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melamori03 [73]
3 years ago
7

. Subtract and simplify. 1 6 m - (-5 8 m + 1 3 )

Mathematics
1 answer:
Diano4ka-milaya [45]3 years ago
3 0

Answer:

<em>74m</em><em> </em><em>-</em><em> </em><em>13</em>

<em><u>Step-by-step explanation:</u></em>

<em>1</em><em>)</em><em> </em><em>Remove</em><em> </em><em>parentheses</em><em>. </em>

<em>16m</em><em> </em><em>+</em><em> </em><em>5</em><em>8</em><em>m</em><em> </em><em>-</em><em> </em><em>13</em>

<em>2</em><em>)</em><em> </em><em>Collect</em><em> </em><em>like</em><em> </em><em>terms</em><em>. </em>

<em>(</em><em>16m</em><em> </em><em>+</em><em> </em><em>58m</em><em>)</em><em> </em><em>-</em><em> </em><em>13</em>

<em>3</em><em>)</em><em> </em><em>Simplify</em><em>.</em>

<em>74m</em><em> </em><em>-</em><em> </em><em>13</em>

<em><u>Therefor</u></em><em><u>,</u></em><em><u> </u></em><em><u>the</u></em><em><u> </u></em><em><u>answer</u></em><em><u> </u></em><em><u>is</u></em><em><u> </u></em><em><u>74m</u></em><em><u> </u></em><em><u>-</u></em><em><u> </u></em><em><u>13</u></em><em><u>.</u></em>

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A high-tech company wants to estimate the mean number of years of college education its employees have completed. A sample of 15
valentinak56 [21]

Answer:

4 - 2.14 \frac{0.7}{\sqrt{15}}=3.61

4 + 2.14 \frac{0.7}{\sqrt{15}}=4.39

The 95% confidence interval is given by (3.61;4.39)

Step-by-step explanation:

Notation and definitions

n=15 represent the sample size

\bar X=4 represent the sample mean  

s=0.7 represent the sample standard deviation

m represent the margin of error

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

Calculate the critical value tc

In order to find the critical value is important to mention that we don't know about the population standard deviation, so on this case we need to use the t distribution. Since our interval is at 95% of confidence, our significance level would be given by \alpha=1-0.95=0.05 and \alpha/2 =0.025. The degrees of freedom are given by:  

df=n-1=15-1=14  

We can find the critical values in excel using the following formulas:  

"=T.INV(0.025,14)" for t_{\alpha/2}=-2.14  

"=T.INV(1-0.025,14)" for t_{1-\alpha/2}=2.14  

The critical value tc=\pm 2.14

Calculate the margin of error (m)

The margin of error for the sample mean is given by this formula:

m=t_c \frac{s}{\sqrt{n}}

m=2.14 \frac{0.7}{\sqrt{15}}=0.387

Calculate the confidence interval  

The interval for the mean is given by this formula:

\bar X \pm t_{c} \frac{s}{\sqrt{n}}

And calculating the limits we got:

4 - 2.14 \frac{0.7}{\sqrt{15}}=3.61

4 + 2.14 \frac{0.7}{\sqrt{15}}=4.39

The 95% confidence interval is given by (3.61;4.39)

4 0
3 years ago
Hello there!
Usimov [2.4K]

firstly chain rule is dy/dx = dy/du * du/dx

a) y = 5u² + u - 1 hence dy/du = 10u +1

u = 3x +1 hence du/dx = 3

so dy/dx = 3(10u +1) {sub in u and expand}

b) y= u^-2 hence dy/du = -2u^-3

u=2x+3 hence du/dx = 2

so dy/dx= 2(-2u^-3) {sub in u}

8 0
3 years ago
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