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dimulka [17.4K]
2 years ago
8

Miguel buys a large bottle and a small bottle of juice. The amount of juice that the manufacturer puts in the large bottle is a

random variable with a mean of 1016 ml and a standard deviation of 8 ml. The amount of juice that the manufacturer puts in the small bottle is a random variable with a mean of 510 ml and a standard deviation of 5ml. If the total amount of juice in the two bottles can be described by a normal model, what’s the probability that the total amount of juice in the two bottles is more than 1540.2 ml?
Mathematics
1 answer:
fomenos2 years ago
7 0

Answer:

Pr(X>1540.2) = 0.0655

Step-by-step explanation:

Expected value of large bottle,

E(Large) = 1016

Expected value of small bottle,

E(small) = 510

Expected value of total

E(total) = 1016 + 510 = 1526

So the new mean is 1526

Find standard deviation of new amount by variance

Variance of large bottle,

v(large) = 8^2 = 64

Variance of small bottle,

v(small) = 5^2 = 25

Variance of total

v(total) = 64+25 = 89

So the new standard deviation

sd(new) = sqrt(89) = 9.434

Find probability using the new mean and s.d.

Pr(X>1540.2)

Z score, z = (x-mean)/sd

= (1540.2 - 1526)/9.434

= 1.505

value in z score

P(z<1.51) = 0.9345

For probability of x > 1540.2

P(z > 1.51) = 1 - 0.9345 = 0.0655

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Triangle ABC was dilated and translated to form similar triangle A'B'C'.
Hunter-Best [27]
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On this exercise is asked to find the scale factor by which the triangle ABC was 
dilated to produce the triangle A'B'C'. 
Dilation is define by the rule (x,y)-- (kx, ky) where k represents the scale factor.

As can be see on the picture the dilation produce was an enlargement meaning that the image is larger that the preimage.
Of this form you can discard the choices A and B as possible solutions.

Lets try 5/2 as the possible scale factor:
(x,y)-- (kx, ky)
A(0,2)--(5/2(0),5/2(2))=A'(0,5)
B(2,2)--(5/2(2),5/2(2))=B'(5,5)
C(2,0)--(5/2(2),5/2(0))=C'(5,0)

Lets try 5/1 or 5 as the scale factor:
A(0,2)--(5(0),5(2))=A'(0,10)
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As said at the beginning of the question the triangle was not only dilated.
After a dilation and a translation, the scale factor of the dilation is letter C or 5/2.  

3 0
2 years ago
Read 2 more answers
According to a Los Angeles Times study of more than 1 million medical dispatches from 2007 to 2012, the 911 response time for me
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Answer:

a) \bar X=10.65

Median =\frac{10.7+10.7}{2}=10.7

Mode= 10.7

b) Range = 11.8-8.3=3.5

s= 0.948

c)IQR = Q_3 -Q_1 = 11.05-10.55=0.5

And we can find the usual limits with:

Lower = Q_1 -1.5 IQR = 10.55 -1.5*0.5=9.8

Upperer = Q_3 +1.5 IQR = 10.55 +1.5*0.5=11.3

And since 8.3 <9.8 we can consider this value too low or as an outlier for this case.

d) The mean for this case was 10.65 and the usual values are between 9.8 and 11.3, so as we can see all are above the specified value of 6 minutes, and we can conclude that the times are not satisfying the quality standards for this case.

And they should be considered apply some strategies to reduce the response time, adding more stations around points selected at the city could be useful in order to reduce the response time.

Step-by-step explanation:

We have the following data:

11.8 10.3 10.7 10.6 11.5 8.3 10.5 10.9 10.7 11.2

Part a

We can calculate the sample mean with the following formula:

\bar X = \frac{\sum_{i=1}^n X_i}{n}

And if we replace we got: \bar X=10.65

For the median we need to sort the values on increasing order and we have:

8.3 10.3 10.5 10.6 10.7 10.7 10.9 11.2 11.5 11.8

Since n =10 we can calculate the median as the average between the 5th and 6th position of the dataset ordered and we got:

Median =\frac{10.7+10.7}{2}=10.7

The mode would be the most repeated value on this case:

Mode= 10.7

Part b

The range is defined as Range =Max-Min and if we replace we got:

Range = 11.8-8.3=3.5

We can calculate the standard deviation with the following formula:

s= sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}

And if we replace we got:

s= 0.948

Part c

For this case we can use the IQR method in order to determine if 8.3 is an outlier or not.

We can calculate the first quartile with these values: 8.3 10.3 10.5 10.6 10.7 10.7 and Q_1= \frac{10.6+10.7}{2}=10.55

And for the Q3 we can use: 10.7 10.7 10.9 11.2 11.5 11.8 and we got Q_3 = \frac{10.9+11.2}{2}=11.05

Then we can find the IQR like this:

IQR = Q_3 -Q_1 = 11.05-10.55=0.5

And we can find the usual limits with:

Lower = Q_1 -1.5 IQR = 10.55 -1.5*0.5=9.8

Upperer = Q_3 +1.5 IQR = 10.55 +1.5*0.5=11.3

And since 8.3 <9.8 we can consider this value too low or as an outlier for this case.

Part d

The mean for this case was 10.65 and the usual values are between 9.8 and 11.3, so as we can see all are above the specified value of 6 minutes, and we can conclude that the times are not satisfying the quality standards for this case.

And they should be considered apply some strategies to reduce the response time, adding more stations around points selected at the city could be useful in order to reduce the response time.

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2 years ago
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Over [174]

Answer:

2 bags of deezznuts and 1 bag of fruit

Step-by-step explanation:

18 bags of nuts

9 bags of fruit

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8 0
2 years ago
a rectangle pyramid fits exactly on top of a rectangular prism. the prism has a length of 18 cm, a width of 6 cm, and a height o
irina1246 [14]
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Now that you have that you need to find the volume of the pyramid using the formula V= L x W x H /3 and when you plug that in it looks like
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Upon simplifying that you get 540.

Now you have the volume of your rectangular prism and your rectangular pyramid add them together and you should get 1512



I recommend drawing it out... it helps a lot:)
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2 years ago
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One centimenter is 0.01 meters. So, you can write the measures as

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T = 152.40+205.10+310.39 = 667.89

8 0
2 years ago
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