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Andru [333]
3 years ago
10

What's the difference between titanium and steel

Mathematics
1 answer:
babunello [35]3 years ago
5 0
Titanium is much harder and lasts longer. meaning it takes longer to oxidate 
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Find the median for the price of movie tickets at various theaters. $3, $4, $4, $4, $5, $5, $5, $5, $5, $6, $7, $7, $8, $9,
enot [183]
<span>$3,$4,$4,$4,$5,$5,$5,$5,$5,$6,$7,$7,$8,$9
Out of all of these the answer is</span>
since,
(5+5)/2=5
so the median is $5

6 0
3 years ago
TRUE/FALSE: The slope of the line that is PEREPNDICULAR to this line is -3/2.
chubhunter [2.5K]

Answer:

True

Step-by Step Explanation:

3 0
3 years ago
What is the solution to the equation 2(2-j)=4(J+10) ?<br><br> j =
sashaice [31]
2(2-j)=4(j+10)\\&#10;4-2j=4j+40\\ 6j=-36\\&#10;j=-6
8 0
3 years ago
Prove the following identities:
dusya [7]

Step-by-step explanation:

Left-hand-side:

\displaystyle \dfrac{1}{sec(x) - tan(x)}

sec(x)−tan(x)

1

\displaystyle = \ \dfrac{1}{sec(x) - tan(x)} * \dfrac{sec(x) + tan(x)}{sec(x) + tan(x)}=

sec(x)−tan(x)

1

∗

sec(x)+tan(x)

sec(x)+tan(x)

\displaystyle = \ \dfrac{sec(x) + tan(x)}{sec^2(x) - tan^2(x)}=

sec

2

(x)−tan

2

(x)

sec(x)+tan(x)

Now use

\displaystyle 1 + tan^2(x) \ = \ sec^2(x)1+tan

2

(x) = sec

2

(x)

and you should be done.....

5 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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