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3 years ago

The volume of a storage tank is 1695.6 m3. The radius is 6m. How tall is the storage tank?

Mathematics

2 answers:

Andreas93 [3]3 years ago

7 0

Answer:

the storage tank is a cylinder.

The formule to calculate the volume is:

$v =\pi{r}^{2} \times h$

$h = v \div \pi {r}^{2}$

$h = 1695.6 \div 36 \times \pi = 14.992$

Ostrovityanka [42]3 years ago

6 0

Answer:

<h3>storage tank is 15 Meter tall</h3>

Step-by-step explanation:

volume = 1695.6 m³ ---- (given)

radius = 6m ---- (given)

volume = π × radius² × height

1695.6 m³ = 3.14 × 6² × height

1695.6 m³ = 113.04 × height

height = 1695.6 ÷ 113.04

height = 15 m

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3 years ago

Membership in Mensa requires an IQ score above 131.5. Nine candidates take an IQ test, and they have a mean IQ score of 133. IQ

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Answer:

a) $P(X\geq 133)=P(\frac{X-\mu}{\sigma}\geq \frac{133-\mu}{\sigma})=P(Z\geq \frac{133-100}{15})=P(Z\geq 2.2)$

And we can find this probability using the complement rule:

$P(Z\geq 2.2)=1- P(z$

b) $P(\bar X\geq 133)=P(\frac{\bar X-\mu}{\sigma_{\bar x}}\geq \frac{133-\mu}{\sigma_{\bar x}})=P(Z\geq \frac{133-100}{5})=P(Z\geq 6.6)$

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$P(Z\geq 6.6)=1- P(z$

c) No. The mean can be lower than 131.5 if we find the probability:

$P(\bar X\leq 133)=P(\frac{\bar X-\mu}{\sigma_{\bar x}}\leq \frac{131.5-\mu}{\sigma_{\bar x}})=P(Z\leq \frac{131.5-100}{5})=P(Z\leq 6.3)$

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Step-by-step explanation:

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Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

A. If 1 person is randomly selected from thegeneral population, find the probability of getting someone with anIQ score of at least 133.

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$X \sim N(100,15)$

Where $\mu=100$ and $\sigma=15$

We are interested on this probability

$P(X\geq 133)$

And the best way to solve this problem is using the normal standard distribution and the z score given by:

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

If we apply this formula to our probability we got this:

$P(X\geq 133)=P(\frac{X-\mu}{\sigma}\geq \frac{133-\mu}{\sigma})=P(Z\geq \frac{133-100}{15})=P(Z\geq 2.2)$

And we can find this probability using the complement rule:

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B. If 9 people are randomly selected,find the probability that their mean IQ score is at least 133.

For this case since the distribution for the random variable X is normal then the distribution for the sample mean is also normal and given by:

$\bar X = \sim N(\mu= 100 ,\sigma_{\bar x}= \frac{15}{\sqrt{9}}=5)$

The new z score is defined as

$z=\frac{x-\mu}{\sigma_{\bar x}}$

If we apply this formula to our probability we got this:

$P(\bar X\geq 133)=P(\frac{\bar X-\mu}{\sigma_{\bar x}}\geq \frac{133-\mu}{\sigma_{\bar x}})=P(Z\geq \frac{133-100}{5})=P(Z\geq 6.6)$

And we can find this probability using the complement rule:

$P(Z\geq 6.6)=1- P(z$

And the probability is calculated from the normal standard table or with excel.

C. Although the results are available,the individual IQ test scores have been lost. Can it be concluded that all 9 candidates have IQ scores above 131.5 so that they allare eligible for Mensa membership?

No. The mean can be lower than 131.5 if we find the probability:

$P(\bar X\leq 133)=P(\frac{\bar X-\mu}{\sigma_{\bar x}}\leq \frac{131.5-\mu}{\sigma_{\bar x}})=P(Z\leq \frac{131.5-100}{5})=P(Z\leq 6.3)$

$P(Z\leq 6.3) \approx 1$

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3 years ago

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