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3 years ago

What's the difference between 1261⁄4 and 78 2⁄3?

Mathematics

2 answers:

lubasha [3.4K]3 years ago

4 0

126 1/4 - 78 2/3 =
505/4 - 236/3 =
1515/12 - 944/12 =
571/12 =
47 7/12 <==

Alisiya [41]3 years ago

4 0

Answer: The answer is (B) $47\dfrac{7}{12}$ .

Step-by-step explanation: We are given to find the difference between the following two numbers:

$n_1=126\dfrac{1}{4},~~~n_2=78\dfrac{2}{3}.$

To find the difference, we need to subtract the smaller number from the larger one.

Since, $n_1>n_2$ , so the difference is given by

$d\\\\=n_1-n_2\\\\=126\dfrac{1}{4}-78\dfrac{2}{3}\\\\=\dfrac{505}{4}-\dfrac{236}{3}\\\\=\dfrac{505\times3-236\times4}{12}\\\\=\dfrac{1515-944}{12}\\\\=\dfrac{571}{12}\\\\=47\dfrac{7}{12}.$

Thus, (B) $47\dfrac{7}{12}$ is the correct option.

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The answer to your question is:

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Answer:

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Step-by-step explanation:

Data given

$\bar X_{1}=810$ represent the mean for sample 1

$\bar X_{2}=770$ represent the mean for sample 2

$s_{1}=67$ represent the sample standard deviation for 1

$s_{2}=56$ represent the sample standard deviation for 2

$n_{1}=13$ sample size for the group 2

$n_{2}=19$ sample size for the group 2

$\alpha=0.05$ Significance level provided

t would represent the statistic (variable of interest)

Concepts and formulas to use

We need to conduct a hypothesis in order to check if the mean for category 1 is higher than the mean for category 2, the system of hypothesis would be:

Null hypothesis: $\mu_{1}-\mu_{2}\leq 0$

Alternative hypothesis: $\mu_{1} - \mu_{2}> 0$

We don't have the population standard deviation's, so for this case is better apply a t test to compare means, and the statistic is given by:

$t=\frac{(\bar X_{1}-\bar X_{2})-\Delta}{\sqrt{\frac{\sigma^2_{1}}{n_{1}}+\frac{\sigma^2_{2}}{n_{2}}}}$ (1)

And the degrees of freedom are given by $df=n_1 +n_2 -2=13+19-2=30$

t-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.

With the info given we can replace in formula (1) like this:

$t=\frac{(810-770)-0}{\sqrt{\frac{67^2}{13}+\frac{56^2}{19}}}}=1.771$

P value

Since is a right tailed test the p value would be:

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