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2 years ago

Find the mean absolute deviation of the setback of data. 9, 12, 3, 5, 8

Mathematics

1 answer:

hodyreva [135]2 years ago

7 0

Answer:

3.5

Step-by-step explanation:

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Solve for x. Your answer must be simplified. x/24>5

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Answer:

x > 120

Step-by-step explanation:

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2 years ago

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Answer:

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3 years ago

Please help, I have so much work to do, and I have to fit in time to study for a huge test they will determine my future

worty [1.4K]

Answer:

Step-by-step explanation:

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2 years ago

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A car has mass 1500 kg and is traveling at a speed of 35 miles/hour. what is its kinetic energy in joules? (Be sure to convert m

photoshop1234 [79]

Answer:

Factor by which kinetic energy increase = 4 times

Step-by-step explanation:

Given,

Mass of the car, v1 = 1500 kg

initial speed of car = 35 miles/h

$=\dfrac{35\times 1609.34}{3600}\ m/s$

= 15.64 m/s

Initial kinetic energy of the car is given by,

$k_1\ =\ \dfrac{1}{2}.m.v_1^2$

$=\ \dfrac{1}{2}\times 1500\times (15.64)^2\ joule$

= 183606.46 J

Final velocity of car v2 = 70 miles/hour

$=\dfrac{70\times 1609.34}{3600}$

= 31.29 m/s

So, final kinetic energy of car is given by

$k_2\ =\ \dfrac{1}{2}.m.v_2^2$

$=\ \dfrac{1}{2}\times 1500\times (31.29)^2$

= 734425.84 J

Now, the ratio of final to initial kinetic energy can be given by,

$\dfrac{k_2}{k_1}=\ \dfrac{734425.84}{183606.46}$

$=>\ k_2\ =\ 4k_1$

Hence, the kinetic energy will increase by 4 times.

7 0

2 years ago

1000 households were surveyed. 275 households own a desktop computer, 455 households own a DVD player, 405 households own two ca

icang [17]

Answer:

Step-by-step explanation:

From the given information,

Suppose

X represents the Desktop computer

Y represents the DVD Player

Z represents the Two Cars

Given that:

n(X)=275

n(Y)=455

n(Z)=405

n(XUY)=145

n(YUZ)=195

n(XUZ)=110

n((XUYUZ))=265

n(X ∩ Y ∩ Z) = 1000-265

n(X ∩ Y ∩ Z) = 735

n(X ∪ Y) = n(X)+n(Y)−n(X ∩ Y)

145 = 275+455 - n(X ∩ Y)

n(X ∩ Y) = 585

n(Y ∪ Z) = n(Y) + n(Z) − n(Y ∩ Z)

195 = 455+405-n(Y ∩ Z)

n(Y ∩ Z) = 665

n(X ∪ Z) = n(X) + n(Z) − n(X ∩ Z)

110 = 275+405-n(X ∩ Z)

n(X ∩ Z) = 570

a. n(X ∪ Y ∪ Z) = n(X) + n(Y) + n(Z) − n(X ∩ Y) − n(Y ∩ Z) − n(X ∩ Z) + n(X ∩ Y ∩ Z)

n(X ∪ Y ∪ Z) = 275+455+405-585-665-570+735

n(X ∪ Y ∪ Z) = 50

c. n(X ∪ Y ∪ C') = n(X ∪ Y)-n(X ∪ Y ∪ Z)

n(X ∪ Y ∪ C') = 145-50

n(X ∪ Y ∪ C') = 95

6 0

3 years ago