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2 years ago

The gestation period of rhinos (487 days) is _____ percent longer than the gestation period of bears (220 days).

Mathematics

1 answer:

Talja [164]2 years ago

8 0

Answer:

The gestation period of rhinos (487 days) is 121. 3 percent longer than the gestation period of bears (220 days).

Step-by-step explanation:

Given:

Gestation period of rhinos = 487 days

Gestation period of bears = 220 days

To Find:

Percentage increase = ?

Solution:

The percentage increase = $\frac{ \text{new value} - \text{final value} }{\text{final value}} \times 100$

On substituting the values,

= $\frac{ 487 - 220}{220}\times 100$

= $\frac{267}{220}\times 100$

=> $1.213 \times 100$

=> 121.3 %

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A random sample of 36 observations has been drawn from a normal distribution with mean 50 and standard deviation 12. Find the pr

Gre4nikov [31]

Answer:

$z= \frac{47 -50}{\frac{12}{\sqrt{36}}}=-1.5$

$z= \frac{53 -50}{\frac{12}{\sqrt{36}}}=1.5$

And using a calculator, excel ir the normal standard table we have that:

$P(47 \leq \bar X \leq 53) =P(-1.5 \leq Z \leq 1.5)$

And we can calculate the probability like this: $P(-1.5 \leq Z \leq 1.5) = P(z$ Step-by-step explanation:

A random sample of 36 observations has been drawn from a normal distribution with mean 50 and standard deviation 12. Find the probability that the sample mean is in the interval 47<=X<53. Is the assumption of normality important. Why?

Previous concepts

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".

Solution to the problem

Let X the random variable that represent the variable of interest of a population, and for this case we know the distribution for X is given by:

$X \sim N(50,12)$

Where $\mu=50$ and $\sigma=12$

Since the distribution for X is normal then we know that the distribution for the sample mean $\bar X$ is given by:

$\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

We can find the probability required like this:

$z= \frac{47 -50}{\frac{12}{\sqrt{36}}}=-1.5$

$z= \frac{53 -50}{\frac{12}{\sqrt{36}}}=1.5$

And using a calculator, excel ir the normal standard table we have that:

$P(47 \leq \bar X \leq 53) =P(-1.5 \leq Z \leq 1.5)$

And we can calculate the probability like this:

$P(-1.5 \leq Z \leq 1.5) = P(z$

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2 years ago

A gallon of Moo Milk costs \$5.12$5.12dollar sign, 5, point, 12. What is the price, in dollars, of an 888 ounce glass of Moo Mil

Sladkaya [172]

The correct statement is:
A gallon of Moo Milk costs $5.12 What is the price, in dollars, of an 8 ounce glass of Moo Milk? There are 128 ounces in 1 gallon.

Solution:
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1 gallon = 128 ounces, so we can write:

Cost of 128 ounces of Moo Milk = $ 5.12

Cost of 1 ounce of Moo Milk = $ 5.12/128 = $ 0.04

Cost of 8 ounces of Moo Milk = $ 0.04 x 8 = $ 0.32

Thus, 8 ounces of Moo Milk will cost $ 0.32

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2 years ago

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Tom [10]

Answer:

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Step-by-step explanation:

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P(x =x) = (e^-λ * λ^x) ÷ x!

Given that :