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dolphi86 [110]
3 years ago
6

Solve using this number 1,,3,5,7,9,11,13,15 ( + + +=30) ​

Mathematics
1 answer:
cricket20 [7]3 years ago
6 0

Step-by-step explanation: 15 + 13 + 2

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what is the sum of the first 25 terms of an arithematic sequence if the sum of its 8th and 18th terms is 72?​
max2010maxim [7]

Answer:

900

Step-by-step explanation:

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Find the surface area of a sphere that has a volume of 288 pi cu. in.​
icang [17]

Answer:

144

Step-by-step explanation:

The surface area of a sphere that has a volume of 288π cu in is 144 pi.

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3 years ago
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The distance on the map between Yappy city and Quiet city is 23 cm.
jekas [21]

The actual distance between Yappy city and Quiet city when it is 23 cm in map is 9.2 miles.

What is scaling ?

It is the technique to resize drawing with comparing to actual size of object which uses proportion or ratio. For example, Mapping of earth, drawing prototype of automobiles.

It is given that :

The distance on the map between Yappy city and Quiet city is 23 cm

and,

5cm = 2 miles

now, for 1 cm on map it is equal to 2/5 miles of actual distance.

Therefore to find the actual distance of 23 cm in map we simply multiply 23 with 2/5 :

actual distance :

2/5 x 23 = 9.2 miles

Therefore, the actual distance between Yappy city and Quiet city is 9.2 miles.

Read more about ratio at:

brainly.com/question/17869111

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5 0
1 year ago
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lapo4ka [179]
The second choice i believe
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3 years ago
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The probability that a professor arrives on time is 0.8 and the probability that a student arrives on time is 0.6. Assuming thes
saul85 [17]

Answer:

a)0.08  , b)0.4  , C) i)0.84  , ii)0.56

Step-by-step explanation:

Given data

P(A) =  professor arrives on time

P(A) = 0.8

P(B) =  Student aarive on time

P(B) = 0.6

According to the question A & B are Independent  

P(A∩B) = P(A) . P(B)

Therefore  

{A}' & {B}' is also independent

{A}' = 1-0.8 = 0.2

{B}' = 1-0.6 = 0.4

part a)

Probability of both student and the professor are late

P(A'∩B') = P(A') . P(B')  (only for independent cases)

= 0.2 x 0.4

= 0.08

Part b)

The probability that the student is late given that the professor is on time

P(\frac{B'}{A}) = \frac{P(B'\cap A)}{P(A)} = \frac{0.4\times 0.8}{0.8} = 0.4

Part c)

Assume the events are not independent

Given Data

P(\frac{{A}'}{{B}'}) = 0.4

=\frac{P({A}'\cap {B}')}{P({B}')} = 0.4

P({A}'\cap {B}') = 0.4 x P({B}')

= 0.4 x 0.4 = 0.16

P({A}'\cap {B}') = 0.16

i)

The probability that at least one of them is on time

P(A\cup B) = 1- P({A}'\cap {B}')  

=  1 - 0.16 = 0.84

ii)The probability that they are both on time

P(A\cap  B) = 1 - P({A}'\cup {B}') = 1 - [P({A}')+P({B}') - P({A}'\cap {B}')]

= 1 - [0.2+0.4-0.16] = 1-0.44 = 0.56

6 0
3 years ago
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