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Feliz [49]
3 years ago
13

Students at university x must be in one of the class ranks, freshman, sophomore, junior, or senior. at university x, 35% of the

students are freshmen and 30% are sophomores. if a student is selected at random, the probability he or she is either a junior or a senior is
Mathematics
1 answer:
miskamm [114]3 years ago
5 0
A student has to be in one of those four ranks.
If a student is not a freshman or sophomore, he must be a junior or a senior.
All students make up 100% of the total number of students.
Since the freshmen are 35% of the students, and the sophomores are 30% of the students, then they make up 65% of all students.
100% - 65% = 35%.
That means the juniors and seniors combined account for 35% of all students.
A student selected randomly has a 35% probability of being a junior or a senior.
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Answer:

not sure but this should be it

8 0
3 years ago
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A shopkeeper bought some eggs at ¢15 each. Six of them were broken while the rest were sold at ¢20 each. If he made a profit of
zvonat [6]
In the given problem it is already stated that the cost of each eggs is 15 cent. Out the total eggs bought 6 were broken. The rest of the eggs were sold for 20 cents each and the profit made was $4.80.
Cost of the 6 broken eggs = (6 * 0.15)
                                           = 0.9 cents
Let us assume that the number of eggs bought = x
Then we can write the equation as
0.2x - (0.15x - 0.9) = 4.80
0.2x - 0.15x + 0.9 = 4.80
0.05x = 4.80 - 0.9
0.05x = 3.9
x = 3.9/0.05
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So the number of eggs bought by him is 78
6 0
3 years ago
Assume that military aircraft use ejection seats designed for men weighing between 141.8 lb and 218 lb. If​ women's weights are
Mariulka [41]

Answer:

P(141.8

And we can find this probability with this difference and using the normal standard table:

P(-0.639

Then the answer would be approximately 55.3% of women between the specifications. And that represent more than the half of women

Step-by-step explanation:

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

X \sim N(173.6,49.8)  

Where \mu=173.6 and \sigma=49.8

We are interested on this probability

P(141.8

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(141.8

And we can find this probability with this difference and using the normal standard table:

P(-0.639

Then the answer would be approximately 55.3% of women between the specifications. And that represent more than the half of women

8 0
3 years ago
Given ΔABC with A(–4, –2), B(4, 4), and C(18, –8), write the equation for the line containing median segment The term is line se
Monica [59]
Here we want to find the equation of the line containing the median CP.

P, being the midpoint of AB can be found using the midpoint formula as:

\displaystyle{ P=( \frac{-4+4}{2} ,  \frac{-2+4}{2} )=(0, 1).


The slope m of the line through CP can be found by the slope formula using points C(18, -8) and P(0, 1):

\displaystyle{ m= \frac{y_2-y_1}{x_2-x_1}= \frac{-8-1}{18-0}= \frac{-9}{18}= \frac{-1}{2}.


Now, we can write the equation of the line with slope -1/2, passing through
P(0, 1):

y-1= \frac{-1}{2}(x-0)\\\\y= -\frac{1}{2}x+1.


Answer: y=-\frac{1}{2}x+1
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Show on a number line 2/3 's?​
Alexxandr [17]
It would be In between 1/2 and 2
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