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kolbaska11 [484]
3 years ago
6

molly was packing books in a box to send to a friend.the box cannot weigh more than 2kg.if each book has a mass of 200g, what is

the maximum number of books she can send?
Mathematics
1 answer:
Lapatulllka [165]3 years ago
4 0
Molly can only send a maximum of 10 books
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Please help me out with this
andre [41]

Answer:

\huge\boxed{\sqrt[3]{c^4}=c^\frac{4}{3}}

Step-by-step explanation:

\sqrt[n]{a^m}=a^\frac{m}{n}\\\\\text{therefore}\\\\\sqrt[3]{c^4}=c^\frac{4}{3}

5 0
2 years ago
Suppose 6 quarts of a solution that is 52% antifreeze is mixed with 10 quarts of a solution that is 32% antifreeze. (A) how many
Vadim26 [7]
The total amount of the resulting mixture can be calculated by adding up the volume of the given substances assuming that volume addition is applicable given the properties of the fluids used. 

That is,
   T = 6 quarts + 10 quarts = 16 quarts

The total volume of the resulting mixture is 16.

Then, we do the component (antifreeze) balance by adding up the resulting antifreeze from the substances to the total. We let x be the percentage of antifreeze in the final mixture.

    6(0.52) + 10(0.32) = 16(x)

The value of x from the equation is 0.395.

Therefore, the answer to this item is 39.5%. 
4 0
2 years ago
Find the missing side of each triangle. Leave your answers in simplest radical form.
Vladimir79 [104]

Answer:

Solution given:

p=10mi

b=6mi

h=x.

by using Pythagoras law

h²=p²+b²

x²=10²+6²

x=\sqrt{136}=2\sqrt{34}

3 0
2 years ago
Read 2 more answers
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
2 years ago
Which number has a 3 that 10 times greater than the 3 in 513?
cestrela7 [59]

The 3 in "513" is only 3. it's 1, 2, 3. If we do 3 x 10 it's 30

3 0
2 years ago
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