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Ostrovityanka [42]
3 years ago
12

Mr. Mckay spent $2.60 for a box of crackers and then divided them evenly between the s students in

Mathematics
1 answer:
Nookie1986 [14]3 years ago
4 0

Answer:

You would go $2.60/s

then when you have your total number of students you take the total cost and divide that by the number of students that you have to figure out the cost of crackers for each student

Step-by-step explanation:

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Plsssssss helppp ;w;w;
drek231 [11]

<u>Answer:</u>

Given below.

<u>Explanation:</u>

Coordinates given: W( -4, -3 ) , X( 1 , -2 ), Y(2, -7 ) , Z( -3, -8)

To find the distance we will use the formula: distance=√((x2-x1)²+(y2-y1)²)

WX = √((1--4)²+(-2--3)²)

      = √26

YZ = √((-3-2)²+(-8--7)²)

     = √26

XY = √((2-1)²+(-7--2)²)

     = √26

WZ = √(-3--4)²+(-8--3)²

      = √26

WY =  √(2--4)²+(-7--3)²

     = 2√13

XZ = √(-3-1)²+(-8--2)²

     = 2√13

7 0
2 years ago
Read 2 more answers
What is the slope of (-13,8) and (3,12)
love history [14]

Answer:

Slope = 1/4

Step-by-step explanation:

Y2 - Y1 / X2 - X1

8 0
3 years ago
Solve for x. Round to the nearest tenth.
ivanzaharov [21]

x ≈ 59°

Work is on piece of paper, yet I don't have a phone. I'm certain that is the answer.

5 0
3 years ago
There are 10 cards. Each card has one number between 1 and 10, so that every number from 1 to 10 appears once.
pychu [463]

Answer:

When a card is chosen at random with replacement five times, X is the number of times a prime number is chosen.          Here the card is chosen with replacement.  This implies probability for choosing a prime number remains the same as the previously drawn card is replaced.

The sample space= {1,2,3,4,5,6,7,8,9,10}

Prime numbers = {2,3,5,7}

Prob for drawing prime number = 4/10 = 0.4

is the same when replacement is done.

Also there are two outcomes either prime or non prime.  Hence in this case, X the no of times a prime number is chosen, is binomial with p =0.4 and q = 0.6 and n=5


When a card is chosen at random without replacement three times, X is the number of times an even number is chosen.

Prob for an even number = 0.5

But after one card drawn say odd number next card has prob for even number as 5/9 hence each draw is not independent of the other.  Hence not binomial.

When a card is chosen at random with replacement six times, X is the number of times a 3 is chosen.

Here since every time replacement is done, probability of drawing a 3 remains constant = 1/10 = 0.3

i.e. each draw is independent of the other and there are only two outcomes , 3 or non 3. Hence here X is binomial.

When a card is chosen at random with replacement multiple times, X is the number of times a card is chosen until a 5 is chosen

Here X is the number of times a card is chosen with replacement till 5 is chosen.  This is not binomial.  Here probability for drawing nth time correct 5 is  P(non 5 in the first n-1 draws)*P(5 in nth draw) = 0.1^(n-1) (0.9)

Because nCr is not appearing i.e. 5 cannot appear in any order but only in the last draw, this is not binomial.

Step-by-step explanation:


6 0
3 years ago
See attached picture
yanalaym [24]

Answer:

\frac{f(x+h)-f(x)}{h}=2x + h

Step-by-step explanation:

Given

f(x)= x^2 + 2

Required

Determine: \frac{f(x+h)-f(x)}{h}

First, we calculate f(x + h)

f(x)= x^2 + 2

f(x+h) = (x+h)^2+2

f(x+h) = x^2+2xh+h^2+2

So, we have:

\frac{f(x+h)-f(x)}{h} = \frac{x^2 + 2xh + h^2 + 2 - x^2 - 2}{h}

\frac{f(x+h)-f(x)}{h}= \frac{x^2 - x^2+ 2xh + h^2 + 2  - 2}{h}

\frac{f(x+h)-f(x)}{h} = \frac{2xh + h^2}{h}

\frac{f(x+h)-f(x)}{h}=2x + h

8 0
3 years ago
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