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3 years ago

Decimals between 10 15 with an interval of 0.75 between each pair of decimals

Mathematics

2 answers:

Nonamiya [84]3 years ago

7 0

Answer:

The decimals between 10 and 15 with an interval of 0.75 between each pair of decimals are 10.75, 11.5, 12.25, 13.0, 13.75,14.5.

Step-by-step explanation:

To find : Decimals between 10 and 15 with an interval of 0.75 between each pair of decimals ?

Solution :

According to question,

We have to add 0.75 in each number starting from 10 to 15.

So, $10+0.75=10.75$

$10.75+0.75=11.5$

$11.5+0.75=12.25$

$12.25+0.75=13.0$

$13+0.75=13.75$

$13.75+0.75=14.5$

Therefore, The decimals between 10 and 15 with an interval of 0.75 between each pair of decimals are 10.75, 11.5, 12.25, 13.0, 13.75,14.5.

aleksley [76]3 years ago

5 0

Should be 10.75, 11.5, 12.25, 13, 13.75, 14.5

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Lemur [1.5K]

Answer:

a) $P(20 \leq X \leq 30) = P(20-0.5 \leq X \leq 30+0.5)$

$P(19.5 \leq X \leq 30.5) = P(\frac{19.5-25}{5} \leq Z \leq \frac{30.5 -25}{5})=P(-1.1 \leq Z \leq 1.1)$

And we can find this probability like this:

$P(-1.1 \leq Z \leq 1.1)= P(Z\leq 1.1) -P(Z\leq -1.1) = 0.864-0.136= 0.728$

b) $P(X \leq 30)= P(X$

And using the z score we got:

$P(X$

c) $P(X$

And if we use the continuity correction we got:

$P(X$

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".

Continuity correction means that we need to add and subtract 0.5 before standardizing the value specified.

Part a

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:

$X \sim N(25,5)$

Where $\mu=25$ and $\sigma=5$

Part a

For this case we want to find this probability:

$P(20 \leq X \leq 30) = P(20-0.5 \leq X \leq 30+0.5)$

And if we use the z score given by:

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

We got this:

$P(19.5 \leq X \leq 30.5) = P(\frac{19.5-25}{5} \leq Z \leq \frac{30.5 -25}{5})=P(-1.1 \leq Z \leq 1.1)$

And we can find this probability like this:

$P(-1.1 \leq Z \leq 1.1)= P(Z\leq 1.1) -P(Z\leq -1.1) = 0.864-0.136= 0.728$

Part b

For this case we want this probability:

$P(X \leq 30)$

And if we use the continuity correction we got:

$P(X \leq 30)= P(X$

And using the z score we got:

$P(X$

Part c

For this case we want this probability:

$P(X$

And if we use the continuity correction we got:

$P(X$

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zvonat [6]

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Answer:

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