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nikitadnepr [17]
3 years ago
9

A pizza Parlor offers five different toppings over the plain cheese pizza base that includes mushrooms ,pepperoni.sausage,chicke

n and olives . How many different pizzas can one order
Mathematics
1 answer:
DochEvi [55]3 years ago
5 0

Answer:

<em>Maximum no. of toppings over the plain cheese pizza base can be 3</em>

Step-by-step explanation:

So for adding different toppings  it is up to the customer whether he/she wants full topping pizza or less.

Therefore  5 pizza toppings can be added as 5 single toppings

Hopefully this might help you.

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Write the equation....
BaLLatris [955]
1-0/-1-0 = -1
Y- 0 = -(x-0)
Y=-x
Y - 1 = -(x+1)
Y-1= -x-1
Y = -x
5 0
3 years ago
What is the product of 84,90?
MAVERICK [17]
84•90 is 7,560.

ndheisn
3 0
4 years ago
There has been much media coverage of the high cost of medicinal drugs in the United States. One concern is the large variation
umka21 [38]

Data for the question :

102.05 99.85 112.3 97.15 111.23 105.37 105.64 106.5 102.97 107.82 106.36 111.24 107.28 114.14 106.28 106.96 98.25 111.55 107.75 101.02 101.12 97.7 97.66 100.54 115.77 112.91 111.04 112.15 102.87 101.14 107.13 108.56 109.56 103.57 108.68 104.59 116.74 116.22 100.22 103.97 111.2 109.34 115.78 101.59 107.93 104.23 96.25 103.84 102.47 102.96 99.26 101.42 108.58 107.69 99.88 102.71 111.25 99.4 117.04 106.35 110.44 102.34 107.25 107.63 105.2 109.14 115.54 101.51 108.49 112.32 109.27 97.54 102.46 105.94 109.42 111.05 102.63 106.99 102.03 108.84 118.8 108.64 95.35 105.47 104.45 102.15 111.4 108.27 104.82 108.4 109.05 116.11 103.7 121.2 99.62 102.81 109.56 103.35 113.02 103.79

Answer:

Range = 25.35

Variance = 29.46

Standard deviation = 5.43

The variation in price of Prozac is high

Step-by-step explanation:

The range of the data :

Maximum - Minimum.

121.2 - 95.35 = 25.35

The variance, s :

s² = Σ(X - m²) / n - 1

Mean, m = Σx / n

X = individual data point

m = mean of data

n = sample size

Using a calculator of save time and ensure accuracy :

s² = 29.45522

The standard deviation, s

s = sqrt(variance)

s = sqrt(s²)

s = sqrt(29.45522)

s = 5.42726.

The range, variance and standard deviation, all measure the degree of variation in a dataset. The values of these statistical measure obtain for the price of 1 product across different pharmaceutical stores, suggests thatvthe variation in price is high;

With a range of about 25.35 and standard deviation of 5.43

4 0
3 years ago
Match the systems of linear equations with their solutions.
OleMash [197]

Answer:

The solutions of linear equations in the procedure

Step-by-step explanation:

Part 1) we have

x+y=-1 ----> equation A

-6x+2y=14 ----> equation B

Solve the system by elimination

Multiply the equation A by 6 both sides

6*(x+y)=-1*6

6x+6y=-6 -----> equation C

Adds equation C and equation B

6x+6y=-6

-6x+2y=14

-------------------

6y+2y=-6+14

8y=8

y=1

Find the value of x

substitute in the equation A

x+y=-1 ------> x+1=-1 ------> x=-2

The solution is the point (-2,1)

Part 2) we have

-4x+y=-9 -----> equation A

5x+2y=3 ------> equation B

Solve the system by elimination

Multiply the equation A by -2 both sides

-2*(-4x+y)=-9*(-2)

8x-2y=18 ------> equation C

Adds equation B and equation C

5x+2y=3

8x-2y=18

----------------

5x+8x=3+18

13x=21

x=21/13

Find the value of y

substitute in the equation A

-4x+y=-9 ------> -4(21/13)+y=-9 ----> y=-9+84/13 -----> y=-33/13

The solution is the point (21/13,-33/13)

Part 3) we have

-x+2y=4 ------> equation A

-3x+6y=11 -----> equation B

Multiply the equation A by 3 both sides

3*(-x+2y)=4*3 ------> -3x+6y=12

so

Line A and Line B are parallel lines with different y-intercept

therefore

The system has no solution

Part 4) we have

x-2y=-5 -----> equation A

5x+3y=27 ----> equation B

Solve the system by elimination

Multiply the equation A by -5 both sides

-5*(x-2y)=-5*(-5)

-5x+10y=25 -----> equation C

Adds equation B and equation C

5x+3y=27

-5x+10y=25

-------------------

3y+10y=27+25

13y=52

y=4

Find the value of x

Substitute in the equation A

x-2y=-5 -----> x-2(4)=-5 -----> x=-5+8 ------> x=3

The solution is the point (3,4)

Part 5) we have

6x+3y=-6 ------> equation A

2x+y=-2 ------> equation B

Multiply the equation B by 3 both sides

3*(2x+y)=-2*3

6x+3y=6

so

Line A and Line B is the same line

therefore

The system has infinite solutions

Part 6) we have

-7x+y=1 ------> equation A

14x-7y=28 -----> equation B

Solve the system by elimination

Multiply the equation A by 7 both sides

7*(-7x+y)=1*7

-49x+7y=7 -----> equation C

Adds equation B and equation C

14x-7y=28

-49x+7y=7

------------------

14x-49x=28+7

-35x=35

x=-1

Find the value of y

substitute in the equation A

-7x+y=1  -----> -7(-1)+y=1 ----> y=1-7 ----> y=-6

The solution is the point (-1,-6)

6 0
3 years ago
Solve for x. Leave your answer in simplest radical form.
strojnjashka [21]

There are two Right angled triangles in the given figure, with two missing sides, let's name the other missing side be " y "

According to Pythagoras theorem,

\boxed{ \large \boxed{h {}^{2}  =  {b}^{2} + p{}^{2}  }}

note :

  • p = perpendicular / opposite side
  • b = base / adjacent side
  • h = hypotenuse

let's solve for " y² " first ( square of other missing side )

➢ \:  \:  {y}^{2}  =  {8}^{2}  + 3 {}^{2}

➢ \:  \:  {y}^{2}  = 64 + 9

➢ \:  \:  {y}^{2}  = 73

now let's solve for " x " , by using pythagorous theorem

➢ \:  \:  {x}^{2}  =  {y}^{2}  +  {6}^{2}

➢ \:  \:  {x}^{2}  = 73 + 36

➢ \:  \:x {}^{2}  =  109

➢ \:  \: x =  \sqrt{109}

Approximate :

➢ \:  \: 10.44 \:  \: units

\mathrm{✌TeeNForeveR✌}

5 0
3 years ago
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