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3 years ago

Each time the carousel runs, 11 children

Mathematics

2 answers:

STALIN [3.7K]3 years ago

6 0

The answer is 5 times

Dmitriy789 [7]3 years ago

4 0

Answer: 5 times

Step-by-step explanation: 11 goes into 50 four times. That means that the first four times the carousel runs, 11 kids will be on it. But since six ungrateful children were left out, the carousel will need to run one extra time so those brats can get a turn.

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10=c+d c is one more than d what is the value of c? with working out

Leya [2.2K]

C= 5.5 I'm pretty sure that's the answer

d= 4.5 and c+ 5.5
Because 4.5 plus 1.0 = 5.5
since c is one more than d

5.5 plus 4.5 = 10

5 0

3 years ago

What value of x is in the solution set of 4x – 12 ≤ 16 + 8x?Which is a correct first step in solving the inequality –4(2x – 1) &

Anna71 [15]

Value of x is -1
The correct first step to solving the inequality is distributing the -4 into the para thesis

8 0

3 years ago

Read 2 more answers

Convert 1111+102 into quinary number

Kipish [7]

Answer:

13421+402

Step-by-step explanation:

5 0

2 years ago

Suppose that the store manager of a small rural pharmacy is doing a linear regression of daily sales of over-the-counter (OTC) d

Doss [256]

Answer:

3.83

Step-by-step explanation:

Mean of x = Σx / n

Mean of x = (14 + 19 + 13 + 6 + 9) / 5 = 12.2

Sum of square (SS) :

(14-12.2)^2 + (19-12.2)^2 + (13-12.2)^2 + (6-12.2)^2 + (9-12.2)^2 = 98.8

Mean of y = Σy / n

Mean of y = (101 + 89 + 48 + 21 + 47) / 5 = 61.2

Σ(y - ybar)² = (101-61.2)^2 + (89-61.2)^2 + (48-61.2)^2 + (21-61.2)^2 + (47-61.2)^2 = 4348.8

df = n - 2 = 5 - 2 = 3

Σ(y - ybar)² / df = 4348.8 / 3 = 1449.6

√(Σ(y - ybar)² / df) = √1449.6 = 38.074

Standard Error = √(Σ(y - ybar)² / df) / √SS

Standard Error = 38.074 / √98.8

Standard Error = 3.83

4 0

2 years ago

(x^2y+e^x)dx-x^2dy=0

klio [65]

It looks like the differential equation is

$\left(x^2y + e^x\right) \,\mathrm dx - x^2\,\mathrm dy = 0$

Check for exactness:

$\dfrac{\partial\left(x^2y+e^x\right)}{\partial y} = x^2 \\\\ \dfrac{\partial\left(-x^2\right)}{\partial x} = -2x$

As is, the DE is not exact, so let's try to find an integrating factor µ(x, y) such that

$\mu\left(x^2y + e^x\right) \,\mathrm dx - \mu x^2\,\mathrm dy = 0$

*is* exact. If this modified DE is exact, then

$\dfrac{\partial\left(\mu\left(x^2y+e^x\right)\right)}{\partial y} = \dfrac{\partial\left(-\mu x^2\right)}{\partial x}$

We have

$\dfrac{\partial\left(\mu\left(x^2y+e^x\right)\right)}{\partial y} = \left(x^2y+e^x\right)\dfrac{\partial\mu}{\partial y} + x^2\mu \\\\ \dfrac{\partial\left(-\mu x^2\right)}{\partial x} = -x^2\dfrac{\partial\mu}{\partial x} - 2x\mu \\\\ \implies \left(x^2y+e^x\right)\dfrac{\partial\mu}{\partial y} + x^2\mu = -x^2\dfrac{\partial\mu}{\partial x} - 2x\mu$

Notice that if we let µ(x, y) = µ(x) be independent of y, then ∂µ/∂y = 0 and we can solve for µ :

$x^2\mu = -x^2\dfrac{\mathrm d\mu}{\mathrm dx} - 2x\mu \\\\ (x^2+2x)\mu = -x^2\dfrac{\mathrm d\mu}{\mathrm dx} \\\\ \dfrac{\mathrm d\mu}{\mu} = -\dfrac{x^2+2x}{x^2}\,\mathrm dx \\\\ \dfrac{\mathrm d\mu}{\mu} = \left(-1-\dfrac2x\right)\,\mathrm dx \\\\ \implies \ln|\mu| = -x - 2\ln|x| \\\\ \implies \mu = e^{-x-2\ln|x|} = \dfrac{e^{-x}}{x^2}$