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qwelly [4]
3 years ago
14

If I paid $9.00 for 4 lb of candies, how much candies can I buy for $12.00?

Mathematics
1 answer:
Lina20 [59]3 years ago
3 0

Answer:

We are given that 4 lb of candies are bought for $9.00

Therefore, the price of 1 lb of candies is:

\frac{9}{4}=2.25

Now the amount of candies that can bought for $12.00 is:

\frac{12}{2.25} =5.3333 lb

Therefore, 5.3333 lb of candies can be bought for $12.00


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If (x + yi) + (4(x + yi) + 6i) = 10, what is x + yi?
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A population of values has a normal distribution with μ=204.9μ=204.9 and σ=81.9σ=81.9. You intend to draw a random sample of siz
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7. μ=204.9 and σ=5.4968

8. μ=75.9 and σ=0.7136

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7. - Given that the population mean =204.9 and the standard deviation is 81.90 and the sample size n=222.

-The sample mean,\mu_xis calculated as:

\mu_x=\mu=204.9, \mu_x=sample \ mean

-The standard deviation,\sigma_x is calculated as:

\sigma_x=\frac{\sigma}{\sqrt{n}}\\\\=\frac{81.9}{\sqrt{222}}\\\\=5.4968

8. For a random variable X.

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\sigma_x=\frac{\sigma}{\sqrt{n}}\\\\=\frac{9.6}{\sqrt{181}}\\\\=0.7136

9. Given the population mean, μ=135.7 and σ=88 and n=59

#We calculate the sample mean;

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#Sample standard deviation:

\sigma_x=\frac{\sigma}{\sqrt{n}}\\\\=\frac{88}{\sqrt{59}}\\\\=11.4566

#The sample size, n=59 is at least 30, so we apply Central Limit Theorem:

P(\bar X>117.4)=P(Z>\frac{117.4-\mu_{\bar x}}{\sigma_x})\\\\=P(Z>\frac{117.4-135.7}{11.4566})\\\\=P(Z>-1.5973)\\\\=1-0.05480 \\\\=0.9452

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