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IceJOKER [234]
2 years ago
12

Sean is a bookworm. On average, he reads pages in minutes.

Mathematics
1 answer:
yulyashka [42]2 years ago
8 0
For example:
5 pages in 10 minutes

Per minute: 5/10 = 0.5 pages in 1 minute or half a page
You might be interested in
Herky was given the following system of equations to solve. 3 x minus 2 y = 17. x minus 3 y = 1.
ioda

Answer:

(7, 2 )

Step-by-step explanation:

Given the 2 equations

3x - 2y = 17 → (1)

x - 3y = 1 → (2)

Rearrange (2) expressing x in terms of y by adding 3y to both sides

x = 1 + 3y → (3)

Substitute x = 1 + 3y in (1)

3(1 + 3y) - 2y = 17 ← distribute and simplify left side

3 + 9y - 2y = 17

3 + 7y = 17 ( subtract 3 from both sides )

7y = 14 ( divide both sides by 7 )

y = 2

Substitute y = 2 into (3) for corresponding value of x

x = 1 + 3(2) = 1 + 6 = 7

Solution is (7, 2 )

5 0
2 years ago
Read 2 more answers
Which value of x is in the solution set of 2x-3>11-5x
Ierofanga [76]

7x>14,x=2። is right answer i think

3 0
1 year ago
Read 3 more answers
Joe receives an average of 780 emails in his personal account and 760 emails in his work account each month. After changing his
Semmy [17]

Answer:

702 emails

Step-by-step explanation:

<h2>This problem bothers on depreciation of value, in this context it is Joe's email that has depreciated by 10%.</h2>

Given data

Average personal emails received monthly = 780 emails

Average work emails received monthly= 760 emails

     

      We are required to solve for the new amount of emails Joe will be receiving after changing his address, to find this value we need to solve for the depreciation of his personal mails.

      After solving for the depreciation , we then need to subtract the depreciation from the initial number of mails to get the new number of mails.

let us solve for 10% depreciation.

depreciation= \frac{10}{100} *780\\depreciation=0.1*780= 78 emails

The new number of mails

= initial number of mail- depreciation\\ =780-78= 702 emails

Joe will be receiving an average of 702 emails in his personal account monthly

7 0
2 years ago
In choice situations of this type, subjects often exhibit the "center stage effect," which is a tendency to choose the item in t
victus00 [196]

Complete Question

The complete question is shown on the first uploaded image

Answer:

a

The probability that he or she would choose the pair of socks in the center position is   p =\frac{1}{5}

The correct answer choice is

X has a binomial distribution with parameters n=100 and p=1/5  

b

The mean is  \mu = 20

The standard deviation is \sigma=4

c

The probability, P =0.0002

d

The correct answer is

The experiment supports the center stage effect. If participants were truly picking the socks at random, it would be highly unlikely for 34 or more to choose the center pair.

Using the R the probability Pe = 0.0003

The probabilities P \approx Pe

Step-by-step explanation:

Since the person selects his or her desired pair of socks at random , then the probability that the person would choose the pair of socks in the center position from all the five identical pair is mathematically evaluated as

                  p =\frac{1}{5}

                    =0.2

The mean of this distribution is mathematical represented as

           \mu = np

substituting the value

         \mu = 100 * 0.2

             \mu = 20

The standard deviation is mathematically represented as

         \sigma = \sqrt{np (1-p)}

substituting the value

           = \sqrt{100 * 0,2 (1-0.2)}

           \sigma=4

Applying normal approximation the probability that 34 or more subjects would choose the item in the center if each subject were selecting his or her preferred pair of socks at random would be mathematically represented as

               P=P(X \ge 34 )

By standardizing the normal approximation we have that

              P(X \ge 34) \approx P(Z \ge z)

Now z is mathematically evaluated as

               z = \frac{x-\mu}{\sigma }

Substituting values

             z = \frac{34-20}{4}

               =3.5

So  using the z table the P(Z \ge 3.5) is  0.0002

The probability P and Pe that 34 or more subject would choose the center pair is very small  So

The correct answer is

The experiment supports the center stage effect. If participants were truly picking the socks at random, it would be highly unlikely for 34 or more to choose the center pair.

 

6 0
2 years ago
A dairy sells $3 and $5 ice creams. In one day they sell 50 ice creams earning a total of $180. How many of each type of ice cre
murzikaleks [220]
We are going to make simultaneous equations.
x will be our $3 ice cream and y will be our $5 ice cream

Equation1 ----            x + y = 50   (the sum of all the ice creams they sell)
Equation 2 ----          3x + 5y = 180  Sum of all the $3 and $5 ice creams is $180
Since we can't solve for both variables we will put one of the variables in terms of the other.
Take x+y=50 and subtract y from both sides.  (I could have done subtracted x - it did not matter).       Now we have x= ₋ y +50  (negative y +50)
Now I am going to take equation 2 and replace the x with -y +50

3 (-y +50) + 5y = 180   
Now I will use the distributive law on the 3 and what's in the parentheses:
-3y + 150 + 5y = 180
Now I will combine like terms (the -3y and the 5y)
2y + 150 = 180
Now subtract 150 from both sides of the equation
2y = 30
Divide both sides by 2
and get y= 15 They sold 15 ice creams that cost $5 each
Since equation 1  is  x+y=50 we can replace y with 15
x + 15 = 50    Now subtract 15 from both sides  x = 35
Since x represents the $3 ice creams, they sold 35 of those.
Check:
35 X 3 = $105
15 x 5  = +  <u>75
</u>               $180  

8 0
2 years ago
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