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3 years ago

2h - (6h - 5 ) = -1

Mathematics

2 answers:

Tcecarenko [31]3 years ago

6 0

Answer:

1.5

Step-by-step explanation:

2h-(6h-5)=-1

2h-6h+5=-1

-4h+5=-1

-4h=-1-5

-4h=-6

h=-6/-4

h=6/4

h=3/2 or 1.5

Ostrovityanka [42]3 years ago

5 0

Answer:

exact form: 3/2

decimal form: h= 1.5

mixed number form: h= 1 1/2

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In a large section of a statistics class, the points for the final exam are normally distributed, with a mean of 71 and a stan

kumpel [21]

Answer:

The lowest score on the final exam that would qualify a student for an A is 80.

The lowest score on the final exam that would qualify a student for a B is 74.68.

The lowest score on the final exam that would qualify a student for a C is 67.33.

The lowest score on the final exam that would qualify a student for a D is 62.

Step-by-step explanation:

When the distribution is normal, we use the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Mean of 71 and a standard deviation of 7.

This means that $\mu = 71, \sigma = 7$

Grades are assigned such that the top 10% receive A's, the next 20% received B's, the middle 40% receive C's, the next 20% receive D's, and the bottom 10% receive F's.

This means that:

90th percentile and above: A

70th percentile and below 90th: B

30th percentile to the 70th percentile: C

10th percentile to the 30th: D

Lowest score for an A:

Top 10% receive A, which means that the lowest score that would qualify a student for an A is the 100 - 10 = 90th percentile, which is X when Z has a pvalue of 0.9, so X when Z = 1.28.

$Z = \frac{X - \mu}{\sigma}$

$1.28 = \frac{X - 71}{7}$

$X - 71 = 7*1.28$

$X = 80$

The lowest score on the final exam that would qualify a student for an A is 80.

Lowest score for a B:

70th percentile, which is X when Z has a pvalue of 0.7, so X when Z = 0.525.

$Z = \frac{X - \mu}{\sigma}$

$0.525 = \frac{X - 71}{7}$

$X - 71 = 7*0.525$

$X = 74.68$

The lowest score on the final exam that would qualify a student for a B is 74.68.

Lowest score for a C:

30th percentile, which is X when Z has a pvalue of 0.3, so X when Z = -0.525.

$Z = \frac{X - \mu}{\sigma}$

$-0.525 = \frac{X - 71}{7}$

$X - 71 = 7*(-0.525)$

$X = 67.33$

The lowest score on the final exam that would qualify a student for a C is 67.33.

Lowest score for a D:

10th percentile, which is X when Z has a pvalue of 0.1, so X when Z = -1.28.

$Z = \frac{X - \mu}{\sigma}$

$-1.28 = \frac{X - 71}{7}$

$X - 71 = 7*(-1.28)$

$X = 62$

The lowest score on the final exam that would qualify a student for a D is 62.

4 0

3 years ago

What is the equation of the line that is perpendicular to the given line and passes through point (2,5)?

ad-work [718]

Answer:

$\large\boxed{y=-\dfrac{1}{3}x+1}$

Step-by-step explanation:

$\text{The slope-intercept form of an equation of a line:}\\\\y=mx+b\\\\m-slope\\b-y-intercept\to(0,\ b)$

$\text{Let}\ k:y=m_1x+b_1\ \text{and}\ l:y=m_2x+b_2,\ \text{then}\\\\l\ \parallel\ k\iff m_1=m_2\\\\l\ \perp\ k\iff m_1m_2=-1\to m_2=-\dfrac{1}{m_1}\\=======================\\\\\text{We must find the slope of the given line.}\\\\\text{The formula of a slope:}\\\\m=\dfrac{y_2-y_1}{x_2-x_1}\\\\\text{Substitute the coordinates of the given points from the graph (0, 1) and (-3, 2)}\\\\m_1=\dfrac{2-1}{-3-0}=\dfrac{1}{-3}=-\dfrac{1}{3}$

$\text{We have an equation:}\\\\y=-\dfrac{1}{3}x+b\\\\\text{From the coordinates of the point (0, 1) we have the y-intercept:}\ b=1.\\\\\text{Therefore we have the equation:}\\\\y=-\dfrac{1}{3}x+1$