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IceJOKER [234]
4 years ago
12

The Island of Knights and Knaves has two types of inhabitants: Knights, who always tell the truth, and Knaves, who always lie. A

s you are exploring the Island of Knights and Knaves you encounter two people named A and B. A tells you "I am a Knave, but B isn’t". B says nothing. Determine the nature of A and B, if you can.
Mathematics
1 answer:
grandymaker [24]4 years ago
3 0

Answer: First, suppose that A is a knight. then when he says "i am a knave" he would be lying, so you have a logical failure because knights can't lie.

if A is a knave and says "I am a Knave, but B isn’t" then he would be telling a truth in the first part.

now you have two paths to tink itm as A said a truth, he can't be a knave. but if you consider the whole sentence can be splitted in two sentences.

I am a Knave ----- wealready know that will be a truth.

but B isn’t----- and now, as the first sentence is true, this must be false, so the sum of both sentences is false.

so A is a knave and B is a knave.

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Answers to questions A and B
Elena L [17]

Answer:

A. m(arc BC) = 16°

B. m(arc BAC) = 344°

Step-by-step explanation:

From the picture attached,

AB is the diameter of the circle.

Since, measure of an arc is always same as the central angle of arc.

Therefore, m(arc AC) = m(∠ADC) = 164°

m(arc AB) = 180°

m(arc BC) = 360° - m(arc AB) + m(arc AC)

                = 360° - (180° + 164°)

                = 360° - 344°

                = 16°

m(arc BAC) = m(arc AB) + m(arc AC)

                   = 180° + 164°

                   = 344°

6 0
3 years ago
A new process for producing a type of resin is supposed to have a mean cycle time of 3.5 hours
jenyasd209 [6]

Answer:

a) Assuming 95% of confidence and \alpha=0.05 we can use the t distribution with 5 degrees of freedom in order to calculate a critical value that accumulates 0.05 of the area on the right tail of the distribution. We can use excel and the code to do this is given by: "=T.INV(1-0.05,5)". And we got the critical value t_{\alpha/2}=2.015.

Since our calculates value < critical value. We fail to reject the null hypothesis, and we can say that at 5% of significance we don't have enough evidence to conclude that the true mean is highr than 3.5 hours.

b) p_v =P(t_{5}>0.204)=0.423  

If we compare the p value and a significance level assumed \alpha=0.05 we see that p_v>\alpha so we can conclude that we FAIL to reject the null hypothesis, and the actual true mean for the time is not significantly higher than 3.5 hours at 5% of significance.  

Step-by-step explanation:

Data given and notation

We can calculate the sample mean and standard deviation with these formulas:

\bar X =\frac{\sum_{i=1}^n X_i}{n}

s =\sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}

The results obtained are given below:  

\bar X=3.507 represent the sample mean  

s=0.084 represent the standard deviation for the sample

n=6 sample size  

\mu_o =3.5 represent the value that we want to test  

\alpha represent the significance level for the hypothesis test.  

t would represent the statistic (variable of interest)  

p_v represent the p value for the test (variable of interest)  

State the null and alternative hypotheses to be tested  

We need to conduct a hypothesis in order to determine if the mean producing a type of resin is higher than 3.5 hours, the system of hypothesis would be:  

Null hypothesis:\mu \leq 3.5  

Alternative hypothesis:\mu > 3.5  

Compute the test statistic  

We don't know the population deviation, so for this case is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:  

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}} (1)  

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

We can replace in formula (1) the info given like this:  

t=\frac{3.507-3.5}{\frac{0.084}{\sqrt{6}}}=0.204  

Now we need to find the degrees of freedom for the t distirbution given by:

df=n-1=6-1=5

What do you conclude?  

a. Use the critical value approach.

Assuming 95% of confidence and \alpha=0.05 we can use the t distribution with 5 degrees of freedom in order to calculate a critical value that accumulates 0.05 of the area on the right tail of the distribution. We can use excel and the code to do this is given by: "=T.INV(1-0.05,5)". And we got the critical value t_{\alpha/2}=2.015.

Since our calculates value < critical value. We fail to reject the null hypothesis, and we can say that at 5% of significance we don't have enough evidence to conclude that the true mean is highr than 3.5 hours.

b. Use the p-value approach

Since is a one right tailed test the p value would be:  

p_v =P(t_{5}>0.204)=0.423  

If we compare the p value and a significance level assumed \alpha=0.05 we see that p_v>\alpha so we can conclude that we FAIL to reject the null hypothesis, and the actual true mean for the time is not significantly higher than 3.5 hours at 5% of significance.  

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Identifying roots - Pre calculus stuff
umka2103 [35]

The answers are 0 and 2

This can be found by looking at the x axis and finding every point where the line crosses

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4 years ago
If the price of 23 toys is Rs. 276, then what will the price (Rs.) of 12 toys?
nata0808 [166]

Answer:

price of 23 toys=Rs276

price of 1 toys=Rs276/23 =Rs 12

price of 12 toys = Rs 12*12 =Rs 144

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3 years ago
Which mode shows two equal expressions when X = 3 ?
mario62 [17]
Model H shows equal expressions
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3 years ago
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