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3 years ago

A teacher orders homework books for her class of 30 students.

Mathematics

2 answers:

nexus9112 [7]3 years ago

8 0

Total value: $144
Explanation: each book cost $4.50

Hunter-Best [27]3 years ago

6 0

Answer:

The total value of the order now is $142.

Step-by-step explanation:

assuming all books are at the same price

135 ÷30=4.5

2x4.5=7

135+7=142

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A potter works 4 days a week, makes 15

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It is 3.75 for the answer

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How do you this?? Scientific notation

Airida [17]

So Scientific Notation is a "simplified version" of a number. For example, is your number is 2,345, the scientific notation would be 2.345 * 10³.

If it was 0.023, then the scientific notation would be 2.3 * 10^-2.

Scientific notation is found by moving the decimal place to the left or right, until only one digit is left to the left of the decimal; it has to be greater than 1, but less than ten. If you move the decimal place to the right, it's a negative power of 10. If you move it to the left, it is a positive power of 10.

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3 years ago

Suiting at 6 a.m., cars, buses, and motorcycles arrive at a highway loll booth according to independent Poisson processes. Cars

dem82 [27]

Answer:

Step-by-step explanation:

From the information given:

the rate of the cars = $\dfrac{1}{5} \ car / min = 0.2 \ car /min$

the rate of the buses = $\dfrac{1}{10} \ bus / min = 0.1 \ bus /min$

the rate of motorcycle = $\dfrac{1}{30} \ motorcycle / min = 0.0333 \ motorcycle /min$

The probability of any event at a given time t can be expressed as:

$P(event \ (x) \ in \ time \ (t)\ min) = \dfrac{e^{-rate \times t}\times (rate \times t)^x}{x!}$

∴

(a)

$P(2 \ car \ in \ 20 \ min) = \dfrac{e^{-0.20\times 20}\times (0.2 \times 20)^2}{2!}$

$P(2 \ car \ in \ 20 \ min) =0.1465$

$P ( 1 \ motorcycle \ in \ 20 \ min) = \dfrac{e^{-0.0333\times 20}\times (0.0333 \times 20)^1}{1!}$

$P ( 1 \ motorcycle \ in \ 20 \ min) = 0.3422$

$P ( 0 \ buses \ in \ 20 \ min) = \dfrac{e^{-0.1\times 20}\times (0.1 \times 20)^0}{0!}$

$P ( 0 \ buses \ in \ 20 \ min) = 0.1353$

Thus;

P(exactly 2 cars, 1 motorcycle in 20 minutes) = 0.1465 × 0.3422 × 0.1353

P(exactly 2 cars, 1 motorcycle in 20 minutes) = 0.0068

(b)

the rate of the total vehicles = 0.2 + 0.1 + 0.0333 = 0.3333

the rate of vehicles with exact change = rate of total vehicles × P(exact change)

$= 0.3333 \times \dfrac{1}{4}$

= 0.0833

∴

$P(zero \ exact \ change \ in \ 10 minutes) = \dfrac{e^{-0.0833\times 10}\times (0.0833 \times 10)^0}{0!}$

P(zero exact change in 10 minutes) = 0.4347

The probability of the 7th motorcycle after the arrival of the third motorcycle is:

$P( 4 \ motorcyles \ in \ 45 \ minutes) =\dfrac{e^{-0.0333\times 45}\times (0.0333 \times 45)^4}{4!}$

$P( 4 \ motorcyles \ in \ 45 \ minutes) =0.0469$

Thus; the probability of the 7th motorcycle after the arrival of the third one is = 0.0469

P(at least one other vehicle arrives between 3rd and 4th car arrival)

= 1 - P(no other vehicle arrives between 3rd and 4th car arrival)

The 3rd car arrives at 15 minutes

The 4th car arrives at 20 minutes

The interval between the two = 5 minutes

For Bus:

P(no other vehicle other vehicle arrives within 5 minutes is)

= $\dfrac{6}{12} = 0.5$

For motorcycle:

$= \dfrac{2 }{12} = \dfrac{1 }{6}$

∴

The required probability = $1 - \Bigg ( \dfrac{e^{-0.5 \times 0.5^0}}{0!} \times \dfrac{e^{-1/6}\times (1/6)^0}{0!} \Bigg)$

= 1- 0.5134

= 0.4866

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3 years ago

WILL MARK BRAINLIST

hammer [34]

$2x +y = 20~~....(i)\\\\6x -5y = -4~~....(ii)\\\\\\\text{Multiply equation (i) by 3:}\\\\6x +3y = 60~~...(iii)\\\\(iii)-(ii):\\\\6x+3y - 6x+5y = 60-(-4)\\\\\implies 8y = 60+4 = 64\\\\\implies y = \dfrac{64}8\\\\\implies y = 8\\\\\\\text{Substitute y =8 in equation (i):}\\\\2x+8 =20\\\\\implies 2x = 20-8 = 12\\\\\implies x = \dfrac{12}2 = 6\\\\\text{Hence (x,y) = (6, 8)}$

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DedPeter [7]

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[2x² + 5]² + [2x² + 5] + 2

[2x² + 5]² + [2x² + 7]↓

This is your final answer. You get it?

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3 years ago