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gizmo_the_mogwai [7]
3 years ago
11

More Challenges:

Mathematics
1 answer:
kenny6666 [7]3 years ago
3 0

Answer:

1. -m(-8n^2-7n+1)

2. (2w+7c) (v+7c)

5. (q+3r)(q^2-3qr+9r^2)

7. x= - 4/7,  1/9

See below for additional problems and help.

Step-by-step explanation:

To factor polynomials, look for patterns and greatest common factors. When you remove these factors, write the factor and what remains.

For example:

  1. 8mn^2+7mn-m\\-m(-8n^2-7n+1)

Notice the term (-8n^2-7n+1) is left and is the term when the expression is divided by -m.

     2. Factor by grouping is similar. Pull out factors within pairs of term. Separate the terms by parenthesis. If the quantities in the [parenthesis are the same, the factoring has been successful.

(2vw + 7cv) + (14cw + 49c^2)\\v(2w+7c)+7c(2w+7c)

Notice that (2w+7c) is the same. The factoring is complete. The factors are:

(2w+7c) (v+7c)

3 - 6 is similar using specific forms for factoring. To find the forms, look in your notes or at resources on online. Here is one example.

      4. A sum of cubes has the form

a^3 + b^3 = (a + b)(a^2 -ab + b^2).

To use this form, take the cube root of each term. a = q and b=3r.

The factors are (q+3)(q^2-3qr+9r^2)

7-9 all involve factoring and then solving. You solve by setting the factors equal to 0.

7. (7x+4)(9x-1) = 0

(7x+4) = 0                      (9x-1)=0

7x=-4                               9x = 1

x= - 4/7                             x = 1/9

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Answer:

a) The probability of a pregnancy lasting X days or longer is given by 1 subtracted by the p-value of Z = \frac{X - \mu}{\sigma}, in which \mu is the mean and \sigma is the standard deviation.

b) We have to find X when Z has a p-value of \frac{a}{100}, and X is given by: X = \mu - Z\sigma, in which \mu is the mean and \sigma is the standard deviation.

Step-by-step explanation:

Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the z-score of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

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In this question:

Mean \mu, standard deviation \sigma

a. Find the probability of a pregnancy lasting X days or longer.

The probability of a pregnancy lasting X days or longer is given by 1 subtracted by the p-value of Z = \frac{X - \mu}{\sigma}, in which \mu is the mean and \sigma is the standard deviation.

b. If the length of pregnancy is in the lowest a​%, then the baby is premature. Find the length that separates premature babies from those who are not premature.

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Someone help me with these questions pls-
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When finding the reciprocal, it is basicaly flipping the numerator and denominator of the number. For instance, the reciporcal of 2 is 1/2, because the denominator of 2 would be 1. For an example more straight forward, the reciporacle of 3/5 is 5/3. This applied to the question goes as followed.

1. 3/5 is 5/3

2. 1/4 is 4

3. 1 is 1

Similarly when dividing fractions or an integer to a fraction, the fraction that is dividing the number turns into the reciprocal then you would multiply.

4.   3/(3/4) is 3*4/3= 4

5.   5/(3/4) is 5*(4/3)= 20/3

6.   8/(4/7) is 8*(7/3)= 56/3

7. 6/(3/5) is 6*(5/3)= 10

8. 2/(5/8) is 2*(8/5)= 16/5

9. 4/(8/9) is 4*(9/8)= 9/2

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