Can i have some help please ?!?!

Pls help

KATRIN_1 [288]

Answer:

169.04 in² (nearest hundredth)

Step-by-step explanation:

Surface area of a cone = $\pi$ r² + $\pi$ r $l$

(where r = radius of the base and $l$ = slant height)

Given slant height $l$ = 10 and surface area = 188.5

Surface area = $\pi$ r² + $\pi$ r $l$

188.5 = $\pi$ r² + 10 $\pi$ r

$\pi$ r² + 10 $\pi$ r - 188.5 = 0

r = $\frac{-10\pi +\sqrt{(10\pi )^2-(4\times\pi \times-188.5)} }{2\pi }$ = 4.219621117...

Volume of a cone = (1/3) $\pi$ r²h

(where r = radius of the base and h = height)

We need to find an expression for h in terms of $l$ using Pythagoras' Theorem a² + b² = c², where a = radius, b = height and c = slant height

r² + h² = $l$ ²

h² = $l$ ² - r²

h = √( $l$ ² - r²)

Therefore, substituting found expression for h:

volume of a cone = (1/3) $\pi$ r²√( $l$ ² - r²)

Given slant height $l$ = 10 and r = 4.219621117...

volume = 169.0431969... = 169.04 in² (nearest hundredth)

5 0

2 years ago

Read 2 more answers

A yearbook printer charges based on the number of pages printed. Here is a table that shows the cost of some recent yearbooks:

asambeis [7]

y = 0.012x + 0.65

Number of pages 50.00 100.00 150.00 200.00 
Cost 1.25 1.85 2.45 3.05

x = number of pages
y = cost

intervals of x = 50 pages
intervals of y = 0.60
0.60 / 50 = 0.012

y = 0.012x + 0.65

 x 0.012*x y 
 0.012 50 0.60 0.65 1.25
0.012 100 1.20 0.65 1.85
 0.012 150 1.80 0.65 2.45
 0.012 200 2.40 0.65 3.05

5 0

3 years ago

A stack of books is 21 inches tall. Each book is 1 3/4 inches. How many books are there?

Maru [420]

There are twelve books

3 0

3 years ago

Read 2 more answers

You spin a spinner with four equal sections labeled 1, 2, 3, and 4 and toss a dime.

PIT_PIT [208]

Its : (1, H), (2, H), (3, H), (4, H), (1, T), (2, T), (3, T), (4, T) When you flip a coin, you can result in heads (H) or tails (T) when you spin a spinner with 4 equal sections, you get 1, 2, 3, or 4 (1, 2, 3, 4) Therefore the 8 combinations are above.

Hope this helps :)

8 0

3 years ago

Read 2 more answers

Luda [366]

Answer:

a) H0: There is no association between level of education and TV station preference (Independence)

H1: There is association between level of education and TV station preference (No independence)

b) $\chi^2 = \frac{(15-10)^2}{10}+\frac{(15-20)^2}{20}+\frac{(10-10)^2}{10}+\frac{(5-10)^2}{10}+\frac{(25-10)^2}{10}+\frac{(10-20)^2}{20} =33.75$

c) $\chi^2_{crit}=5.991$

d) Since the p value is lower than the significance level we enough evidence to reject the null hypothesis at 5% of significance, and we can conclude that we have dependence 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:

High school Some College Bachelor or higher Total

Public Broadcasting 15 15 10 40

Commercial stations 5 25 10 40

Total 20 40 20 80

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

Part a

H0: There is no association between level of education and TV station preference (Independence)

H1: There is association between level of education and TV station preference (No independence)

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

The statistic to check the hypothesis is given by:

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

Part b

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:

$E_{1} =\frac{20*40}{80}=10$

$E_{2} =\frac{40*40}{80}=20$

$E_{3} =\frac{20*40}{80}=10$

$E_{4} =\frac{20*40}{80}=10$

$E_{5} =\frac{40*40}{80}=20$

$E_{6} =\frac{20*40}{80}=10$

And the expected values are given by:

High school Some College Bachelor or higher Total

Public Broadcasting 10 20 10 40

Commercial stations 10 10 20 40

Total 20 30 30 80

Part b

And now we can calculate the statistic:

$\chi^2 = \frac{(15-10)^2}{10}+\frac{(15-20)^2}{20}+\frac{(10-10)^2}{10}+\frac{(5-10)^2}{10}+\frac{(25-10)^2}{10}+\frac{(10-20)^2}{20} =33.75$

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

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

Part c

In order to find the critical value we need to look on the right tail of the chi square distribution with 2 degrees of freedom a value that accumulates 0.05 of the area. And this value is $\chi^2_{crit}=5.991$

Part d

And we can calculate the p value given by:

$p_v = P(\chi^2_{3} >33.75)=2.23x10^{-7}$

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

"=1-CHISQ.DIST(33.75,2,TRUE)"

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

7 0

3 years ago