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Butoxors [25]
4 years ago
5

Lisa drinks 0.55 L of milk everyday . How much milk does she drink in 11 days? Write your answer in milliliters

Mathematics
1 answer:
kow [346]4 years ago
6 0

Answer:

She will drink 6050mL of milk in 11 days.

Step-by-step explanation:

Multiply by the amount of milk she drinks and how many days there are.

0.55L x 11 = 6.05L

Convert the answer into milliliters (multiply by 1000 because there are 1000mL in 1L).

6.05L x 1000 = 6050mL

Therefore, Lisa will drink 6050mL of milk in 11 days.

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Which is not possible distance for the flight from Haiti back to Cuba?
lara31 [8.8K]
The option which is not a possible distance for the flight from Haiti back to Cuba is 43, because that number is far too low to be a possible option. The correct answer should be somewhere around 500km, which means that the remaining options are close to it, whereas 43 is not.
8 0
4 years ago
An explosion causes debris to rise vertically with an initial velocity of 160 feet per second. What is the speed of debris when
lisov135 [29]
The question is asking what v_final is, given that v_initial is at 300 feet. and v_initial is at 0 feet.  
We know there will be a constant downward acceleration of 32.15 ft/s^2.
 Use the following equation: 
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 v_final^2 = (160 ft/s)^2 + 2(-32.15 ft/s^2)(300 ft) = 6310 ft^2/s^2
 v_final = (6310 ft^2/s^2)^1/2 = 79.4 ft/s.
5 0
3 years ago
A broker has calculated the expected values of two different financial instruments X and Y. Suppose that E(x)= $100, E(y)=$90 SD
Sveta_85 [38]

Expectation is linear, meaning

E(<em>a X</em> + <em>b Y</em>) = E(<em>a X</em>) + E(<em>b Y</em>)

= <em>a </em>E(<em>X</em>) + <em>b</em> E(<em>Y</em>)

If <em>X</em> = 1 and <em>Y</em> = 0, we see that the expectation of a constant, E(<em>a</em>), is equal to the constant, <em>a</em>.

Use this property to compute the expectations:

E(<em>X</em> + 10) = E(<em>X</em>) + E(10) = $110

E(5<em>Y</em>) = 5 E(<em>Y</em>) = $450

E(<em>X</em> + <em>Y</em>) = E(<em>X</em>) + E(<em>Y</em>) = $190

Variance has a similar property:

V(<em>a X</em> + <em>b Y</em>) = V(<em>a X</em>) + V(<em>b Y</em>) + Cov(<em>X</em>, <em>Y</em>)

= <em>a</em>^2<em> </em>V(<em>X</em>) + <em>b</em>^2 V(<em>Y</em>) + Cov(<em>X</em>, <em>Y</em>)

where "Cov" denotes covariance, defined by

E[(<em>X</em> - E(<em>X</em>))(<em>Y</em> - E(<em>Y</em>))] = E(<em>X Y</em>) - E(<em>X</em>) E(<em>Y</em>)

Without knowing the expectation of <em>X Y</em>, we can't determine the covariance and thus variance of the expression <em>a X</em> + <em>b Y</em>.

However, if <em>X</em> and <em>Y</em> are independent, then E(<em>X Y</em>) = E(<em>X</em>) E(<em>Y</em>), which makes the covariance vanish, so that

V(<em>a X</em> + <em>b Y</em>) = <em>a</em>^2<em> </em>V(<em>X</em>) + <em>b</em>^2 V(<em>Y</em>)

and this is the assumption we have to make to find the standard deviations (which is the square root of the variance).

Also, variance is defined as

V(<em>X</em>) = E[(<em>X</em> - E(<em>X</em>))^2] = E(<em>X</em>^2) - E(<em>X</em>)^2

and it follows from this that, if <em>X</em> is a constant, say <em>a</em>, then

V(<em>a</em>) = E(<em>a</em>^2) - E(<em>a</em>)^2 = <em>a</em>^2 - <em>a</em>^2 = 0

Use this property, and the assumption of independence, to compute the variances, and hence the standard deviations:

V(<em>X</em> + 10) = V(<em>X</em>)  ==>  SD(<em>X</em> + 10) = SD(<em>X</em>) = $90

V(5<em>Y</em>) = 5^2 V(<em>Y</em>) = 25 V(<em>Y</em>)  ==>  SD(5<em>Y</em>) = 5 SD(<em>Y</em>) = $40

V(<em>X</em> + <em>Y</em>) = V(<em>X</em>) + V(<em>Y</em>)  ==>  SD(<em>X</em> + <em>Y</em>) = √[SD(<em>X</em>)^2 + SD(<em>Y</em>)^2] = √8164 ≈ $90.35

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Irina18 [472]

Answer:

$2250

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3 years ago
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STatiana [176]
To calculate this we need to find derivative of C

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Every time you have a function that depends on some variable and you need to calculate minimum or maximum, make derivative of it and set it equal to 0


3 0
3 years ago
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