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4 years ago

8

The price of oil, in dollars per barrel, declined last week by 3.5%. If it started the week at 102.00 per barrel, at what per ba

rrel price did it end the week?

1 answer:

scoray [572]4 years ago

6 0

Answer:

98.57

Step-by-step explanation:

3.5% over 100% X 102=3.43

102-3.43= 98.57

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Identify the reflection rule in the given figure.

saul85 [17]

Answer: A) Reflection along the line y = -1

===========================================================

Explanation:

Point A is at (1,5). Point A' is at (1,-7).

Apply the midpoint formula to find the center of these two points.

You should get (1,-1) as that midpoint.

Follow these steps for B(-1,3) and B'(-1,-5) to get the midpoint (-1,-1)

Those midpoints we found are located on the mirror line. We only need two such points to plot it. Drawing a line through (1,-1) and (-1,-1) graphs the equation y = -1. This is the horizontal line through -1 on the y axis.

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3 years ago

Write the equation of the line perpendicular to 4x – 5y = -10 through the point (2, 3). explain please

Snezhnost [94]

Answer:

y =-5/4x +11/2

Step-by-step explanation:

First we need to find the slope of the line

4x-5y = 10

Solve for y

Subtract 4x from each side

4x-4x-5y =-4x-10

-5y = -4x-10

Divide by -5

-5y/-5 = -4x/-5 -10/-5

y = 4/5x +2

The slope is 4/5

The slope of a line that is perpendicular is the negative reciprocal

m perpendicular = - (5/4)

We know have a slope = -5/4 and a point (2,3)

We can use the point slope form of a line

y-y1 = m(x-x1)

y-3 = -5/4 (x-2)

Distribute

y-3 =-5/4x +10/4

Add 3 to each side

y-3+3 = -5/4 x +5/2 +3

Get a common denominator

y = -5/4x +5/2 +6/2

y =-5/4x +11/2

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3 years ago

Thirty-seven school buses arrived at the zoo, each carrying thirty-seven to forty-three students per bus.

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Answer:

C 1,600

Step-by-step explanation:

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3 years ago

Square root of 1.9321

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The answer to square root of 1.9321 is - 1.39

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4 years ago

Read 2 more answers

An article in Technometrics (Vol. 19, 1977, p. 425) presents the following data on the motor fuel octane ratings of several blen

Slav-nsk [51]

Answer:

$Median = 90.4$

$Q_1 = 88.6$

$Q_3 = 92.2$

Step-by-step explanation:

Given

The above data

Required

- A stem and leaf display

- The median

- The quartiles

First, determine the range of the data

$Smallest = 83.4$

$Highest = 100.3$

Next, group each dataset base on common whole numbers.

So, we have:

$83.4$

$84.3\ 84.3$

$85.3$

$86.7\ 86.7\ 86.7$

$87.4\ 87.5\ 87.6\ 87.7\ 87.8\ 87.9$

$88.2\ 88.3\ 88.3\ 88.3\ 88.4\ 88.5\ 88.5\ 88.6\ 88.6\ 88.7\ 88.9$

$89.0\ 89.2\ 89.3\ 89.3\ 89.6\ 89.7\ 89.8\ 89.8\ 89.9\ 89.9$

$90.0\ 90.1\ 90.1\ 90.1\ 90.3\ 90.4\ 90.4\ 90.4\ 90.5\ 90.6\ 90.7\ 90.8\ 90.9$

$91.0\ 91.0\ 91.0\ 91.1\ 91.1\ 91.1\ 91.2\ 91.2\ 91.2\ \ 91.5\ 91.6\ 91.6\ 91.8\ 91.8$

$92.2\ 92.2\ 92.2\ 92.3\ 92.6\ 92.7\ 92.7\ 92.7$

$93.0\ 93.2\ 93.3\ 93.3\ 93.4\ 93.7$

$94.2\ 94.2\ 94.4\ 94.7$

$96.1\ 96.5$

$98.8$

$100.3$

Next, we construct the stem and leaf plot.

The whole numbers will be the stem while the decimal parts will be the leaf.

So, we have:

$\begin{array}{ccc}{Stem} & {} & {Leaf} & {83} & {|} & {.4} & {84} & {|} & {.3\ .3} & {85} & {|} & {.3} & {86} & {|} & {.7\ .7\ .7} & {87} &{|} & {.4\ .5\ .6\ .7\ .8\ .9} & {88} & {|} & {.2\ .3\ .3\ .3\ .4\ .5\ .5\ .6\ .6\ .7\ .9} &{89} & {|} & {.0\ .2\ .3\ .3\ .6\ .7\ .8\ .8\ .9\ .9} & {90} & {|} &{.0\ .1\ .1\ .1\ .3\ .4\ .4\ .4\ .5\ .6\ .7\ .8\ .9} & {91} &{|}&{.0\ .0\ .0\ .1\ .1\ .1\ .2\ .2\ .2\ .5\ .6\ .6\ .8\ .8} &{92} &{|} &{.2\ .2\ .2\ .3\ .6\ .7\ .7\ .7} \ \end{array}$

$\begin{array}{ccc} {93} & {|} & {.0\ .2\ .3\ .3\ .4\ .7} & {94} &{|} & {.2\ .2\ .4\ .7} &{96} & {|} & {.1\ .5} & {98} & {|} & {.8} & {100} &{|} &{.3} \ \end{array}$

From the above plot,

$n = 83$

The median is calculated as:

$Median = \frac{n+1}{2}th$

$Median = \frac{83+1}{2}th$

$Median = \frac{84}{2}th$

$Median = 42nd$

i.e. the median is at the 42nd position.

From the above stem and leaf plot.

The 42nd position is at stem 90 and the leaf .4

So the median is:

$Median = 90.4$

The lower quartile (Q) is calculated as:

$Q_1 = \frac{n+1}{4}th$

$Q_1 = \frac{83+1}{4}th$

$Q_1 = \frac{84}{4}th$

$Q_1 = 21st$

i.e. the lower quartile is at the 21st position.

From the above stem and leaf plot.

The 42nd position is at stem 88 and the leaf .6

So the lower quartile is:

$Q_1 = 88.6$

The upper quartile (Q3) is calculated as:

$Q_3 = 3 * \frac{n+1}{4}th$

$Q_3 = 3 * \frac{83+1}{4}th$

$Q_3 = 3 * \frac{84}{4}th$

$Q_3 = 3 * 21th$

$Q_3 = 63rd$

i.e. the upper quartile is at the 63rd position.

From the above stem and leaf plot.

The 63rd position is at stem 92 and the leaf .2

So the upper quartile is:

$Q_3 = 92.2$

8 0

3 years ago