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3 years ago

Simplify the following.

Mathematics

1 answer:

GalinKa [24]3 years ago

8 0

Answer:

(2f-17h)/24k

Step-by-step explanation:

We can make the common denominator for 3k and 8k in the form of 24k, so we get (multiply the first fraction by 8 and the second one by 3 to get 24k on each denominator) (8f-32h)/3k-(6f-15h)/24k. Now we can just subtract the numerators :

8f-32h-6f+15h (sign gets flipped on 6f and 15h), so we get

(2f-17h)/24k

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Evaluate the expression when m = -2

yarga [219]

Answer:

-22

Step-by-step explanation:

(Is that Aleks?)

Plug it in.

(-2)^2 + 9(-2) - 8

= 4 + (-18) - 8

= -14 - 8

= -22

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3 years ago

Is this set of ordered pAirs a function <br> (0,2) (3,3) (8,7) (2,2) (3,9)

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No, not a function

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3 years ago

There were 90 boys at the Hammarskjold school dance. They made up 60% of all of the students there. How many total students were

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Step-by-step explanation:

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3 years ago

Select the expression this is equivalent to (x+6)3 + (x +6)2 (x +2) help me asap

marusya05 [52]

Answer:

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Step-by-step explanation:

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3 years ago

Listed below are the number of years it took for a random sample of college students to earn bachelor's degrees (based on data f

schepotkina [342]

Answer:

a) Mean = 6.5, sample standard deviation = 3.50

b) Standard error = 0.7826

c) Point estimate = 6.5

d) Confidence interval: (5.1469 ,7.8531)

Step-by-step explanation:

We are given the following data set for students to earn bachelor's degrees.

4, 4, 4, 4, 4, 4, 4.5, 4.5, 4.5, 4.5, 4.5, 4.5, 6, 6, 8, 9, 9, 13, 13, 15

a) Formula:

$\text{Standard Deviation} = \sqrt{\displaystyle\frac{\sum (x_i -\bar{x})^2}{n-1}}$

where $x_i$ are data points, $\bar{x}$ is the mean and n is the number of observations.

$Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}$

$Mean =\displaystyle\frac{130}{20} = 6.5$

Sum of squares of differences = 6.25 + 6.25 + 6.25 + 6.25 + 6.25 + 6.25 + 4 + 4 + 4 + 4 + 4 + 4 + 0.25 + 0.25 + 2.25 + 6.25 + 6.25 + 42.25 + 42.25 + 72.25 = 233.5

$S.D = \sqrt{\frac{233.5}{19}} = 3.50$

b) Standard Error

$= \displaystyle\frac{s}{\sqrt{n}} = \frac{3.50}{\sqrt{20}} = 0.7826$

c) Point estimate for the mean time required for all college is given by the sample mean.

$\bar{x} = 6.5$

d) 90% Confidence interval:

$\bar{x} \pm t_{critical}\displaystyle\frac{s}{\sqrt{n}}$

Putting the values, we get,

$t_{critical}\text{ at degree of freedom 19 and}~\alpha_{0.10} = \pm 1.729$

$6.5 \pm 1.729(\frac{3.50}{\sqrt{20}} ) = 6.5 \pm 1.3531 = (5.1469 ,7.8531)$

e) No, the confidence interval does not contain the value of 4 years. Thus, confidence interval is not a good estimator as most of the value in the sample is of 4 years. Most of the sample does not lie in the given confidence interval.

5 0

4 years ago