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Morgarella [4.7K]
3 years ago
13

A box weighs 3 7/8 pounds. This box is 3/4

Mathematics
1 answer:
Maurinko [17]3 years ago
6 0

Answer:

3 1/8 pounds

Step-by-step explanation:

rewrite equation into seperate parts: 3+7/8-3/4. then solve fraction parts: 7/8-3/4 lcd would be 8, so 7/8-6/8 would be 1/8. ad the 3 and the answer is 3 1/8

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For each of the following situations, state whether the appropriate test would be a Independent Samples T test or Paired T Test,
Anvisha [2.4K]

Answer:

B. Paired T Test

B. H0: μd = 0 and Ha: μd < 0 where μd= mean difference in blood pressure (calculated as aspirin - placebo)

Step-by-step explanation:

As the experiment is based on the difference between the results of a pair, the most appropiate test is a paired t-test.

If the 80 people group had been dividided in 2 and then compare the means for each group, it would have been a independent samples t-test.

For each pair, a variable d (for difference) is calculated and used as the outcome sample to perform the test.

Then, the null hypothesis will state that the population mean for this difference is not significantly different from 0, and the alternative hypothesis claiming it is signficant different from 0.

The sample means for the treatment group and the placebo is not individually calculated, as only the difference for each pair is used as sample in a paired t-test.

7 0
4 years ago
Find the area of this parallelogram​
mixer [17]

Answer:

Area of this parallelogram​ A = base x height = 11 x 6 = 66 (cm2)

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5 0
3 years ago
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kirill [66]

Answer:

f=9

9 squared = 81

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Step-by-step explanation:

7 0
3 years ago
The given matrix is the augmented matrix for a linear system. Use technology to perform the row operations needed to transform t
shtirl [24]

Answer:

x_{1} = \frac{176}{127} + \frac{71}{127}x_{4}\\\\ x_{2} = \frac{284}{127} + \frac{131}{254}x_{4}\\\\x_{3} = \frac{845}{127} + \frac{663}{254}x_{4}\\

Step-by-step explanation:

As the given Augmented matrix is

\left[\begin{array}{ccccc}9&-2&0&-4&:8\\0&7&-1&-1&:9\\8&12&-6&5&:-2\end{array}\right]

Step 1 :

r_{1}↔r_{1} - r_{2}

\left[\begin{array}{ccccc}1&-14&6&-9&:10\\0&7&-1&-1&:9\\8&12&-6&5&:-2\end{array}\right]

Step 2 :

r_{3}↔r_{3} - 8r_{1}

\left[\begin{array}{ccccc}1&-14&6&-9&:10\\0&7&-1&-1&:9\\0&124&-54&77&:-82\end{array}\right]

Step 3 :

r_{2}↔\frac{r_{2}}{7}

\left[\begin{array}{ccccc}1&-14&6&-9&:10\\0&1&-\frac{1}{7} &-\frac{1}{7} &:\frac{9}{7} \\0&124&-54&77&:-82\end{array}\right]

Step 4 :

r_{1}↔r_{1} + 14r_{2} , r_{3}↔r_{3} - 124r_{2}

\left[\begin{array}{ccccc}1&0&4&-11&:-8\\0&1&-\frac{1}{7} &-\frac{1}{7} &:\frac{9}{7} \\0&0&- \frac{254}{7} &\frac{663}{7} &:-\frac{1690}{7} \end{array}\right]

Step 5 :

r_{3}↔\frac{r_{3}. 7}{254}

\left[\begin{array}{ccccc}1&0&4&-11&:-8\\0&1&-\frac{1}{7} &-\frac{1}{7} &:\frac{9}{7} \\0&0&1&-\frac{663}{254} &:-\frac{1690}{254} \end{array}\right]

Step 6 :

r_{1}↔r_{1} - 4r_{3} , r_{2}↔r_{2} + \frac{1}{7} r_{3}

\left[\begin{array}{ccccc}1&0&0&-\frac{71}{127} &:\frac{176}{127} \\0&1&0&-\frac{131}{254} &:\frac{284}{127} \\0&0&1&-\frac{663}{254} &:\frac{845}{127} \end{array}\right]

∴ we get

x_{1} = \frac{176}{127} + \frac{71}{127}x_{4}\\\\ x_{2} = \frac{284}{127} + \frac{131}{254}x_{4}\\\\x_{3} = \frac{845}{127} + \frac{663}{254}x_{4}\\

6 0
3 years ago
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lianna [129]
3•7=21 if times three times seven it equals twenty one
8 0
3 years ago
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