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3 years ago

Suppose the 13 inch is aware cost $.78 at the same rate how much (in cents) will 33 inches of wire cost

1 answer:

3 0

When comparing to thing with same rate, set up a proportion:
Inches/cost

13/.78=33/x
(13)(x)=(.78)(33) [cross multiplied]
13x=25.74 [simplified]
13/13x=25.74/13 [division property]
x=1.98

33 inches of wire will cost $1.98.

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Ravi made 6 identical necklaces, each having beads and a pendant. The total cost of the beads and pendants for all 6 necklaces w

mart [117]

Answer: $ 2.20

Step-by-step explanation: 31.80-18.60= 13.20

13.20 divided by 6= 2.20

——————-

2.20 x 6= 13.2

13.20+ 18.60= 31.80

4 0

3 years ago

Suppose there is this little town with a finite number of people: (1) No two inhabitants have exactly the same number of hairs.

Vadim26 [7]

The highest possible number of inhabitants in that little town are 743.

<h3>What are inhabitant?</h3>

A person or animal that lives in a place is called as the inhabitant.

Suppose there is this little town with a finite number of people: (1) No two inhabitants have exactly the same number of hairs. (2) No inhabitant has exactly 743 hairs or no hairs at all. (3) There are more inhabitants than there are hairs on the head of any inhabitant.

Let say there are 519 people in the town. and make them stand in a line with increasing number of hairs on their heads. This way, there will be a person on the last, who has no hair.

There are more number of people than the hairs. From bald to 742 hairs, 743 is the limit.

Thus, the highest possible number of inhabitants in that little town are 743.

Learn more about inhabitant.

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5 0

2 years ago

It is known that the life of a particular auto transmission follows a normal distribution with mean 72,000 miles and standard de

scoray [572]

Answer:

a) $P(X$

$P(z$

b) $P(X>65000)=P(\frac{X-\mu}{\sigma}>\frac{65000-\mu}{\sigma})=P(Z>\frac{65000-72000}{12000})=P(z>-0.583)$

$P(z>-0.583)=1-P(Z$

c) $P(X>100000)=P(\frac{X-\mu}{\sigma}>\frac{100000-\mu}{\sigma})=P(Z>\frac{100000-72000}{12000})=P(z>2.33)$

$P(z>2.33)=1-P(Z$

Sicne this probability just represent 1% of the data we can consider this value as unusual.

d) $z=1.28$

And if we solve for a we got

$a=72000 +1.28*12000=87360$

So the value of height that separates the bottom 90% of data from the top 10% is 87360.

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 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".

Part a

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

$X \sim N(72000,12000)$

Where $\mu=72000$ and $\sigma=12000$

We are interested on this probability

$P(X$

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:

$P(X$

And we can find this probability using excel or the normal standard table and we got:

$P(z$

Part b

$P(X>65000)$

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:

$P(X>65000)=P(\frac{X-\mu}{\sigma}>\frac{65000-\mu}{\sigma})=P(Z>\frac{65000-72000}{12000})=P(z>-0.583)$

And we can find this probability using the complement rule and excel or the normal standard table and we got:

$P(z>-0.583)=1-P(Z$

Part c

$P(X>100000)$

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:

$P(X>100000)=P(\frac{X-\mu}{\sigma}>\frac{100000-\mu}{\sigma})=P(Z>\frac{100000-72000}{12000})=P(z>2.33)$

And we can find this probability using the complement rule and excel or the normal standard table and we got:

$P(z>2.33)=1-P(Z$

Sicne this probability just represent 1% of the data we can consider this value as unusual.

Part d

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

$P(X>a)=0.1$ (a)

$P(X$ (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 0.9 of the area on the left and 0.1 of the area on the right it's z=1.28. On this case P(Z<1.28)=0.9 and P(z>1.28)=0.1

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

$P(X$