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exis [7]
3 years ago
9

Express 0.208 (the 8 is repeating) as a fraction in simplest form. 10 POINT AWARD

Mathematics
1 answer:
den301095 [7]3 years ago
5 0
Let x=0.208bar or 0.20888...

1000x=208.888 and 100x=20.888 so

1000x-100x=208.888-20.888

900x=188

x=47/225
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Cone A is divided into Cone B and Frustum C. B A 12cm C 20cm 20cm Express the volume of Frustum C as a fraction of the volume of
zaharov [31]

The volume of the frustum is volume of the whole cone(A) minus the smaller cone(B) which is would give the volume of frustum(C) = 256cm³

<h3>Calculation of a frustum</h3>

The volume of cone A V=πr²h/3

Where radius = 20cm

The volume of cone B = V=πr²h/3

Where radius = 12cm

Therefore volume of frustum =

V=π * 20² * h/3 - π * 12² *h/3

The variables will cancel out each other

V = 20² - 12²

V = 400- 144

V = 256cm³

Therefore, the volume of the frustum(C) = 256cm³

Learn more about cone here:

brainly.com/question/1082469

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7 0
2 years ago
The equation of the graph line is 2X-3Y=12
Klio2033 [76]

Answer:

A. -4

Step-by-step explanation:

For solving for x intercepts analytically. You can set the the y in the equation to 0. So, 2x-3(0)=12, and solving for x will get you -4.

You can also solve graphically by plugging in the equation and looking at where it intercepts the x axis.

3 0
3 years ago
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What is the value of x?<br> (3x+ 30)<br> 20, 15, 25, 10
slamgirl [31]

The value of <em>x </em>in the equation, (3x + 30), is 10.

5 0
3 years ago
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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
Find the vertex of this parabola:<br> y = -2x2 + 4x - 10
amm1812
The parabola equation in its vertex form is y = a(x-h)² + k , where:
a is the same as the a coefficient in the standard form;
h is the x-coordinate of the parabola vertex; and.
k is the y-coordinate of the parabola vertex.
that’s how you find it
3 0
3 years ago
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