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Sloan [31]
3 years ago
12

On a certain hot summers, 256 people used the public swimming pool. The daily prices are $1.75 for children and $2.75 fo adults.

The receipts for admission totaled to $551.50. How many children and adults swam that day? Explain.
Mathematics
1 answer:
antiseptic1488 [7]3 years ago
5 0

Answer:152 children and 104 adults swam.

Step-by-step explanation:

Let x represent the number of children tickets. It also means that x children swam.

Let y represent the number of adult tickets. It also means that y adults swam.

On a certain hot summers, 256 people used the public swimming pool. This means that

x + y = 256

The daily prices are $1.75 for children and $2.75 for adults. The receipts for admission totaled to $551.50. This means that

1.75x + 2.75y = 551.5- - - - - -- - - - - -1

Substituting x = 256 - y into equation 1, it becomes

1.75(256 - y) + 2.75y = 551.5

448 - 1.75y + 2.75y = 551.5

- 1.75y + 2.75y = 551.5 - 448

y = 103.5

Approximately 104 adults

x = 256 - y = 256 - 104

x = 152

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F.DIST.RT(3.67,5.69)

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The rectangle shown as a perimeter of 70 cm and the given area. It’s length is 8 more than twice it’s width. Write and solve a s
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width of the rectangle = b = x

length of the rectangle = l = 8 + 2x

Perimeter of the rectangle = 70cm

Also, perimeter of the rectangle = 2(l + b)

70 = 2[x + (8 + 2x)]

70 = 2(x + 8 + 2x)

70 = 2(3x + 8)

70 = 6x + 16

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54 = 6x

54/6 = x

9 = x

Therefore, b = x

b = 9cm

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I = 8 + 2×9

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mr farmer rectangular land measures 2000 meters by 3000 meters. if he walks around the perimeter of his land, how far does he wa
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10000 meters

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Hope this helps!

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Step-by-step explanation:

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Radioactive Decay
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Answer:

Percentage of (226Ra) after 900 years is 68%

Step-by-step explanation:

Let P(t) be the amount of (226Ra) present at any time t

Half life of (226 Ra) = 1599 years

If P₀ is initial amount of (226 Ra) then after 1599 years

P(1599)=P₀/2

Decay i amount of radioactive substance is related to time t as

\frac{dP}{dt}=kP(t)\\\\\frac{1}{P}\,dP=kdt\\\\Integrating\,\, both\,\,sides\\\\ln|P|=kt+c\\\\P(t)=Ce^{kt}\\\\at \,\, t=0\,\, P(0)=P_{o}\\\\P(0)=Ce^{k0}\\\\P_{o}=C\\\\then\\\\P(t)=P_{o}e^{kt}

To find value of k

at\,\, t=1599\,years\\\\P(1599)=\frac{P_{o}}{2}\\\\then\\\\\frac{P_{o}}{2} =P_{o}e^{k(1599)}\\\\\frac{1}{2} =e^{k(1599)}\\\\ln|\frac{1}{2}|=k(1599)\\\\k=\frac{ln|\frac{1}{2}|}{1599}=-4.3\times 10^{-3}\\\\\implies P(t)=P_{o}e^{-4.3\times 10^{-3}t}\\\\at\,\, t=900 \\\\P(900)=P_{o}e^{-4.3\times 10^{-3}(900)}\\\\P(900)=0.68P_{o}

Percentage of radioactive element is:

Amount after 900 years=\frac{P(900)}{P_{o}}\times 100\\\\=\frac{0.68P_{o}}{P_{o}}\times 100\%\\\\=68\%

3 0
3 years ago
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