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Diano4ka-milaya [45]
3 years ago
7

Pat needs boards that are one half foot long. Which equation shows how many one half foot pieces he can get from a four foot lon

g board
Mathematics
2 answers:
andrezito [222]3 years ago
7 0

Answer:

4/.5 = 8

Step-by-step explanation:

You can solve this by adding up .5 8 times to equal 4. Another way to look at it is that you have 4 individual pieces that are a foot long and you decide to split all of them in half. Let me know if you have any other questions.

Paraphin [41]3 years ago
3 0

Answer:

The equation that shows that is the division between 4 and 0.5 as:

\frac{4}{0.5}

Step-by-step explanation:

First, it is necessary to know that one half foot is equivalent to 0.5 foot.

We can solve this using a rule of three in which we know that 1 piece has one half foot long, then 4 foot long how many pieces have. This is:

1 board -----------  0.5 foot

 X        ------------  4 foot

Where X is the number of pieces that he can get from a four foot long board.

Solving for X, we get:

X  = \frac{4*1}{0.5} = \frac{4}{0.5}

So, the equation that shows how many one half foot pieces he can get from a four foot long board is:

X=\frac{4}{0.5}

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Step-by-step explanation:

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If a test of H subscript 0 colon space mu subscript D equals 0 space v s. space H subscript a colon space space mu subscript D g
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Answer:

p_v =P(t_{(n-1)}>t_{calculated}) =0.0601

The p value on this case is given  by the problem.

If we compare the p value with a significance level assumed \alpha=0.05, we see that p_v > \alpha and we can conclude that we FAIL to reject the null hypothesis that the difference mean between after and before is less or equal than 0.

Step-by-step explanation:

A paired t-test is used to compare two population means where you have two samples in  which observations in one sample can be paired with observations in the other sample. For example  if we have Before-and-after observations (This problem) we can use it.  

Let put some notation  

x=test value before , y = test value after

The system of hypothesis for this case are:

Null hypothesis: \mu_y- \mu_x \leq 0

Alternative hypothesis: \mu_y -\mu_x >0

The first step is calculate the difference d_i=y_i-x_i

The second step is calculate the mean difference  

\bar d= \frac{\sum_{i=1}^n d_i}{n}

The third step would be calculate the standard deviation for the differences, and we got:

s_d =\frac{\sum_{i=1}^n (d_i -\bar d)^2}{n-1}

The 4 step is calculate the statistic given by :

t=\frac{\bar d -0}{\frac{s_d}{\sqrt{n}}}=t_{calculated}

The next step is calculate the degrees of freedom given by:

df=n-1

Now we can calculate the p value, since we have a right tailed test the p value is given by:

p_v =P(t_{(n-1)}>t_{calculated}) =0.0601

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If we compare the p value with a significance level assumed \alpha=0.05, we see that p_v > \alpha and we can conclude that we FAIL to reject the null hypothesis that the difference mean between after and before is less or equal than 0.

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3 years ago
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