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1 year ago

A shoreline is eroding at a rate of 1/3 foot per 2 months how many feet are eroding this year

Mathematics

1 answer:

mylen [45]1 year ago

8 0

Answer:

2 feet

Step-by-step explanation:

The shoreline is eroding at 1/3 foot per 2 months. There are 12 months in a year so you multiply 1/3 x 6 and you get 2.

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6. Let A and B be nxn matrices . Compute (A + B) (A + B). Explain all steps.

Pie

Answer:

(A+B)(A+B)=A.A+B.A+A.B+B.B

Step-by-step explanation:

Given that matrices A and B are nxn matrices

We need to find (A+B)(A+B)

For understanding the multiplication of matrices let'take A is mxn and B is pxq matrices,we can multiple only when n=p,so our Ab matrices will be mxq.

We know that that in matrices AB is not equal to BA.

Now find

(A+B)(A+B)=A.A+B.A+A.B+B.B

So from we can say that (A+B)(A+B) is not equal to A.A+2B.A+B.B because AB is not equal to BA in matrices.

So (A+B)(A+B)=A.A+B.A+A.B+B.B

4 0

2 years ago

-x + 2y = 3 2x – 3y = -6

s2008m [1.1K]

Answer:

x = -3

y = 0

Step-by-step explanation:

Given equations :- 

-x + 2y = 3 ....... ( i )

2x - 3y = -6 ....... ( ii )

From ( i ) 

-x + 2y = 3

-x = 3 - 2y

x = 2y - 3 ........ ( iii )

From ( ii ) 

2x - 3y = -6

2x = -6 + 3y

$x = \frac{ - 6 + 3y}{2}$

........ ( iv )

Equating ( iii ) and ( iv )

x = x

$2y - 3 = \frac{ - 6 + 3y}{2}$

4y - 6 = -6 + 3y

4y - 3y = -6 + 6

y = 0

Putting value of y in ( iii )

x = 2y - 3

x = 2 ( 0 ) - 3

x = -3

4 0

2 years ago

HELP ME PLEASE!! look at screenshot!! (10 pts)

ivolga24 [154]

Answer:

C) 72 square yards

Step-by-step explanation:

I cut it into two shapes. One being the rectangle on top with values 12 yd and 3 yd. the other being the square 6yd by 6yd

12•3= 36

6•6= 36

36+36= 72

72 square yards

hope this helps chu luv <3

5 0

2 years ago

Which Quadrant contains the point named by (4,-8)

ELEN [110]

The quadrants the point (4.-8) is in is the fourth quadrants

5 0

2 years ago

Read 2 more answers

Average precipitation for the first 7 months of the year, the average precipitation in toledo, ohio, is 19.32 inches. if the ave

Colt1911 [192]

Part A:

The probability that a normally distributed data with a mean, μ and standard deviation, σ is greater than a given value, a is given by:

$P(x\ \textgreater \ a)=1-P(x\ \textless \ a)=1-P\left(z\ \textless \ \frac{a-\mu}{\sigma}\right)$

Given that the average precipitation in Toledo, Ohio for the past 7 months is 19.32 inches with a standard deviation of 2.44 inches, the probability that a randomly selected year will have precipitation greater than 18 inches for the first 7 months is given by:

$P(x\ \textgreater \ 18)=1-P(x\ \textless \ 18) \\ \\ =1-P\left(z\ \textless \ \frac{18-19.32}{2.44}\right) \\ \\ =1-P(z\ \textless \ -0.5410) \\ \\ =1-0.29426=\bold{0.7057}$

Part B:

The probability that an n randomly selected samples of a normally distributed data with a mean, μ and standard deviation, σ is greater than a given value, a is given by:

$P(x\ \textgreater \ a)=1-P(x\ \textless \ a)=1-P\left(z\ \textless \ \frac{a-\mu}{\frac{\sigma}{\sqrt{n}}}\right)$

Given that the average precipitation in Toledo, Ohio for the past 7 months is 19.32 inches with a standard deviation of 2.44 inches, the probability that 5 randomly selected years will have precipitation greater than 18 inches for the first 7 months is given by:

 $P(x\ \textgreater \ 18)=1-P(x\ \textless \ 18) \\ \\ =1-P\left(z\ \textless \ \frac{18-19.32}{\frac{2.44}{\sqrt{5}}}\right) \\ \\ =1-P(z\ \textless \ -1.210) \\ \\ =1-0.1132=\bold{0.8868}$

7 0

2 years ago