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lisabon 2012
3 months ago
5

The data set represents the number of rings each person in a room is wearing. 0, 2, 4, 0, 2, 3, 2, 8, 6 What is the interquartil

e range of the data?
2
3
4
6
Mathematics
2 answers:
Leona [12.6K]3 months ago
4 0
D1,..,d9 = 0,0,2,2,2,3,4,6,8 //there are 9 values, presented in ascending order
Q2 (median) = d5 = 2 //middle value
Q1 = (d2+d3) / 2 = (0+2)/2 = 1
(Q1 represents the median of d1,d2,d3,d4, but as there is no singular middle element among four, the average is computed)
Q3 = (d7+d8) / 2 = (4+6)/2 = 5
interquartile range = IQR = Q3 - Q1 = 5 -1 = 4
final answer: 4

Leona [12.6K]3 months ago
4 0

Answer:

Option C.

Step-by-step explanation:

The provided data set is

0, 2, 4, 0, 2, 3, 2, 8, 6

Rearranging the data in ascending order gives:

0, 0, 2, 2, 2, 3, 4, 6, 8

Next, the data set is divided into four equal segments.

(0, 0), (2, 2), 2, (3, 4), (6, 8)

Now, we see:

Q_1=\dfrac{0+2}{2}=1

Q_2=Median=2

Q_3=\dfrac{4+6}{2}=5

The formula for the interquartile range:

IQR=Q_3-Q_1

IQR=5-1

IQR=4

Thus, the interquartile range calculated is 4, making option C correct.

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The manufacturer's suggested retail price (MSRP) for a particular car is $25,425, and it is expected to be
PIT_PIT [12445]

Response:

(a) The linear depreciation function relating to the car's value y after a few years t is given by:

y = 25,425 - 2,739 × t

(b) The projected worth of the car in 7 years amounts to $6,252.

(c) The depreciation is at a rate of $2,739 yearly.

Step-by-step explanation:

The manufacturer's suggested retail price (MSRP) for the vehicle is $25,425.

Its expected valuation in 5 years is $11,730.

(a) The linear depreciation formula is as follows:

Depreciation \ Per \ Year \ = \dfrac{Cost \ of \, Asset - Salvage \ Value}{Life \ of \, Asset \ in \ use}

Where;

Asset Cost = $25,425

Salvage Value = $11,730

Asset Lifespan = 5 years

From this, we obtain;

Depreciation \ Per \ Year \ = \dfrac{\$ 25,425 - \$11,730}{5} = \$2,739/year

Thus, the linear depreciation function can be expressed as:

y - 25,425 = -2,739×(t - 0)

y = -2,739·t + 0 + 25,425 = 25,425 -2,739·t

y = 25,425 - 2,739 × t

Where;

y = The estimated worth of the car after a specified duration

t = The years calculated for depreciation

(b) The value of the vehicle in 7 years can be calculated as below;

For t = 7, we find;

y = 25,425 -2,739·t = y = 25,425 -2,739 × 7 = $6,252

The expected car value after 7 years is $6,252.

(c) Depreciation \ Per \ Year \ = \dfrac{\$ 25,425 - \$11,730}{5} = \$2,739/year

The car depreciates at a yearly rate of $2,739.

3 0
2 months ago
Student researchers Haley, Jeff, and Nathan saw an article on the Internet claiming that the average vertical jump for teens was
tester [12383]

Answer:

At the α = 0.10 level, there is no substantial evidence indicating that the average vertical jump for students at this school differs from 15 inches.

Step-by-step explanation:

A hypothesis test is necessary to verify the assertion that the average vertical jump of students diverges from 15 inches.

The null and alternative hypotheses are:

H_0: \mu=15\\\\H_a: \mu\neq15

The significance level is set at 0.10.

The sample mean recorded is 17, and the sample standard deviation is 5.37.

The degrees of freedom are calculated as df=(20-1)=19.

The t-statistic is:

t_{19}=\frac{M-\mu}{s_M/\sqrt{n}} =\frac{17-15}{5.37/\sqrt{20}}=\frac{2}{5.37/4.47} =2/1.20=1.67

The two-tailed P-value corresponding to t=1.67 is P=0.11132.

<pSince this P-value exceeds the significance level, the result is not significant. Therefore, the null hypothesis remains unchallenged.

At the α = 0.10 level, there is no compelling evidence that the average vertical jump of students at this school deviates from 15 inches.

7 0
3 months ago
Alejandro made an error in the steps below when determining the equation of the line that is perpendicular to the line 4x – 3y =
Svet_ta [12734]
The equation of the perpendicular line can be identified by determining its slope and applying the given point within the standard formula.

Standard equation: y-y1 = m(x-x1)

m*m'=-1
where m' indicates the slope of the perpendicular line
m denotes the slope of the original line

m = -coefficient of x/coefficient of y = -4/-3 = 4/3
m' = -3/4

Substituting the point (3, -2):
y+2 = -3/4*(x-3)
4y+8 = -3x+9

Thus, the equation of the perpendicular line is: 3x+4y-1=0
7 0
2 months ago
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You can upgrade lighting at your factory to LED bulbs that cost $6.95 each and last an average of 5 years. It costs $3 in labor
Inessa [12570]
Each LED bulb, along with installation labor, is priced at
.. $6.95 +$3 = $9.95

For 100 bulbs over a span of 10 years, that equals (100*10) = 1000 bulb·years. At $9.95 per bulb, 5 bulb·years are obtained, and thus the projected total cost for 1000 bulb·years is
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In summary, for a decade, the installation and changes of 200 bulbs in 100 lamps amount to $1990. Therefore, the yearly cost is...
.. $1990/(10 yr) = $199/yr
3 0
3 months ago
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