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Inga [223]
2 years ago
13

An albatross is a large bird that can fly 400 kilometers in 8 hours at a constant speed. Using d for distance in kilometers and

t for number of hours, an equation that represents this situation is d=50t. What are two constants of proportionality for the relationship between distance in kilometers and number of hours? What is the relationship between these two values?
Mathematics
1 answer:
uranmaximum [27]2 years ago
7 0

We are given relation between distance( in kilometers) travelled and number of hours taken by an Albatross bird.

Distance in kilometers taken by variable d.

Number of hours by t.

And situation is given d=50t.

We know, direct relation formula y= kx read as y is directly proportional to x. And k is the constants of proportionality.

In the given fuction y is taken by variable d and x is taken by variable t.

If we compare d=50t by y = mx, the value of k is 50.

Therefore, constant of proportionality is 50.

Constant is a number that never change it's value.

Therefore, constant of proportionality is just 50.

But we can say variables d and t are two variables those are directly proportional to each other.

Constant of proportionality is actually unit rate of change.

For the given situation constant of proportionality is the number of km travelled in per unit time (in 1 hour.)

If we divide 400km 8 hours, we get 400/8 = 50 km per hour.

Therefore, variables of proporation are d and t and relation between them is " they are directly proportional to each other".

And the constants of proportionality is just 50.

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Alina [70]

Answer:

a) P(6.4

b) a=7 +1.036*0.333=7.345

So the value of bread-making machine that separates the bottom 85% of data from the top 15% is 7.345.

Step-by-step explanation:

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 central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

Let X the random variable life of a bread making machine. We know from the problem that the distribution for the random variable X is given by:

X\sim N(\mu =7,\sigma =1)

We take a sample of n=9 . That represent the sample size.

From the central limit theorem we know that the distribution for the sample mean \bar X is also normal and is given by:

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

\bar X \sim N(\mu=7, \frac{1}{\sqrt{9}})

Solution to the problem

Part a

(a) the probability that the mean life of a random sample  of 9 such machines falls between 6.4 and 7.2

In order to answer this question we can use the z score in order to find the probabilities, the formula given by:

z=\frac{\bar X- \mu}{\frac{\sigma}{\sqrt{n}}}

The standard error is given by this formula:

Se=\frac{\sigma}{\sqrt{n}}=\frac{1}{\sqrt{9}}=0.333

We want this probability:

P(6.4

Part b

b) The value of x to the right of which 15% of the  means computed from random samples of size 9 would fall.

For this part we want to find a value a, such that we satisfy this condition:

P(\bar X>a)=0.15   (a)

P(\bar X   (b)

Both conditions are equivalent on this case. We can use the z score again in order to find the value a.  

As we can see on the figure attached the z value that satisfy the condition with 0.85 of the area on the left and 0.15 of the area on the right it's z=1.036. On this case P(Z<1.036)=0.85 and P(Z>1.036)=0.15

If we use condition (b) from previous we have this:

P(X  

P(z

But we know which value of z satisfy the previous equation so then we can do this:

z=1.036

And if we solve for a we got

a=7 +1.036*0.333=7.345

So the value of bread-making machine that separates the bottom 85% of data from the top 15% is 7.345.

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