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Naily [24]
4 years ago
8

List S and list T each contain 5 positive integers, and for each list the average (arithmetic mean) of the integers in the list

is 40. If the integers 30, 40, and 50 are in both lists, is the standard deviation of the integers in list S greater than the standard deviation of the integers in list T? (1) The integer 25 is in list S. (2) The integer 45 is in list T.
Mathematics
1 answer:
gtnhenbr [62]4 years ago
3 0

Answer:

Yes, SDS > SDT

Step-by-step explanation:

List S

25, 30, 40, 50, XS

Average list S = 40

So, we could write,

Average list S = 40 = (25 + 30 + 40 + 50 + XS) / 5

Solving for XS

XS = 40 x 5 – 25 – 30 – 40 – 50 = 200 – 145 = 55  

SDs = SD (25, 30, 40, 50, 55) = 12.74

List T

30, 40, 45, 50, XT

Average list T = 40

So, we could write,

Average list T = 40 = (30 + 40 + 45 + 50 + XT) / 5

Solving for XT

XT = 40 x 5 – 30 – 40 – 45 – 50 = 200 – 165 = 35  

SDT = SD (30, 35, 40, 45, 50) = 7.1

Even at first sight SDS > SDT  because 25 is out of the range 30-50, while 45 is within that range.

List S is more spread than list T.

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What is the property of (4 + 7) + 8 = 4 + (7 +8) ?
nasty-shy [4]

Answer:

Commutative

Step-by-step explanation:

2 + 4 = 4 + 2 as an example

4 0
2 years ago
Is 8.012 greater than 8.03
morpeh [17]
Answer: No, 8.03 is greater than 8.012
8 0
2 years ago
Read 2 more answers
Nigel is planning his training schedule for a marathon over a 4-day period. He is uncertain how many miles he will run on two da
KengaRu [80]

Answer:

12+17+y+z or 29+y+z.

Step-by-step explanation:

We have been given that Nigel is planning his training schedule for a marathon over a 4-day period. He is uncertain how many miles he will run on two days. One expression for the total miles he will run is 12+y+17+z.

The Commutative Property of Addition states that we can add numbers in any order. For example a and b be two numbers.

According to Commutative Property of Addition a+b=b+a.

Similarly, we can write an equivalent expression to our given expression as:

12+17+y+z

We can simplify our expression as:

29+y+z.

Therefore, our required expression would be 12+17+y+z or 29+y+z.

4 0
3 years ago
Litter such as leaves falls to the forest floor, where the action of insects and bacteria initiates the decay process. Let A be
Travka [436]

Answer:

D = L/k

Step-by-step explanation:

Since A represents the amount of litter present in grams per square meter as a function of time in years, the net rate of litter present is

dA/dt = in flow - out flow

Since litter falls at a constant rate of L  grams per square meter per year, in flow = L

Since litter decays at a constant proportional rate of k per year, the total amount of litter decay per square meter per year is A × k = Ak = out flow

So,

dA/dt = in flow - out flow

dA/dt = L - Ak

Separating the variables, we have

dA/(L - Ak) = dt

Integrating, we have

∫-kdA/-k(L - Ak) = ∫dt

1/k∫-kdA/(L - Ak) = ∫dt

1/k㏑(L - Ak) = t + C

㏑(L - Ak) = kt + kC

㏑(L - Ak) = kt + C'      (C' = kC)

taking exponents of both sides, we have

L - Ak = e^{kt + C'} \\L - Ak = e^{kt}e^{C'}\\L - Ak = C"e^{kt}      (C" = e^{C'} )\\Ak = L - C"e^{kt}\\A = \frac{L}{k}  - \frac{C"}{k} e^{kt}

When t = 0, A(0) = 0 (since the forest floor is initially clear)

A = \frac{L}{k}  - \frac{C"}{k} e^{kt}\\0 = \frac{L}{k}  - \frac{C"}{k} e^{k0}\\0 = \frac{L}{k}  - \frac{C"}{k} e^{0}\\\frac{L}{k}  = \frac{C"}{k} \\C" = L

A = \frac{L}{k}  - \frac{L}{k} e^{kt}

So, D = R - A =

D = \frac{L}{k} - \frac{L}{k}  - \frac{L}{k} e^{kt}\\D = \frac{L}{k} e^{kt}

when t = 0(at initial time), the initial value of D =

D = \frac{L}{k} e^{kt}\\D = \frac{L}{k} e^{k0}\\D = \frac{L}{k} e^{0}\\D = \frac{L}{k}

4 0
3 years ago
What number is 60 percent of 29
nevsk [136]
60% is 0.60 so you do 0.60 x 29 which = to 17.4
so your answer is 17.40 or 17.4 is 60% of 29
7 0
4 years ago
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