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sineoko [7]
3 years ago
15

Find the magnitude of the vectors {-4, 2} to the nearest tenth

Mathematics
1 answer:
Dimas [21]3 years ago
5 0
\bf \begin{array}{rllll}
\ \textless \ -4&,&2\ \textgreater \ \\
a&&b
\end{array}\implies 
\begin{array}{llll}
||\ \textless \ a,b\ \textgreater \ ||=\sqrt{a^2+b^2}\\\\
||\ \textless \ -4,2\ \textgreater \ ||=\sqrt{(-4)^2+2^2}
\end{array}
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Which of the following shows the polynomial below written in descending order ?
Igoryamba

Answer:

A. 4x¹¹ + x⁷ + 3x³ - 5x + 9

Step-by-step explanation:

order the exponents from largest to smallest. variables always come first; therefore, -5x before 9

4 0
2 years ago
Read 2 more answers
I need help with this please
Ierofanga [76]
C is the correct answer


we should find the area of the square and then divide it by 24.
(360/15=24)


6 \times 6 \times \pi \times  \frac{1}{24}  = 36 \times \pi \times  \frac{1}{24}   \\  = 113.04 \times  \frac{1}{24}  = 4.71



good luck
8 0
3 years ago
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I need help on this may you help
HACTEHA [7]

Answer:

should be 63

Step-by-step explanation:

6 0
1 year ago
The average of four numbers is 98. If three of the numbers are 86,87,and 91 what is the fourth number?
timurjin [86]
To find the average, you have to add all the numbers up and then divide by the amount of numbers you added together.

Equation: (86+87+91+x)/4=98
Solve for x:
86+87+91+x=98*4
264+x=392
x=392-264
x=128

The fourth number is 128
7 0
3 years ago
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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