Answer:
t=3
Step-by-step explanation:
5t-4=|11
+4 |+4
5t= | 15
÷5 | ÷5
t = 3
Answer:
-14a^2b-42ab^2+56abc
Step-by-step explanation:
You can use the FOIL method
multiply the first numbers
then inner
then outer
then last
Which of these is logically equivalent to the given statement? (1 point)
1. If Amelia did not go to the park, then she did not finish her homework.
2. If Amelia did not finish her homework, then she will go to the park.
3. If Amelia goes to the park, then she did not finish her homework.
4. If Amelia finishes her homework, then she cannot go to the park.
Hello there.
In this problem, we can use our intuition of logic, but I will show a proof of the result in a truth table later. Then, let's get started!
Given:
→ Amelia finishes the homework (sentence H, can be True or False)
→ Amelia goes to the park (P, true or false)
Then, we have: If H, then P. Logically:
H ⇒ P
Then we can think: everytime she does the homework, she goes to the park. Therefore, if she did not go to the park, she will not have finished the homework (It is an equivalent sentence).
Alternative 1.
==========
Now, let's prove that (H ⇒P) is equivalent to (¬P ⇒ ¬H), via the truth table:
H P ¬H ¬P (H ⇒ P) (¬P ⇒ ¬H)
T T F F T T
T F F T F F
F T T F T T
F F T T T T
As we can see, the results are identical, therefore, the sentences are indeed equivalent.
I hope it hepls :)
Answer:
Step-by-step explanation:
Solution:-
State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: The distribution of severity of psoriasis cases at the end and prior are same.
Alternative hypothesis: The distribution of severity of psoriasis cases at the end and prior are different.
Formulate an analysis plan. For this analysis, the significance level is 0.05. Using sample data, we will conduct a chi-square goodness of fit test of the null hypothesis.
Analyze sample data. Applying the chi-square goodness of fit test to sample data, we compute the degrees of freedom, the expected frequency counts, and the chi-square test statistic. Based on the chi-square statistic and the degrees of freedom, we determine the P-value.
DF = k - 1 = 4 - 1
D.F = 3
(Ei) = n * pi
Category observed Num expected num [(Or,c -Er,c)²/Er,c]
Remission 380 20 6480
Mild
symptoms 520 770 81.16883117
Moderate
symptoms 95 160 24.40625
Severe
symptom 5 50 40.5
Sum 1000 1000 6628.075081
Χ2 = Σ [ (Oi - Ei)2 / Ei ]
Χ2 = 6628.08
Χ2Critical = 7.81
where DF is the degrees of freedom, k is the number of levels of the categorical variable, n is the number of observations in the sample, Ei is the expected frequency count for level i, Oi is the observed frequency count for level i, and Χ2 is the chi-square test statistic.
The P-value is the probability that a chi-square statistic having 3 degrees of freedom is more extreme than 6628.08.
We use the Chi-Square Distribution Calculator to find P(Χ2 > 19.58) =less than 0.000001
Interpret results. Since the P-value (almost 0) is less than the significance level (0.05), we cannot accept the null hypothesis.
We reject H0, because 6628.08 is greater than 7.81. We have statistically significant evidence at alpha equals to 0.05 level to show that distribution of severity of psoriasis cases at the end of the clinical trial for the sample is different from the distribution of the severity of psoriasis cases prior to the administration of the drug suggesting the drug is effective.
The chi-square test is a statistical method that determines if there's a significant difference between observed and expected frequencies in different categories, such as symptom status in this clinical trial. Without post-treatment numbers, we can't run the exact test. However, if the test statistic exceeded the critical value, we could conclude that the drug significantly affected symptom statuses.
This question pertains to the use of a chi-squared test, which is a statistical method used to determine if there's a significant difference between observed frequencies and expected frequencies in one or more categories. For this case, the categories are the symptom statuses (remission, mild, moderate, and severe).
To conduct a chi-square test, you first need to know the observed frequencies (the initial percentages given in the question) and the expected frequencies (the percentages after treatment). As the question doesn't provide the numbers after treatment, I can't perform the exact chi-square test.
If the post-treatment numbers were provided, you would compare them to the pre-treatment numbers using the chi-squared formula, which involves summing the squared difference between observed and expected frequencies, divided by expected frequency, for all categories. The result is a chi-square test statistic, which you would then compare to a critical value associated with a chosen significance level (commonly 0.05) to determine if the treatment has a statistically significant effect.
To interpret a chi-square test statistic, if the calculated test statistic is larger than the critical value, it suggests that the drug made a significant difference in the distribution of symptom statuses. If not, we can't conclude the drug was effective.
#SPJ3
Answer:
D is the answer to your question
Answer:
3.8
Step-by-step explanation:
Rounding to the nearest tenth means that eliminating the digits in the hundredths and thousandths place and if the hundredths place is greater than 5, then you also have to add a 1 to the tenths place.
So, The number becomes:
=> 3.8 (Hundredths place have 6 which is greater than 5)
Answer:
3.8
Step-by-step explanation:
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