Construct two chords BC and DE that intersect in the interior of a circle at point F

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Evgesh-ka [11]

There seems to be a flaw with this question because it says that there are five x-intercepts but the given information only gives you 4 x-intercepts to work with.

Even means the graph is symmetric about the y-axis

The best answer is <span>A.(–6, 0), (–2, 0), and (0, 0)

because you do not have to worry about another point (0,0). Plus we need (-6,0) for it to be symmetric with (6,0).

Consider function f(x) = x²(x-6)(x+6)(x+2)</span>²(x-2)<span>². It is even and fits these conditions as it has x-intercepts at (6,0), (-6,0), (-2,0), (2,0), and (0,0). again, the question does not tell us the fifth x-intercept, so we need to assume that there is another one that needs to be there...and so (-2,0) must have (2,0) for it to be even as well.</span>

4 0

2 years ago

Read 2 more answers

Some scientists believe alcoholism is linked to social isolation. One measure of social isolation is marital status. A study of

frez [133]

Answer:

1) H0: There is independence between the marital status and the diagnostic of alcoholic

H1: There is association between the marital status and the diagnostic of alcoholic

2) The statistic to check the hypothesis is given by:

$\sum_{i=1}^n \frac{(O_i -E_i)^2}{E_i}$

3) $\chi^2 = \frac{(21-33.143)^2}{33.143}+\frac{(37-41.429)^2}{41.429}+\frac{(58-41.429)^2}{41.429}+\frac{(59-46.857)^2}{46.857}+\frac{(63-58.571)^2}{58.571}+\frac{(42-58.571)^2}{58.571} =19.72$

4) $df=(rows-1)(cols-1)=(3-1)(2-1)=2$

And we can calculate the p value given by:

$p_v = P(\chi^2_{2} >19.72)=5.22x10^{-5}$

And we can find the p value using the following excel code:

"=1-CHISQ.DIST(19.72,2,TRUE)"

Since the p value is lower than the significance level so then we can reject the null hypothesis at 5% of significance, and we can conclude that we have association between the two variables analyzed.

Step-by-step explanation:

A chi-square goodness of fit test "determines if a sample data matches a population".

A chi-square test for independence "compares two variables in a contingency table to see if they are related. In a more general sense, it tests to see whether distributions of categorical variables differ from each another".

Assume the following dataset:

Diag. Alcoholic Undiagnosed Alcoholic Not alcoholic Total

Married 21 37 58 116

Not Married 59 63 42 164

Total 80 100 100 280

Part 1

We need to conduct a chi square test in order to check the following hypothesis:

H0: There is independence between the marital status and the diagnostic of alcoholic

H1: There is association between the marital status and the diagnostic of alcoholic

The level os significance assumed for this case is $\alpha=0.05$

Part 2

The statistic to check the hypothesis is given by:

$\sum_{i=1}^n \frac{(O_i -E_i)^2}{E_i}$

Part 3

The table given represent the observed values, we just need to calculate the expected values with the following formula $E_i = \frac{total col * total row}{grand total}$

And the calculations are given by:

$E_{1} =\frac{80*116}{280}=33.143$

$E_{2} =\frac{100*116}{280}=41.429$

$E_{3} =\frac{100*116}{280}=41.429$

$E_{4} =\frac{80*164}{280}=46.857$

$E_{5} =\frac{100*164}{280}=58.571$

$E_{6} =\frac{100*164}{280}=58.571$

And the expected values are given by:

Diag. Alcoholic Undiagnosed Alcoholic Not alcoholic Total

Married 33.143 41.429 41.429 116

Not Married 46.857 58.571 58.571 164

Total 80 100 100 280

And now we can calculate the statistic:

$\chi^2 = \frac{(21-33.143)^2}{33.143}+\frac{(37-41.429)^2}{41.429}+\frac{(58-41.429)^2}{41.429}+\frac{(59-46.857)^2}{46.857}+\frac{(63-58.571)^2}{58.571}+\frac{(42-58.571)^2}{58.571} =19.72$

Part 4

Now we can calculate the degrees of freedom for the statistic given by:

$df=(rows-1)(cols-1)=(3-1)(2-1)=2$

And we can calculate the p value given by:

$p_v = P(\chi^2_{2} >19.72)=5.22x10^{-5}$

And we can find the p value using the following excel code:

"=1-CHISQ.DIST(19.72,2,TRUE)"

Since the p value is lower than the significance level so then we can reject the null hypothesis at 5% of significance, and we can conclude that we have association between the two variables analyzed.

7 0

3 years ago

The mayor is interested in finding a 98% confidence interval for the mean number of pounds of trash per person per week that is

saul85 [17]

Solution :

Given :

Sample mean, $\overline X = 34.2$

Sample size, n = 129

Sample standard deviation, s = 8.2

a. Since the population standard deviation is unknown, therefore, we use the t-distribution.

b. Now for 95% confidence level,

α = 0.05, α/2 = 0.025

From the t tables, T.INV.2T(α, degree of freedom), we find the t value as

t =T.INV.2T(0.05, 128) = 2.34

Taking the positive value of t, we get

Confidence interval is ,

$\overline X \pm t \times \frac{s}{\sqrt n}$

$34.2 \pm 2.34 \times \frac{8.2}{\sqrt {129}}$

(32.52, 35.8)

95% confidence interval is (32.52, 35.8)

So with $95 \%$ confidence of the population of the mean number of the pounds per person per week is between 32.52 pounds and 35.8 pounds.

c. About $95 \%$ of confidence intervals which contains the true population of mean number of the pounds of the trash that is generated per person per week and about $5 \%$ that doe not contain the true population of mean number of the pounds of trashes generated by per person per week.

4 0

2 years ago

Wendell is looking over some data regarding the strength, measured in Pascals (Pa), of some building materials and how the stren

Slav-nsk [51]

The logarithmic model for the length when the strength is of 8 Pascals is given by:

$f^{-1}(8) = \log_{2}{8} = \log_2{2^3} = 3$
That is, the length is of 3 units.

<h3>What is the function?</h3>

The strength in Pascals for a building of length x is given by:

$f(x) = 2^x$

To find the length given the strength, we apply the inverse function, that is:

$2^y = x$

$\log_{2}{2^y} = \log_2{x}$

$y = \log_2{x}$

Hence, when the strength is of 8 Pascals, $x = 8$ , and the length is given by:

$f^{-1}(8) = \log_{2}{8} = \log_2{2^3} = 3$

You can learn more about logarithmic functions at brainly.com/question/25537936

6 0

2 years ago

Aneesha travels at a rate of 50 miles per hour. Morris is traveling 3 feet per second less than Aneesha. Which is the best estim

goldfiish [28.3K]

If Aneesha travels at a rate of
50 miles............................1h
and 1mile=5280ft, therefore 50 miles = 264000 ft
and 1 h =3600 seconds we can say that Aneesha travels

264,000 ft in 3600 seconds or
264,000/3600= 73.33 ft/s

Morris travels 3 ft per second less than her
so 73.33-3=70.33 ft/s for Morris

3 0

3 years ago