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Blizzard [7]
3 years ago
11

What is 13 divided by 1997?

Mathematics
1 answer:
Eva8 [605]3 years ago
6 0
13 divided by 1997 is 0.0065097646
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Marty’s dog can run 3 miles in 20 minutes. What is the dog’s speed in miles per hour?
stich3 [128]
9mph because 20 minutes time 3 is one hour so multiply 3 by 3 to get 9
7 0
3 years ago
Read 2 more answers
HELPPP <br> Solve the equation <br> log (2x-3)=2
MaRussiya [10]

Answer:

2.5 or 5/2

Step-by-step explanation:

2x-3=2

2x=5 (both sides plus 3)

x=5/2 or 2.5(divid by 2 on both sides)

8 0
3 years ago
Determine the solution set for the inequality statement? + 4 &lt; 9 y &gt; -10 y &lt; -10 y &gt; - 5/2 y &lt; - 5/2
Igoryamba

Answer:

Sorry im not sure on this one...

Step-by-step explanation:

4 0
3 years ago
Members of the band sold 1150 worth of packages of pens and markers. They sold a total of 310 packages, including 10 more packag
gayaneshka [121]

Answer:

Each package of pen costs 2.5

Each package of marker costs 5

Step-by-step explanation:

Please refer to the attached image for explanations

3 0
3 years ago
The mean annual income for people in a certain city is 37 thousand dollars, with a standard deviation of 28 thousand dollars. A
Aloiza [94]

Answer:

P( 31 < \bar X< 41)

And we can ue the z score formula given by:

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

And using this formula we got for the limits:

z = \frac{31-37}{\frac{28}{\sqrt{50}}}= -1.515

z = \frac{41-37}{\frac{28}{\sqrt{50}}}= 1.01

So we want to find this probability:

P(-1.515

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".  

Solution to the problem

Let X the random variable that represent the annual income of a population, and for this case we know the following info:

\mu=37 and \sigma=28  and we are omitting the zeros from the thousand to simplify calculations

We select a sample size of n=50>30.

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".

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

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

And we want to find this probability:

P( 31 < \bar X< 41)

And we can ue the z score formula given by:

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

And using this formula we got for the limits:

z = \frac{31-37}{\frac{28}{\sqrt{50}}}= -1.515

z = \frac{41-37}{\frac{28}{\sqrt{50}}}= 1.01

So we want to find this probability:

P(-1.515

4 0
3 years ago
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