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Eduardwww [97]
3 years ago
11

What is the mean of the data set? I got 177 when i added up the numbers but what would i divide 177 by?

Mathematics
1 answer:
Damm [24]3 years ago
7 0
If you would like to know the mean of the data set, you can calculate this using the following steps:

The <span>mean </span><span>of the data set is the sum of all the data values divided by the number of these values.
</span><span>  </span>
<span>12, 17, 16, 10, 20, 13, 14, 14, 12, 12, 19, 18
</span>
<span>12 + 17 + 16 + 10 + 20 + 13 + 14 + 14 + 12 + 12 + 19 + 18 = 177
</span>the number of data values: 12
177 / 12 = 14.75

The correct result would be B. 14.75.
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The attendance at the art museum at the New Year's opening was 250 people. The attendance has been increasing at a rate of 3% ea
erastovalidia [21]

Given:

Initially the number of people at art museum = 250

Rate of increase = 3% per month

To find:

The number of people at art museum in the end of a year.

Solution:

The general exponential growth model is:

y=a\left(1+\dfrac{r}{100}\right)^t

Where, a is the initial value, r is the rate of increase in % and t is the time period.

We know that, 1 year = 12 months.

Putting a=250, r=3 and t=12, we get

y=250\left(1+\dfrac{3}{100}\right)^{12}

y=250\left(1+0.03\right)^{12}

y=250\left(1.03\right)^{12}

y=250(1.42576)

y=356.44

y\approx 356

The number of people at the end of a year is 356. Therefore, the correct option is D.

3 0
3 years ago
What is circumference and pi ?<br>how to solve ?​
alexandr402 [8]

Hi :)

<u>===============================================================</u>

CIRCUMFERENCE

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3 0
1 year ago
In a class of 140 students, 3\5 of them are girls. how many girls are in the class?
Mila [183]

Answer:

84

Step-by-step explanation:

3/5 of 140

=>3/5 x 140

=>3 x 28

=>84

6 0
3 years ago
A random sample of 28 statistics tutorials was selected from the past 5 years and the percentage of students absent from each on
SVEN [57.7K]

Answer:

Step-by-step explanation:

Hello!

X: number of absences per tutorial per student over the past 5 years(percentage)

X≈N(μ;σ²)

You have to construct a 90% to estimate the population mean of the percentage of absences per tutorial of the students over the past 5 years.

The formula for the CI is:

X[bar] ± Z_{1-\alpha /2} * \frac{S}{\sqrt{n} }

⇒ The population standard deviation is unknown and since the distribution is approximate, I'll use the estimation of the standard deviation in place of the population parameter.

Number of Absences 13.9 16.4 12.3 13.2 8.4 4.4 10.3 8.8 4.8 10.9 15.9 9.7 4.5 11.5 5.7 10.8 9.7 8.2 10.3 12.2 10.6 16.2 15.2 1.7 11.7 11.9 10.0 12.4

X[bar]= 10.41

S= 3.71

Z_{1-\alpha /2}= Z_{0.95}= 1.645

[10.41±1.645*(\frac{3.71}{\sqrt{28} } )]

[9.26; 11.56]

Using a confidence level of 90% you'd expect that the interval [9.26; 11.56]% contains the value of the population mean of the percentage of absences per tutorial of the students over the past 5 years.

I hope this helps!

7 0
4 years ago
Can someone please answer this question I need it today please show work and please answer it correctly i
svlad2 [7]

Answer:

24) $495

25) 14%

26) 25/X = 83/100

27) 0.7p

28) x + .085x and 1.085x

29) $221.90

30) $24.10

31) $6.13

32) 40%

Step-by-step explanation:

24) 600 - (600 × 0.25) = 450

450 × 1.10 = 495

25) (106 - 93) ÷ 93 = 0.13978

0.13978 × 100 = 13.978 ~ 14

27) 1.0 - 0.3 = 0.7

28) 1.00 + 0.085 = 1.085

29) 100% - 15% = 85%

240 × 0.85 = 204

204 × 1.0875 = 221.85

30) 25.89 × 4 = 103.56

103.56 + 179.99 = 283.55

283.55 × 0.085 = 24.10175

31) 8.75 × 0.70 = 6.125

32) 80 - (80 × 0.40) = 48

4 0
4 years ago
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