Can someone please help me with this math question?

List all prime numbers between 27 and 35.

LiRa [457]

Answer:

29 and 31 are the only primes.

5 0

3 years ago

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Which is true about the completely simplified difference of the polynomials 6x6 − x3y4 − 5xy5 and 4x5y + 2x3y4 + 5xy5?

Galina-37 [17]

<span>Let be A= 6x6 − x3y4 − 5xy5 and B= 4x5y + 2x3y4 + 5xy5 and when we do their difference, it is A - B = 6x6 − x3y4 − 5xy5 -( 4x5y + 2x3y4 + 5xy5)=6x6 − x3y4 − 5xy5 - 4x5y - 2x3y4 - 5xy5 = 6x6 - x3y4 - 2x3y4 - 5xy5 -4xy5 -5xy5=6x6 - 3x3y4 -14xy5, so the final solution is A - B =6x6 - 3x3y4 -14xy5, the degree of this is equal to the degree of - 3x3y4, and it is 3+4=7, the answer is The difference has 3 terms and a degree of 7.</span>

8 0

3 years ago

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The score on an exam from a certain MAT 112 class, X, is normally distributed with μ=78.1 and σ=10.8.

salantis [7]

a) $X$

b) 0.1539

c) 0.1539

d) 0.6922

Step-by-step explanation:

a)

In this problem, the score on the exam is normally distributed with the following parameters:

$\mu=78.1$ (mean)

$\sigma = 10.8$ (standard deviation)

We call X the name of the variable (the score obtained in the exam).

Therefore, the event "a student obtains a score less than 67.1) means that the variable X has a value less than 67.1. Mathematically, this means that we are asking for:

$X$

And the probability for this to occur can be written as:

$p(X$

b)

To find the probability of X to be less than 67.1, we have to calculate the area under the standardized normal distribution (so, with mean 0 and standard deviation 1) between $z=-\infty$ and $z=Z$ , where Z is the z-score corresponding to X = 67.1 on the s tandardized normal distribution.

The z-score corresponding to 67.1 is:

$Z=\frac{67.1-\mu}{\sigma}=\frac{67.1-78.1}{10.8}=-1.02$

Therefore, the probability that X < 67.1 is equal to the probability that z < -1.02 on the standardized normal distribution:

$p(X$

And by looking at the z-score tables, we find that this probability is:

$p(z$

And so,

$p(X$

c)

Here we want to find the probability that a randomly chosen score is greater than 89.1, so

$p(X>89.1)$

First of all, we have to calculate the z-score corresponding to this value of X, which is:

$Z=\frac{89.1-\mu}{\sigma}=\frac{89.1-78.1}{10.8}=1.02$

Then we notice that the z-score tables give only the area on the left of the values on the left of the mean (0), so we have to use the following symmetry property:

$p(z>1.02) =p(z$