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Mrrafil [7]
3 years ago
13

A circle has a radius of 7x 9 y 5 cm. The area of a circle can be found using A = πr 2 . What is the area of this circle in squa

re centimeters.
Mathematics
1 answer:
Natalka [10]3 years ago
7 0

Answer:

hi

Step-by-step explanation:

Area of a circle  

A

=

π

r

2

Given  

r

=

6

x

9

y

5

c

m

A

=

π

(

6

x

9

y

5

)

2

=

36

π

x

18

y

10

s

q

c

m

A

=

113.0973

x

18

y

10

c

m

2

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25

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Find the slope of the graph of the following: 9x - 3y =15 a. -3 b. 3 c. -1/3 d. 1/3 ***
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To solve this you just need to convert this equation to slope intercept form to do this you just need to isolate y. To isolate -3y subtract 9x from both sides and your problem should look something like this 3y = -9x + 15 the -9 is the slope but you need to simplify that in order to do that divide both sides of the problem by 3 now your problem should look like this y = 3x + 5. The answer is B or positive 3
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A person drilled a hole in a die and filled it with a lead​ weight, then proceeded to roll it 200 times. Here are the observed f
Alona [7]

Solution :

The objective of the study is to test the claim that the loaded die behaves at a different way than a fair die.

Null hypothesis, $H_0:p_1=p_1=p_3=p_4=p_5=p_6=\frac{1}{6}$

That is the loaded die behaves as a fair die.

Alternative hypothesis, $H_a$ : loaded die behave differently than the fair die.

Number of attempts , n = 200

Expected frequency, $E_i=np_i$

                                        $=200 \times \frac{1}{6} = 33.333$

Test statistics, $x^2= \sum^6_{i=1} \frac{(O_i-E_i)^2}{E_i} $

                            $=\frac{(28-33.333)^2}{33.333}+\frac{(29-33.333)^2}{33.333}+\frac{(40-33.333)^2}{33.333}+\frac{(41-33.333)^2}{33.333}+\frac{(28-33.333)^2}{33.333}+$$\frac{(34-33.333)^2}{33.333}$

≈ 5.8

Degrees of freedom, df = n - 1

                                       = 6 - 1

                                       = 5

Level of significance, α = 0.10

At α = 0.10 with df = 5, the critical value from the chi square table

$x^2_{\alpha}= \text{chi inv}(0.10,5)$

     = 9.236

Thus the critical value is $x_{\alpha}^2=9.236$

$P \text{ value} = P[x^2_{df} \geq x^2]$

             $=P[x^2_5\geq 5.80]$

             = chi dist (5.80, 5)

             = 0.3262

Decision : The value of test statistics 5.80 is not greater than the critical value 9.236, thus fail to reject $H_o$ at 10% LOS.

Conclusion : There is no enough evidence to support the claim that the loaded die behave in a different than a fair die.

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olchik [2.2K]

Answer:

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Step-by-step explanation:

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