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Dmitriy789 [7]
3 years ago
10

2/13 divided by 1/6 need help

Mathematics
2 answers:
tia_tia [17]3 years ago
8 0

Dividing by a fraction means the same as multiplying by its reciprocal.

In other words, change the division sign to multiplication

and flip or take the reciprocal of the second fraction.

So we can rewrite 2/13 ÷ 1/6 as 2/13 × 6/1.

Now multiply across the numerators and denominators to get 12/13.

inessss [21]3 years ago
4 0

Answer:

Step-by-step explanation:

KCF = Keep Change Flip

2/3 divided by 1/6

K        C             F

2/3 you keep

divided you change it to Multiplication X

And Flip 1/6 to 6/1

then Multiply

2/3 x 6/1 =

12/3 = 4

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Answer:

There is enough evidence to claim that population mean 1 is less than population 2  at the 0.05 significance level.

Step-by-step explanation:

We are given the following in the question:

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Group 2:

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Alpha, α = 0.05

First, we design the null and the alternate hypothesis

H_{0}: \mu_1 \geq \mu_2\\H_A: \mu_1 < \mu_2

Since, the population variances are equal and that the two populations are normally distributed, we use t-test(pooled test) for difference of two means.

Formula:

Pooled standard deviation

s_p = \sqrt{\displaystyle\frac{(n_1-1)\sigma_1^2 + (n_2-1)\sigma_2^2 }{n_1 + n_2 - 2}}

t_{stat} = \displaystyle\frac{\mu_1-\mu_2}{s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}

\text{Degree of freedom} = n_1 + n_2 - 2

Putting all the values we get:

s_p = \sqrt{\displaystyle\frac{(11-1)(3.7)^2 + (13-1)(6.6)^2 }{11 + 13 - 2}} = \sqrt{29.9827} = 5.48

t_{stat} = \displaystyle\frac{63.3-70.2}{5.48\sqrt{\frac{1}{11}+\frac{1}{13}}} = -3.073

\text{Degree of freedom} = 11 + 13 - 2 = 22

Now,

t_{critical} \text{ at 0.05 level of significance, 22 degree of freedom } = -1.717

Since,

t_{stat} < t_{critical}

We fail to accept the null hypothesis and reject the null hypothesis. We accept the alternate hypothesis.

We conclude that the mean score for Group 1 is significantly lower than the mean score for Group 2.

There is enough evidence to claim that population mean 1 is less than population 2  at the 0.05 significance level.

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