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3 months ago

. About 400,000 people visited an art museum in December what could be the exact number of people who visit the art museum

1 answer:

5 0

Considering the visitor count is likely rounded to the nearest hundred thousand, the precise figure could range from 350,000 to 449,999. If rounded to the nearest ten thousand, it would be between 395,000 and 404,999.

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Quiz 3

AnnZ [12381]

Answer: observational study

Step-by-step explanation:

4 0

2 months ago

Use the normal approximation to the binomial distribution to answer this question. Fifteen percent of all students at a large un

AnnZ [12381]

Answer: 0.1289

Step-by-step explanation:

Given: The proportion of students absent on Mondays at a large university.: $p=0.15$

Sample size: $n=12$

Mean: $\mu=np=12\times0.15=1.8$

Standard deviation = $\sigma=\sqrt{np(1-p)}$

$\Rightarrow\ \sigma=\sqrt{12(0.15)(1-0.15)}=1.23693168769\approx1.2369$

Let x represent a binomial variable.

Referencing the standard normal distribution table,

$P(x=4)=P(x\leq4)-P(x\leq3)$ (1)

Z score for normal distribution:-

$z=\dfrac{x-\mu}{\sigma}$

For x=4

$z=\dfrac{4-1.8}{1.2369}\approx1.78$

For x=3

$z=\dfrac{3-1.8}{1.2369}\approx0.97$

Thus, from (1)

$P(x=4)=P(z\leq1.78)-P(z\leq0.97)\\\\=0.962462-0.8339768\approx0.1289$

Consequently, the likelihood of four students being absent = $0.1289$

3 0

3 months ago

Give an equation which describes the intersection of this sphere with the plane z=0z=0.

Svet_ta [12734]

Center coordinates: (x0, y0, z0) and radius r.

The equation of the sphere is:
(x - x0)^2 + (y - y0)^2 + (z - z0)^2 = r^2

4 0

3 months ago

Doug purchased land for $8,000 in 1995. The year value of the land depreciated by 4% each year thereafter. Use an exponential fu

Leona [12618]

Answer:

Choice C. $6,012

Step-by-step explanation:

We know that

The formula to find the depreciated value is given by

$V=P(1-r)^{t}$

where

V stands for the depreciated value

P represents the original value

r corresponds to the depreciation rate in decimal

t refers to the Number of Time Periods

In this scenario, we have

t = 7 years

P = $8,000

r = 0.04

Plug these into the formula mentioned earlier

$V = 8,000(1-0.04)^{7} = 6,012$

Hope this assists you:)

4 0

3 months ago

Suppose that the weights of airline passenger bags are normally distributed with a mean of 47.88 pounds and a standard deviation

Zina [12379]

Answer:

There is a probability of 24.51% that the weight of a bag exceeds the maximum permitted weight of 50 pounds.

Step-by-step explanation:

Problems dealing with normally distributed samples can be addressed using the z-score formula.

For a set with the mean $\mu$ and a standard deviation $\sigma$ , the z-score for a measure X is calculated by

$Z = \frac{X - \mu}{\sigma}$

Once the Z-score is determined, we consult the z-score table to find the related p-value for this score. The p-value signifies the likelihood that the measured value is less than X. Since all probabilities total 1, calculating 1 minus the p-value gives us the probability that the measure exceeds X.

For this case

Imagine the weights of passenger bags are normally distributed with a mean of 47.88 pounds and a standard deviation of 3.09 pounds, thus $\mu = 47.88, \sigma = 3.09$

What probability exists that a bag’s weight will surpass the maximum allowable of 50 pounds?

That translates to $P(X > 50)$

Thus

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{50 - 47.88}{3.09}$

$Z = 0.69$

$Z = 0.69$ has a p-value of 0.7549.

<pthis indicates="" that="" src="https://tex.z-dn.net/?f=P%28X%20%5Cleq%2050%29%20%3D%200.7549" id="TexFormula10" title="P(X \leq 50) = 0.7549" alt="P(X \leq 50) = 0.7549" align="absmiddle" class="latex-formula">.

Additionally, we have that

$P(X \leq 50) + P(X > 50) = 1$

$P(X > 50) = 1 - 0.7549 = 0.2451$

There is a probability of 24.51% that the weight of a bag will exceed the maximum allowable weight of 50 pounds.

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6 0

3 months ago