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Yakvenalex [24]
2 years ago
10

3. A fisherman scoops water out of his boat after a rainstorm. If the container he uses to scoop water can hold 2.1 liters, whic

h is closest to the total amount of water, in liters, that he will remove with 8 scoops? 4 6 10 16 PREVIOUS
Mathematics
1 answer:
Scorpion4ik [409]2 years ago
6 0
Answer: 16
explanation: 2.1 x 8 = 16.8
16 is the closest to 16.8
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A or \frac{1}{2}x +2

Step-by-step explanation:

Remember y=mx+b | m = slope & b = y intercept

Think back of rise over run. Where the rise is 1 and 2 is run. Now Also, lets not forget the y intercept which you can easily obtain by looking where the line intercepts the y axis. Which in this case it would be 2.

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Help me please Asap
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it is the last one {-4, -1, 0, 1, 2}

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Find the area of each figure in square units.
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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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Molodets [167]

Answer:

65 degrees

Step-by-step explanation:

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