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almond37 [142]
2 years ago
6

Determine two numbers that have a product of 40 but have a sum of 13.

Mathematics
1 answer:
Paha777 [63]2 years ago
8 0

Answer:

The two numbers are 8 and 5.

Step-by-step explanation:

Let consider x and y to be these two numbers.

So;

(x) * (y) = 40 ---- (1)

x + y     = 13 ----- (2)

From (2), let x =  13 - y

Then;

(13 - y) * y = 40

13y - y² = 40

13y - y² - 40 = 0

-y² + 13y - 40 = 0

Multiply through by (-)

y² - 13y + 40 = 0

By factorization;

(y - 8) (y - 5) = 0

y = 8 or  y = 5

CHECK:

8 * 5 = 40

8 + 5 = 13

Therefore, the two numbers are 8 and 5.

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nordsb [41]
LEAST COMMON MULTIPLE is 56

hope this helped☺

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What is the equation of the linear function represented by the table?
Korolek [52]
Hello!

You can put x into the answer choices to see if we get y
----------------------------------------------------------------------------------------------------
y = -x + 9
y = -(-5) + 9
y = 5 + 9
y = 14

This could be a answer

y = -(-2) + 9
y = 2 + 9
y = 11

y = -(1) + 9
y = -1 + 9
y = 8

y = -(4) + 9
y = -4 + 9
y = 5

we got y every time so the answer is the first one

The answer is A) y = -x + 9

Hope this helps!
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3 years ago
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The population of a species of rabbit triples every year. This can be modeled by f(x) = 4(3)x and f(5) = 972. What does the 972
Alona [7]
972 represent the total population of the rabbits after five yours. 
Answer: B
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3 years ago
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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

8 0
3 years ago
PLEASE HELP!!!!!!!!!!!!!!!!!!!!!!!!
charle [14.2K]

Answer:

I think it is C.2920

3 0
2 years ago
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