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Leno4ka [110]
3 years ago
14

A data set is shown below. Which is the outlier? 56, 64, 73, 59, 98, 65, 59.

Mathematics
1 answer:
dlinn [17]3 years ago
8 0

Most of the numbers are in the 50’s and 60’s so the outlier would be 98

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What is the value of the expression below? (4/5+3/5)+3.5*5
statuscvo [17]

Step-by-step explanation:

(4/5+3/5)+3.5*5

= 7/5+7/2*5

= 7/5+35/2

= 14/10+175/10

= 189/10

= 18.9

8 0
4 years ago
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Will give brainliest and 100 points
____ [38]

Answer:

Kayla plays on average 0.2 more chess games than Becker

Step-by-step explanation:

The first thing we must do is calculate the mean of each one.

Becker:

(5 + 2 + 4 + 1 + 1 + 4 + 5 + 3 + 2 + 1) / 10 = 2.8

Kayla:

(2 + 3 + 1 + 1 + 4 + 1 + 5 + 3 + 5 + 5) / 10 = 3

if we subtract these averages, we have to:

3 - 2.8 = 0.2

which means that Kayla plays on average 0.2 more chess games than Becker. Since it is not even a complete game, the difference between the two is very small and almost irrelevant.

7 0
3 years ago
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A dog walking service charges a $19 one-time fee, plus $40 per week. The
rusak2 [61]

Answer:

Kevin's estimate is incorrect. When x=4, the value of y is $116, which is not close to

$180.

Step-by-step explanation:

7 0
3 years ago
The sum of two consecutive odd integers is at most 166. What are the greatest consecutive odd integers?
bagirrra123 [75]
. x + (x + 2)≤166
. 2x+2≤166
. x≤83 - 1
. x≤82
. x=81
. 81+(81+2)=164;  164<166
answer: 81 and 83
4 0
4 years ago
A sample of a radioactive isotope had an initial mass of 360 mg in the year 1998 and
RUDIKE [14]

Answer:

193 mg

Step-by-step explanation:

Exponential decay formula:

  • A_t = A_0e^r^t
  • where Aₜ = mass at time t, A₀ = mass at time 0,  r = decay constant (rate), t = time  

Our known variables are:

  • 1998 to the year 2004 is a total of t = 6 years.
  • The sample of radioactive isotope has an initial mass of A₀ = 360 mg at time 0 and a mass of Aₜ = 270 mg at time t.

Let's solve for the decay constant of this sample.

  • 270=360e^-^r^(^6^)
  • 270=360e^-^6^r
  • \frac{3}{4} =e^-^6^r
  • \text{ln} (\frac{3}{4} )= \text{ln}(e^-^6^r)
  • \text{ln} (\frac{3}{4} )=-6r
  • r=-\frac{\text{ln}\frac{3}{4} }{6}
  • r=0.04794701

Using our new variables, we can now solve for Aₜ at t = 7 years, since we go from 2004 to 2011.

Our new initial mass is A₀ = 270 mg. We solved for the decay constant, r = 0.04794701.

  • A_t=270e^-^(^0^.^0^4^7^9^4^8^0^1^)^(^7^)
  • A_t=270e^-^0^.^3^3^5^6^2^9^0^7
  • A_t=193.01982213

The expected mass of the sample in the year 2011 would be 193 mg.

3 0
3 years ago
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