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arlik [135]
3 years ago
10

HELP PLEASE

Mathematics
1 answer:
oee [108]3 years ago
6 0
Using a graph tool

case 1) 
2x-y=-13
y=x+9

the solution is the point (-4,5)
see the attached figure N 1

case 2)
y=3x-7
y=2x-5

the solution is the point (2,-1)
see the attached figure N 2

case 3)
3x+2y=10
6x-y=10

the solution is the point (2,2)
see the attached figure N 3

case 4)
y=6
x=-5
the solution is the point (-5,6)
see the attached figure N 4

case 5)
4x-3y=5
3x+2y=-9
the solution is the point (-1,-3)


case 6)
x+y=7
x-y=-1
the solution is the point (3,4)

the answer in the attached figure



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Working together, it takes two different sized hoses 20 minutes to fill a small swimming pool. If it takes 30 minutes for the la
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60 minutes for the larger hose to fill the swimming pool by itself

Step-by-step explanation:

It is given that,

Working together, it takes two different sized hoses 20 minutes to fill a small swimming pool.

takes 30 minutes for the larger hose to fill the swimming pool by itself

Let x be the efficiency to fill the swimming pool by larger hose

and y be the efficiency to fill the swimming pool by larger hose

<u>To find LCM of 20 and 30</u>

LCM (20, 30) = 60

<u>To find the efficiency </u>

Let x be the efficiency to fill the swimming pool by larger hose

and y be the efficiency to fill the swimming pool by larger hose

x = 60/30 =2

x + y = 60 /20 = 3

Therefore efficiency of y = (x + y) - x =3 - 2 = 1

so, time taken to fill the swimming pool by small hose = 60/1 = 60 minutes

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Slices of pizza for a certain brand of pizza have a mass that is approximately normally distributed with a mean of 67.7 grams an
Alexxx [7]

Answer:

<u>a. s.e. = 0.338</u>

<u>b. The  probability of finding a random slice of pizza with a mass of less than 67.2 grams is 0.5871 or 58.71%</u>

<u>c. The probability of finding a 20 random slices of pizza with a mean mass of less than 67.2 grams is 0.0694 or 6.94%</u>

d. <u>The sample mean of 62.75 would represent the 15th percentile</u>

Step-by-step explanation:

1. Let's review the information provided to us to answer the question properly:

μ of the mass of slices of pizza for a certain brand = 67.7 grams

σ of the mass of slices of pizza for a certain brand = 2.28 grams

2. For samples of size 20 pizza slices, what is the standard deviation for the sampling distribution of the sample mean?

Let's recall that the standard deviation of the sampling distribution of the mean is called the standard error of the mean and its formula is:

μσs.e.= √σ/n, where n is the sample size.

s.e. = √2.28/20

<u>s.e. = 0.338</u>

3. What is the probability of finding a random slice of pizza with a mass of less than 67.2 grams?

Let's find the z-score for X = 67.2, this way:

z-score = (X - μ)/σ

z-score = (67.2 - 67.7)/2.28

z-score = 0.5/2.28

z-score = 0.22 (rounding to the next hundredth)

Now, using the z-table, let's find p, the probability:

p (z = 0.22) = 0.5871

<u>The  probability of finding a random slice of pizza with a mass of less than 67.2 grams is 58.71%</u>

4.  What is the probability of finding a 20 random slices of pizza with a mean mass of less than 67.2 grams?

Let's use the central limit theorem to find the z-score, this way:

z-score = X - μ/s.e

z-score = 67.2 - 67.7/0.338

z-score = -0.5/0.338

z-score = - 1.48

Now, using the z-table, let's find p, the probability:

<u>p (z = -1.48) = 0.0694</u>

5. What sample mean (for a sample of size 20) would represent the bottom 15% (the 15th percentile)?

For p = 0.150, let's find the z-score:

z-score = - 2.17

z-score = X - μ/s.e

- 2.17 = (X - 67.7)/2.28

- 4.95 = X - 67.7

X = 67.7 - 4.95

X = 62.75 (rounding to the next hundredth)

<u>The sample mean of 62.75 would represent the 15th percentile</u>

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