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3 years ago

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There are 1000 students in a college.Out of 20000 in the whole university in a study of 200 were found to be smokers in the coll

ege and 1000 in whole university. Is there any significant difference between the proportion of smokers in college and university

1 answer:

vagabundo [1.1K]3 years ago

4 0

Answer:

1000 students in college

2000 students in University

200 out of 2000 are smokers

200 out 1000 are smokers

200 : 2000

1 :10

200 : 1000

1 : 5

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PLEASE HELP WITH MATH EASY

Marat540 [252]

Answer:

(2,4)

Step-by-step explanation:

the point where the lines intercept is the solution, so it is (2,4)

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2 years ago

Read 2 more answers

Test scores of the student in a school are normally distributed mean 85 standard deviation 3 points. What's the probability that

Mrrafil [7]

Answer:

The probability that a random selected student score is greater than 76 is $\\ P(x>76) = 0.99865$ .

Step-by-step explanation:

The Normally distributed data are described by the normal distribution. This distribution is determined by two <em>parameters</em>, the <em>population mean</em> $\\ \mu$ and the <em>population standard deviation</em> $\\ \sigma$ .

To determine probabilities for the normal distribution, we can use <em>the standard normal distribution</em>, whose parameters' values are $\\ \mu = 0$ and $\\ \sigma = 1$ . However, we need to "transform" the raw score, in this case <em>x</em> = 76, to a z-score. To achieve this we use the next formula:

$\\ z = \frac{x - \mu}{\sigma}$ [1]

And for the latter, we have all the required information to obtain <em>z</em>. With this, we obtain a value that represent the distance from the population mean in standard deviations units.

<h3>The probability that a randomly selected student score is greater than 76</h3>

To obtain this probability, we can proceed as follows:

First: obtain the z-score for the raw score x = 76.

We know that:

$\\ \mu = 85$

$\\ \sigma = 3$

$\\ x = 76$

From equation [1], we have:

$\\ z = \frac{76 - 85}{3}$

Then

$\\ z = \frac{-9}{3}$

$\\ z = -3$

Second: Interpretation of the previous result.

In this case, the value is <em>three</em> (3) <em>standard deviations</em> <em>below</em> the population mean. In other words, the standard value for x = 76 is z = -3. So, we need to find P(x>76) or P(x>-3).

With this value of $\\ z = -3$ , we can obtain this probability consulting <em>the cumulative standard normal distribution, </em>available in any Statistics book or on the internet.

Third: Determination of the probability P(x>76) or P(x>-3).

Most of the time, the values for the <em>cumulative standard normal distribution</em> are for positive values of z. Fortunately, since the normal distributions are <em>symmetrical</em>, we can find the probability of a negative z having into account that (for this case):

$\\ P(z>-3) = 1 - P(z>3) = P(z$

Then

Consulting a <em>cumulative standard normal table</em>, we have that the cumulative probability for a value below than three (3) standard deviations is:

$\\ P(z$

Thus, "the probability that a random selected student score is greater than 76" for this case (that is, $\\ \mu = 85$ and $\\ \sigma = 3$ ) is $\\ P(x>76) = P(z>-3) = P(z$ .

As a conclusion, more than 99.865% of the values of this distribution are above (greater than) x = 76.

<em>We can see below a graph showing this probability.</em>

As a complement note, we can also say that:

$\\ P(z3)$

$\\ P(z3)$

Which is the case for the probability below z = -3 [P(z<-3)], a very low probability (and a very small area at the left of the distribution).

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3 years ago

A central angle of a circle measures 1.5 radians. If the radius of the circle is 3 cm, what is the area of the related sector?

soldi70 [24.7K]

Answer:

Option B. $6.75\ cm^{2}$

Step-by-step explanation:

we know that

The area of a circle is equal to

$A=\pi r^{2}$

we have

$r=3\ cm$

substitute

$A=\pi (3^{2})=9 \pi\ cm^{2}$

Remember that

$2\pi$ radians subtends the complete circle of area $9 \pi\ cm^{2}$

so

by proportion

Find the area of the related sector for a central angle of $1.5$ radians

Let

x------> the area of the related sector

$\frac{9 \pi}{2\pi}\frac{cm^{2}}{radians} =\frac{x}{1.5}\frac{cm^{2}}{radians}\\ \\x=9*1.5/2\\ \\x= 6.75\ cm^{2}$

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3 years ago

Does some body knows how to do this ?

lakkis [162]

It's asking for the "perimeter" of the molding as far as I can tell,
namely, how long is the border of that molding
notice the picture here

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3 years ago

Nutka1998 [239]

Answer:

$m\angle MEJ=25^{\circ}.$

Step-by-step explanation:

The arcs LY, KM, KL and MJ together form the full revolution angle, thus

$4x+50^{\circ}+6x+x+10^{\circ}+4x=360^{\circ},\\ \\15x=300^{\circ},\\ \\x=20^{\circ}.$

Note that

$m\angle MOJ=4x=80^{\circ},$

then

$m\angle MLJ=\dfrac{1}{2}\cdot 80^{\circ}=40^{\circ}.$

So,

$m\angle ELM=180^{\circ}-40^{\circ}=140^{\circ}.$

Also

$m\angle LOK=30^{\circ},$

so

$m\angle KML=\dfrac{1}{2}\cdot 30^{\circ}=15^{\circ}.$

In triangle EML,

$m\angle MEL+m\angle EML+m\angle ELM=180^{\circ},\\ \\m\angle MEL=180^{\circ}-15^{\circ}-140^{\circ}=25^{\circ}.$

Thus, $m\angle MEJ=25^{\circ}.$

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3 years ago