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Oduvanchick [21]
3 years ago
6

The sum of three consecutive even numbers is one hundred sixty-two. What is the smallest of the three numbers?

Mathematics
2 answers:
notsponge [240]3 years ago
8 0
The first guy is correct
zepelin [54]3 years ago
3 0
Since the numbers are consecutive, they must be near each other in size, and therefore near one-third of 168. And since the lower number is -2 off the middle, and the upper number is +2 from the middle, the middle number must be exactly one-third of 168—otherwise the sum of the three could never be 168. Labelling the three consecutive numbers as x, y, and z:

x + y + z = 168

(y-2) + y + (y+2) = 168.

3y = 168

y = 56

therefore the smallest number (x) = 54.
436 viewsView upvotes

1




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(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.

7 0
3 years ago
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