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xz_007 [3.2K]
3 years ago
6

-7(-5v-3u 4) use the distributive property to remove the parentheses

Mathematics
1 answer:
Marrrta [24]3 years ago
3 0
Distributeiv poepty
a(b+c)=ab+ac

remember that (-) times (-)=(+)
and (+) TIMES (-)=(-)

SO
<span>-7(-5v-3u+4)=(-7)(-5v)+(-7)(-3u)+(-7)(4)=35v+21u-28</span>
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Jobisdone [24]

Answer:

x = 2

y = 3

Step-by-step explanation:

\left \{ {{0.4x + 0.3y = 1.7} \atop {0.7x - 0.2y = 0.8}} \right.

\left \{ {{2(0.4x + 0.3y) = 2 * 1.7} \atop {3(0.7x - 0.2y) = 3 * 0.8}} \right.

\left \{ {{0.8x + 0.6y = 3.4} \atop {2.1x - 0.6y) = 2.4}} \right.

2.9x = 5.8

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0.3y + 0.4 * 2 = 1.7

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4 0
3 years ago
Which quadratic function is represented by the graph? 3 O y = 0.5(x + 2)2 + 4 O y = 0.5(x + 3)2 -0.5 O y = 0.5(x - 3)2 -0.5 O y
nadya68 [22]

Answer:

No solution is possible from the given information.

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Step-by-step explanation:

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3 years ago
Single-sample t test and paid days off: The number of paid days off (e.g., vacation, sick leave) taken by eight employees at a s
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Answer:

A. µ is the population average number of paid days off taken by employees that the company.

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D. t=-0.79178; df=7

Step-by-step explanation:

For the null and alternate hypotheses, you want to first define what µ represents. Next, you state them, the null hypothesis being equal to the previously known average, and the alternate hypothesis being greater than, less than, or not equal to the previous average, selecting one depends on the context in the problem.

A one-mean t-test would be used as we are looking to compare the mean of a data set, as well as the fact that we do not know the population standard deviation.

When conducting these tests, we need to ensure that three conditions are being met. The first is random, which means that the data set is randomly gathered, or not, the data here does not seem to be random, which may be concerning. Next is independence, this is done when we survey less than 10% of the overall population, in this case it is a small company, so we do not know if it is less than 10% of the population. Last is normality, the data set is not sufficiently large (greater than 30 people surveyed) so we cannot use the central limit theorem to justify that the data is normal. We can use a normality plot, but when the data is placed on a normality plot most of the data appears to be linear, but the 27 day data point does not seem to be normal, so we cannot fully ensure that it is normal. Based on the data not following these conditions, we have concerns about proceeding with the test, we will therefore have to proceed with caution.

For the last part, use that T-test function on a calculator with statistics functions. Remember to include the degrees of freedom in the answer. (The degrees of freedom is one less than the sample size).

6 0
4 years ago
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