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MA_775_DIABLO [31]
3 years ago
9

A set of data with a mean of 45 and a standard deviation of 8.3 is normally distributed. Find the value that is +1 standard devi

ation away from the mean.
Mathematics
2 answers:
Anuta_ua [19.1K]3 years ago
7 0

Answer:

  53.3

Step-by-step explanation:

Add one standard deviation to the mean:

  45 +1(8.3) = 53.3

Sliva [168]3 years ago
4 0

Answer:

53.3

Step-by-step explanation:

+1 standard deviation away from the mean is 45+1(8.3)

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What’s the slope of the line
Softa [21]

Answer:    -1, 1

-1 = x & 1 = y

7 0
1 year ago
Read 2 more answers
A. 36<br> B. 12<br> C. 80<br> D. 60<br> E. 48<br> F. 20
Art [367]

Answer:

it's d because it's correct

Step-by-step explanation:

6 0
3 years ago
I need help i dont understand this​
mel-nik [20]
140 tickets sold at the door

Step by Step Explanation:

x = tickets sold in advance
y = tickets sold at the door

We use the formula ax+by=c and a+b=c

x + y = 514
17x + 20y = 9158

y = 514 - x
17x + 20(514 - x)
17x + 10,280 - 20x = 9,158
-3x + 10,280 = 9,158
-3x = -11222
x = 374

This is for tickets sold in advance so we need to find y

374 + y = 514
y = 140

140 tickets were sold at the door

8 0
3 years ago
if there are two containers of sugar solution, the first is 4% concentration and second is 8% concentration. how much of each sh
Elina [12.6K]

Answer:

<u>30 gallons solution 4% and 10 gallons solution 8%</u>

Step-by-step explanation:

<h3>x•4%+y•8%=40•5% </h3>

<em><u>4x+8y=40•5 and x+y=40</u></em>

x=40-y

4(40-y)+8y=200

160-4y+8y=200

4y=200-160

4y=40 => y=10 (gallons solution 8%)

=> x=40-10=30 (gallons solution 4%)

30 gallons solution 4% and 10 gallons solution 8%

5 0
3 years ago
The radioactive isotope radium has a half-life of 1,600 years. The mass of a 100-gram sample of radium will be______ grams after
Olenka [21]
To solve this we are going to use the half life equation N(t)=N_{0} e^{( \frac{-0.693t}{t _{1/2} }) }
Where:
N_{0} is the initial sample
t is the time in years
t_{1/2} is the half life of the substance
N(t) is the remainder quantity after t years 

From the problem we know that:
N_{0} =100
t=200
t_{1/2} =1600

Lets replace those values in our equation to find N(t):
N(200) =100e^{( \frac{(-0.693)(200)}{1600}) }
N(200)=100e^{( \frac{-138.6}{1600} )}
N(200)=100e^{-0.086625}
N(200)=91.7

We can conclude that after 1600 years of radioactive decay, the mass of the 100-gram sample will be 91.7 grams.

8 0
3 years ago
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