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jeka94
3 years ago
12

If it rained 17 out of the last 60 days.What was the percentage of rainy days ?

Mathematics
2 answers:
Aloiza [94]3 years ago
7 0
28.33%
to find the percent you have to divide the quantity by the whole then times your quotient by 100
Fantom [35]3 years ago
4 0
Out of the 60 days, it rained 10.2% of the time.
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Write equations for the vertical and horizontal lines passing through the point . (7,5)
Leya [2.2K]

A horizontal line is a line where all of the y values are the same. In this case, \boxed{y=5}, so that is the equation.

A vertical line is where all of the x values are the same.  Here, \boxed{x=7}, so that's the equation.

4 0
4 years ago
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Answer ASAP THANKKK YOUUUUU
alexandr402 [8]

Answer:

D. 40

Step-by-step explanation:

Interquartile range is the difference between the upper quartile value (Q3) and the lower quartile value (Q1).

In a box plot, Q1 is located at the beginning of the edge of the rectangular box from our left, while the Q3 is located at the end of the edge of the rectangular box to our right.

Interquartile range for City A = 70 - 40 = 30

Interquartile range for City B = 80 - 40

Therefore, city B has greater variability. The interquartile range is 40.

7 0
3 years ago
What type of angle is formed by the base and height of a triangle?​
Harrizon [31]

Answer:

the Angle is an Acute

Step-by-step explanation:you can't but a little box in it and it it not a 90 degree angle

3 0
3 years ago
Random digits 0,1,2,3,4,5,6,7,8,9. what is the probability that a multiple of 4 is picked?
Anton [14]

|\Omega|=10\\|A|=3\\\\P(A)=\dfrac{3}{10}=30\%

4 0
3 years ago
You believe the population is normally distributed. Find the 80% confidence interval. Enter your answer as an open-interval (i.e
victus00 [196]

You intend to estimate a population mean μ from the following sample. 26.2 27.7 8.6 3.8 11.6 You believe the population is normally distributed. Find the 80% confidence interval.  Enter your answer as an open-interval (i.e., parentheses) accurate to twp decimal places.

Answer:

The  Confidence interval = (8.98 , 22.18)

Step-by-step explanation:

From the given information:

mean = \dfrac{ 26.2+ 27.7+ 8.6+ 3.8 +11.6 }{5}

mean = 15.58

the standard deviation \sigma = \sqrt{\dfrac{\sum(x_i - \mu)^2 }{n}}

the standard deviation = \sqrt{\dfrac{(26.2  - 15.58)^2 +(27.7 - 15.58)^2 +(8.6  - 15.58)^2 + (3.8  - 15.58)^2  + (11.6  - 15.58)^2  }{5 }  }

standard deviation = 9.62297

Degrees of freedom df = n-1

Degrees of freedom df = 5 - 1

Degrees of freedom df = 4

For df  at 4 and 80% confidence level, the critical value t from t table  = 1.533

The Margin of Error M.O.E = t \times \dfrac{\sigma}{\sqrt{n}}

The Margin of Error M.O.E = 1.533 \times \dfrac{9.62297}{\sqrt{5}}

The Margin of Error M.O.E = 1.533 \times 4.3035

The Margin of Error M.O.E = 6.60

The  Confidence interval = ( \mu  \pm M.O.E )

The  Confidence interval = ( \mu  +  M.O.E , \mu - M.O.E )

The  Confidence interval = ( 15.58 - 6.60 , 15.58 + 6.60)

The  Confidence interval = (8.98 , 22.18)

7 0
3 years ago
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