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2 years ago

5. The half-life of Po-210 is 140 days. If the initial mass of the sample is 5

Mathematics

1 answer:

Anit [1.1K]2 years ago

6 0

Answer:

1.25kg

Step-by-step explanation:

useing the problem the half life equation will be:

f(x)=5(0.5)^(x/140)

x is amount of days

so plugging in 280 for x will get 1.25kg

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What is 362,000 rounded to the nearest ten thousand

Anon25 [30]

Answer:

360,000

Step-by-step explanation:

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2 years ago

(a) Find the size of each of two samples (assume that they are of equal size) needed to estimate the difference between the prop

zalisa [80]

Answer:

(a) The sample sizes are 6787.

(b) The sample sizes are 6666.

Step-by-step explanation:

(a)

The information provided is:

Confidence level = 98%

MOE = 0.02

n₁ = n₂ = n

$\hat p_{1} = \hat p_{2} = \hat p = 0.50\ (\text{Assume})$

Compute the sample sizes as follows:

$MOE=z_{\alpha/2}\times\sqrt{\frac{2\times\hat p(1-\hat p)}{n}$

$n=\frac{2\times\hat p(1-\hat p)\times (z_{\alpha/2})^{2}}{MOE^{2}}$

$=\frac{2\times0.50(1-0.50)\times (2.33)^{2}}{0.02^{2}}\\\\=6786.125\\\\\approx 6787$

Thus, the sample sizes are 6787.

(b)

Now it is provided that:

$\hat p_{1}=0.45\\\hat p_{2}=0.58$

Compute the sample size as follows:

$MOE=z_{\alpha/2}\times\sqrt{\frac{\hat p_{1}(1-\hat p_{1})+\hat p_{2}(1-\hat p_{2})}{n}$

$n=\frac{(z_{\alpha/2})^{2}\times [\hat p_{1}(1-\hat p_{1})+\hat p_{2}(1-\hat p_{2})]}{MOE^{2}}$

$=\frac{2.33^{2}\times [0.45(1-0.45)+0.58(1-0.58)]}{0.02^{2}}\\\\=6665.331975\\\\\approx 6666$

Thus, the sample sizes are 6666.

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2 years ago

For which acute angles is the following statement true? The complement of the angle is smaller in measure than either of the two

Hitman42 [59]

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Its complement is 90 - 2x.

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Therefore, 90 - 2x < x

90 < 3x

or, x > 30.

2x > 60

Therefore, the statement is true for all angles that are greater than 60 degrees.

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3 years ago

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Answer:

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2 years ago

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mario62 [17]

65.5 divided by 2.5=26.2
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3 years ago

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