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2 years ago

Which Of the following could not be points on the units circle ?

Mathematics

2 answers:

ycow [4]2 years ago

8 0

Answer:

Options a and c

Step-by-step explanation:

A unit circle is one which has origin as centre and radius as 1.

The equation of the unit circle would be

$x^2+y^2 =1$

Thus we find that any point satisfying the above equation lies on the unit circle

Let us try the given points one by one

a) $(0.8)^2+(-0,6)^2 =1$ . Hence lies on the circle

b) (1,1)

Cannot lie on circle since $1^2+1^2 \neq 1$

c) $(\frac{-2}{3} ,\frac{\sqrt{5} }{3} )\\(\frac{-2}{3})^2+(\frac{\sqrt{5} }{3} )^2 =1$ . Hence lies on the circle

d) $(\frac{\sqrt{3} }{2} ,\frac{1}{3} )\\(\frac{\sqrt{3} }{2})^2+(\frac{1}{3} )^2\\\neq 1$

Hence does not lie on the circle

Only option A,C are right

Serjik [45]2 years ago

3 0

answer A , that could not be point

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A. 2^5 

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D. 2^6

Step-by-step explanation:

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Answer:

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Step-by-step explanation:

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2 years ago

A restaurant chain has two locations in a medium-sized town and, believing that it has oversaturated the market for its food, is

scoray [572]

Answer:

Yes, since the test statistic value is greater than the critical value.

Step-by-step explanation:

Data given and notation

$\bar X_{1}=360000$ represent the mean for the downtown restaurant

$\bar X_{2}=340000$ represent the mean for the freeway restaurant

$s_{1}=50000$ represent the sample standard deviation for downtown

$s_{2}=40000$ represent the sample standard deviation for the freeway

$n_{1}=36$ sample size for the downtown restaurant

$n_{2}=36$ sample size for the freeway restaurant

t would represent the statistic (variable of interest)

$\alpha=0.05$ significance level provided

Develop the null and alternative hypotheses for this study?

We need to conduct a hypothesis in order to check if the true mean of revenue for downtown is higher than for freeway restaurant, the system of hypothesis would be:

Null hypothesis: $\mu_{1}\leq \mu_{2}$

Alternative hypothesis: $\mu_{1} > \mu_{2}$

Since we don't know the population deviations for each group, for this case is better apply a t test to compare means, and the statistic is given by:

$t=\frac{\bar X_{1}-\bar X_{2}}{\sqrt{\frac{s^2_{1}}{n_{1}}+\frac{s^2_{2}}{n_{2}}}}$ (1)

t-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.

Determine the critical value.

Based on the significance level $\alpha=0.05$ and $\alpha/2=0.025$ we can find the critical values from the t distribution dith degrees of freedom df=36+36-2=55+88-2=70, we are looking for values that accumulates 0.025 of the area on the right tail on the t distribution.

For this case the value is $t_{1-\alpha/2}=1.66$

Calculate the value of the test statistic for this hypothesis testing.

Since we have all the values we can replace in formula (1) like this:

$t=\frac{360000-340000}{\sqrt{\frac{50000^2}{36}+\frac{40000^2}{36}}}}=1.874$

What is the p-value for this hypothesis test?

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

$p_v =P(t_{70}>1.874)=0.033$

Based on the p-value, what is your conclusion?

Comparing the p value with the significance level given $\alpha=0.05$ we see that $p_v$ so we can conclude that we reject the null hypothesis, and the true mean for the downtown revenue restaurant seems higher than the true mean revenue for the freeway restaurant.

The best option would be:

Yes, since the test statistic value is greater than the critical value.

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2 years ago

I’m supposed to find the slope intercept form. The x intercept is (2,0) and the y intercept is (0,6). I keep getting 6/2. But, t

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2 years ago

I would like to know the value of b and c

Ratling [72]

Answer:

a = 3, b = 8, c = 14

Step-by-step explanation:

First, we can organize this, putting lower values first.

2, 3, 3, 5, 9, 9, 11, 12

a , b , c somewhere there

Given this information, one thing we can look at first is the mode. The mode is 3, so there must be more 3s than any other number. Currently, there are 2 3s and 2 9s, so there must be at least one more 3 and no more 9s to make that true. Therefore, a, b, or c is 3. Therefore, we have

2, 3, 3, 3, 5, 9, 9, 11, 12

2 of a, b, c somewhere in there

Next, the median is 8. In our current state, the median is 5. There are 9 numbers, with 11 total including the 2 remaining values. Because there will be an odd amount of values, the median must be a number on the list. Therefore, our list is

2, 3, 3, 3, 5, 8, 9, 9, 11, 12

1 of a, b, c somewhere in there

There are 4 numbers above the median and 5 numbers below right now. To balance this out, there must be another number above the median. As a consequence, the remaining value must be greater than 8.

Finally, we know that the range is 12, so maximum - minimum = 12. Because the remaining number must be greater than 8, the minimum number is 2, no matter what. Therefore,

maximum - 2 = 12

add 2 to both sides to isolate maximum

maximum = 14

There is no 14 currently on the list, so the remaining value must be 14.

Our a, b, and c are as follows, in order from smallest to largest:

3, 8, 14

8 0

1 year ago