the expected frequencies is:

$e_{i}= n * P_{i}$

Where:

$e_{i}$ is the expected value of the i-category.

$P_{i}$ is the probability for the i-category under the null hypothesis.

n= sample size.

Observed frequencies:

Well: 62

City: 43

Stream: 31

Total: 136

Expected frequencies:

$e_{W}= n * P_{W} = 136*1/3 = 45.33$

$e_{C}= n * P_{C}= 136 * 1/3= 45.33$

$e_{S}= n * P_{S}= 136 * 1/3= 45.33$

Typ, if the expected frequencies are correctly calculated, the sum of them must give the total of the sample (although depending on the type of approximation it can vary in one or two decimals)

45.33 * 3= 135.99 ≅ 136

I hope it helps!

5 0

2 years ago

In January of 2003(group 1), 1188 out of 1500 spots were bare ground (no vegetation). Find the sample proportion of bare ground

Nikitich [7]

Answer:

$n= 1500$ represent the random sample selected

$X= 1188$ represent the number of pots that were bare ground (no vegetation

$\hat p=\frac{X}{n}$

And replacing we got:

$\hat p=\frac{1188}{1500}= 0.792$

So then the sample proportion of bare ground spots is 0.792 for this sample

Step-by-step explanation:

We have the following info given from the problem:

$n= 1500$ represent the random sample selected

$X= 1188$ represent the number of pots that were bare ground (no vegetation)

And for this case if we want to find the sample proportion of bare ground spots we can use this formula:

$\hat p=\frac{X}{n}$

And replacing we got:

$\hat p=\frac{1188}{1500}= 0.792$

So then the sample proportion of bare ground spots is 0.792 for this sample

7 0

2 years ago

In a given course a student receives preliminary examination grades of 81, 85, and 95. The final examination is weighed for one-

Vsevolod [243]

Answer:

B. 96

Step-by-step explanation:

Let x represent score of final third exam.

We have been given that in a given course a student receives preliminary examination grades of 81, 85, and 95.

Let us find average of preliminary scores.

$\frac{81+85+95}{3}=\frac{261}{3}=87$

We have been given that the final examination is weighed for one-third and the average of the preliminary grades is weighed as 2/3 of the final grade. This means that final grade is equally weighed into 3 parts. The two parts are 87 each and third part is x.

Using our given information, we can set an equation as:

$\frac{87+87+x}{3}=90$

$\frac{174+x}{3}=90$

$\frac{174+x}{3}*3=90*3$

$174+x=270$

$174-174+x=270-174$

$x=96$

Therefore, the final examination grade should be 96 for a semester average of 90.

4 0

2 years ago

The initial temperature of a cup of tea is 200ºF. The surrounding temperature is 70ºF, and the value of the constant k is 0.6.

Nezavi [6.7K]

Newton's Law of Cooling states that the change of the temperature of an object is proportional to the difference between its own temperature and the ambient temperature over time.

Therefore when expressed mathematically, this is equivalent to:

dT = - k (T – Ts) dt

dT / (T – Ts) = - k dt

Integrating:

ln [(T2– Ts) / (T1– Ts)] = - k (t2 – t1)

Before we plug in the values, let us first convert the temperatures into absolute values R (rankine) by adding 460.

R = ˚F + 460

T1 = 200 + 460 = 660 R

Ts = 70 + 460 = 530 R

ln [(T2– 530) / (660 – 530)] = - 0.6 (2 - 0)

T2 = 569.16 R

T2 = 109 ºF

Answer: After 2 hours, it will be 109 ºF

6 0

2 years ago

Farmer McCoy can plant either corn or soybeans. The probabilities that the next harvest prices will go up, stay the same, or go

Snezhnost [94]

Answer:

McCoy should plant soybeans

Step-by-step explanation:

Thinking process:

Let the two crops be corn and soybeans as indicated.

The profit margin for corn will be $ 30 000. However, the loss will be:

$ 35 000 - $ 30 000 = $ 5 000

For the soybean, the profit margin will be $ 10 000.

The loss will be $ 5 000

The net balance will be given as: $ 5 000 - $ 10 000 = - $ 5 000. (This represents a gain)

Thus, soybean will be more profitable.

4 0

2 years ago