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tino4ka555 [31]
2 years ago
11

5.

Mathematics
1 answer:
ANTONII [103]2 years ago
7 0

The greatest possible number of club members is 7

<em><u>Solution:</u></em>

Given that, local readers’ club has a set of 49 hardback books and a set of 21 paperbacks

Each set can be divided equally among the club members

To find the greatest possible number of club members, we have to find the greatest common factor of 49 and 21

The greatest number that is a factor of two (or more) other numbers.

When we find all the factors of two or more numbers, and some factors are the same ("common"), then the largest of those common factors is the Greatest Common Factor.

<em><u>Greatest common factor of 49 and 21:</u></em>

The factors of 21 are: 1, 3, 7, 21

The factors of 49 are: 1, 7, 49

Then the greatest common factor is 7

Thus, the greatest possible number of club members is 7

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Answer:

x+2 is a factor of f(x)  = 5x^2 + 13x + 6

Step-by-step explanation:

To check: If (x+2) is a a factor of the polynomial f(x)  = 5x^2 + 13x + 6

We need to check if x  = -2 is the FACTOR of the given polynomial.

⇒ To show:  f (-2)  =  0

Now

f(-2)  = 5(-2)^2 + 13(-2) + 6\\=5(4) -26 + 6\\= 20 + 6 - 26\\= 26 - 26  = 0\\\implies f(-2) = 0

Since, f(-2) = 0

⇒  -2 is the ROOT of the Polynomial f(x).

⇒(x +2) is the factor of the given polynomial.

Hence, x+2 is a factor of f(x)  = 5x^2 + 13x + 6

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A simple random sample of size nequals81 is obtained from a population with mu equals 83 and sigma equals 27. ​(a) Describe the
Ivanshal [37]

Answer:

a) \bar X \sim N (\mu, \frac{\sigma}{\sqrt{n}})

With:

\mu_{\bar X}= 83

\sigma_{\bar X}=\frac{27}{\sqrt{81}}= 3

b) z= \frac{89-83}{\frac{27}{\sqrt{81}}}= 2

P(Z>2) = 1-P(Z

c) z= \frac{75.65-83}{\frac{27}{\sqrt{81}}}= -2.45

P(Z

d) z= \frac{89.3-83}{\frac{27}{\sqrt{81}}}= 2.1

z= \frac{79.4-83}{\frac{27}{\sqrt{81}}}= -1.2

P(-1.2

Step-by-step explanation:

For this case we know the following propoertis for the random variable X

\mu = 83, \sigma = 27

We select a sample size of n = 81

Part a

Since the sample size is large enough we can use the central limit distribution and the distribution for the sampel mean on this case would be:

\bar X \sim N (\mu, \frac{\sigma}{\sqrt{n}})

With:

\mu_{\bar X}= 83

\sigma_{\bar X}=\frac{27}{\sqrt{81}}= 3

Part b

We want this probability:

P(\bar X>89)

We can use the z score formula given by:

z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}

And if we find the z score for 89 we got:

z= \frac{89-83}{\frac{27}{\sqrt{81}}}= 2

P(Z>2) = 1-P(Z

Part c

P(\bar X

We can use the z score formula given by:

z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}

And if we find the z score for 75.65 we got:

z= \frac{75.65-83}{\frac{27}{\sqrt{81}}}= -2.45

P(Z

Part d

We want this probability:

P(79.4 < \bar X < 89.3)

We find the z scores:

z= \frac{89.3-83}{\frac{27}{\sqrt{81}}}= 2.1

z= \frac{79.4-83}{\frac{27}{\sqrt{81}}}= -1.2

P(-1.2

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