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Ratling [72]
3 years ago
12

A water station table at a marathon starts with 200 plastic cups on it. Every minute, 15 cups are taken from the table. How many

minutes could have passed if there are fewer than 50 cups on the table?
Mathematics
1 answer:
Bogdan [553]3 years ago
7 0

Answer:

is dancing with the circus grows and the mongooses you join us for the circus grows and the mongooses you join us for the circus grows and the mongooses you

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Find the slope of the line that passes through (1,3) and (8,9)​
Volgvan

Answer:

6/7

Step-by-step explanation:

4 0
3 years ago
I need help with this please.
mote1985 [20]

Answer:

37.5

Step-by-step explanation:

It uses Sin and the law of Sin says that the ratio of given sides and an angle is always constant

3 0
4 years ago
Read 2 more answers
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.

3 0
3 years ago
Write the word sentence as an equation.
Vitek1552 [10]
Y divided by 6 = 11.
5 0
3 years ago
A simple question I can't answer thnks yall
DENIUS [597]
The factors would be (x-1)(9x+7)

so the answer would be A
6 0
3 years ago
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