Triangle BAC was rotated 90° clockwise and dilated at a scale factor of 2 from the origin to create triangle XYZ. Based on these
1 answer:
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Answer:
a)TRUE
b)FALSE
c)TRUE
d)FALSE
e)TRUE
f)TRUE
g)TRUE
h)FALSE
i)FALSE
j)TRUE
Step-by-step explanation:
a) For every x there is y such that :
TRUE
This statement is true, because for every real number there is a square number of that number, and that square number is also a real number. For example, if we take 6.5, there is a square of that number and it equals 39.0625.
b) For every x there is y such that :
FALSE
For example, if x = -1, there is no such real number so that its square equals -1.
c) There is x for every y such that xy = 0
TRUE
If we put x = 0, then for every y it will be xy=0*y=0
d)There are x and y such that
FALSE
There are no such numbers. If we rewrite the equation we obtain an incorrect statement:
e)For every x, if there is y such that xy=1:
TRUE
The statement is true. If we have a number x, then multiplying x with 1/x (Since x is not equal to 0 we can do this for ever real number) gives 1 as a result.
f)There is x for every y such that if then xy=1.
TRUE
The statement is equivalent to the statement in e)
g)For every x there is y such that x+y = 1
TRUE
The statement says that for every real number x there is a real number y such that x+y = 1, i.e. y = 1-x
So, the statement says that for every real umber there is a real number that is equal to 1-that number
h) There are x and y such that
We have to solve this system of equations.
From the first equation it yields x=2-2y and inserting that into the second equation we have:
Which is obviously false statement, so there are no such x and y that satisfy the equations.
FALSE
i)For every x there is y such that
We have to solve this system of equations.
From the first equation it yields and inserting that into the second equation we obtain:
Inserting that back to the first equation we obtain
So, there is an unique solution to this equations:
x=1 and y=1
The statement is FALSE, because only for x=1 (and not for every x) exists y (y=1) such that
j)For every x and y there is a z such that
TRUE
The statament is true for all real numbers, we can always find such z. z is a number that is halway from x and from y.
Answer:
The probability that Polly's Sample will give a result within 1% of the value 65% is 0.6424
Step-by-step explanation:
The variable that assigns the value 1 if a person had its original major and 0 otherwise is a Bernoulli variable with paramenter 0.65. Since she asked the question to 1850 people, then the number of students that will have their original major is a Binomial random variable with parameters n = 1850, p = 0.65.
Since the sample is large enough, we can use the Central Limit Theorem to approximate that random variable to a Normal random variable, which we will denote X.
The parameters of X are determined with the mean and standard deviation of the Binomal that we are approximating. The mean is np = 1850*0.65 = 1202.5, and the standard deviation is √np(1-p) = √(1202.5*0.35) = 20.5152.
We want to know the probability that X is between 0.64*1850 = 1184 and 0.66*1850 = 1221 (that is, the percentage is between 64 and 66). In order to calculate this, we standarize X so that we can work with a standard normal random variable W ≈ N(0,1). The standarization is obtained by substracting the mean from X and dividing the result by the standard deviation, in other words
The values of the cummulative function of the standard normal variable W, which we will denote are tabulated and they can be found in the attached file.
Now, we are ready to compute the probability that X is between 1184 and 1221. Remember that, since the standard random variable is symmetric through 0, then \phi(-z) = 1-\phi(z) for each positive value z.
Therefore, the probability that Polly's Sample will give a result within 1% of the value 65% is 0.6424.