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3 years ago

Pleaseee helpp i have 20 mins for a unit testt

Mathematics

1 answer:

TEA [102]3 years ago

4 0

Answer: Its c

Step-by-step explanation:

I think

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Which statement is true about the polynomial 3x2y2 − 5xy2 − 3x2y2 + 2x2 after it has been fully simplified? It has 2 terms and a

Nitella [24]

Answer:

<em>B. It has 2 terms and a degree of 3.</em>

Step-by-step explanation:

There's a quizlet that has this exact question.

4 0

3 years ago

Read 2 more answers

PLEASE HELP. Is this graph parallel, perpendicular or neither?

algol [13]

Answer:

The lines are parallel

Step-by-step explanation:

When lines are parallel they travel side-by-side without ever crossing paths.

3 0

3 years ago

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How do I find the solution to this using the substitution method? Please help!

Sonbull [250]

Answer:

y= -1; x=5

Step-by-step explanation:

Replace the x in the first equation with x=1-4y

2(1-4y)+7y=3

2-8y+7y=3

-y=1

y= -1

x=1-4(-1))

x=5

Hope this helps

6 0

4 years ago

Note: Figure is not drawn to scale.

Paraphin [41]

Answer:

Step-by-step explanation:

No explanation

5 0

3 years ago

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The ages of 11 students enrolled in an on-line macroeconomics course are given in the following steam-and-leaf display:

blsea [12.9K]

Answer:

The standard deviation of the age distribution is 6.2899 years.

Step-by-step explanation:

The formula to compute the standard deviation is:

$SD=\sqrt{\frac{1}{n}\sum\limits^{n}_{i=1}{(x_{i}-\bar x)^{2}}}$

The data provided is:

X = {19, 19, 21, 25, 25, 28, 29, 30, 31, 32, 40}

Compute the mean of the data as follows:

$\bar x=\frac{1}{n}\sum\limits^{n}_{i=1}{x_{i}}$

$=\frac{1}{11}\times [19+19+21+...+40]\\\\=\frac{299}{11}\\\\=27.182$

Compute the standard deviation as follows:

$SD=\sqrt{\frac{1}{n}\sum\limits^{n}_{i=1}{(x_{i}-\bar x)^{2}}}$

$=\sqrt{\frac{1}{11-1}\times [(19-27.182)^{2}+(19-27.182)^{2}+...+(40-27.182)^{2}]}}\\\\=\sqrt{\frac{395.6364}{10}}\\\\=6.28996\\\\\approx 6.2899$

Thus, the standard deviation of the age distribution is 6.2899 years.

7 0

3 years ago