Why isn't 20 over 100 proportional

The daily demand for ice cream cones at a price of $1.20 per cone is 50 cones. At a price of $2.20 per cone, the demand is 30 co

Lapatulllka [165]

Answer:

<h2>The demand at a price of $1.50 per cone is 44 cones</h2>

Step-by-step explanation:

let us apply the interpolation formula to solve this problem

we have

$\frac{y-y1}{x-x1} =\frac{y2-y1}{x2-x1}$

Now let us set the values to the notation above

x= $1.20

x1= $2.20

x2= $1.50

y = 50 cones

y1= 30 cones

y2= ?

$y2= \frac{(y-y1)}{(x-x1)} (x2-x1)+y1$

Substituting our given data we have

$y2= \frac{(50-30)}{(1.2-2.2)} (1.5-2.2)+30\\\\y2= \frac{20}{-1} (-0.7)+30\\\\y2= (20*0.7)+30\\y2= 14+30\\y2= 44$

the demand at a price of $1.50 per cone is 44 cones

4 0

3 years ago

If f(x)=-16x + 70x + 5, what is f(3)?

Svetach [21]

Answer :
F(3)=-16(3)+70(3)+5
F(3)= 167
, check using your calculator

4 0

3 years ago

Read 2 more answers

5. A survey of student pizza preferences showed that 43 students preferred cheese, 56 preferred sausage, 39 preferred pepperoni,

cestrela7 [59]

Answer:

P (Cheese) = 0.199, P (Sausage) = 0.259, P (Pepperoni) = 0.181,

P (Supreme) = 0.130, P (Another Kind) = 0.144

and P (Does not like any kind) = 0.088

Step-by-step explanation:

Given:

Number of students who prefer cheese = 43

Number of students who prefer sausage = 56

Number of students who prefer pepperoni = 39

Number of students who prefer supreme = 28

Number of students who prefer another kind = 31

Number of students who did not like any kind = 19

∴ The total number of students surveyed = $43+56+39+28+31+19=216$ The number of students who prefer pizza = $43+56+39+28+31=197$

The probability that a students likes pizza is,

$P(Student\ likes\ pizza)=\frac{No.\ of\ students\ who\ prefer\ pizza}{Total\ no.\ of\ students\ surveyed}$

$=\frac{197}{216} \\=0.912$

The probability that a students does not likes pizza is,

$P(Student\ does\ not\ likes\ pizza)=\frac{No.\ of\ students\ who\ does\ not\ prefer\ pizza}{Total\ no.\ of\ students\ surveyed}$

$=\frac{19}{216} \\=0.088$

The probability distribution of students who prefer different kinds of pizza is:

The probability that a student likes cheese:

$P(A\ Student\ prefers\ cheese)=\frac{No.\ of\ students\ who\ prefer\ cheese}{Total\ no.\ of\ students\ surveyed}$

$=\frac{43}{216}\\=0.199$

The probability that a student likes sausage:

$P(A\ Student\ prefers\ sausage)=\frac{No.\ of\ students\ who\ prefer\ sausage}{Total\ no.\ of\ students\ surveyed}$

$=\frac{56}{216}\\=0.259$

The probability that a student likes pepperoni:

$P(A\ Student\ prefers\ pepperoni)=\frac{No.\ of\ students\ who\ prefer\ pepperoni}{Total\ no.\ of\ students\ surveyed}$

$=\frac{39}{216}\\=0.181$

The probability that a student likes supreme:

$P(A\ Student\ prefers\ supreme)=\frac{No.\ of\ students\ who\ prefer\ supreme}{Total\ no.\ of\ students\ surveyed}$

$=\frac{28}{216}\\=0.130$

The probability that a student likes another kind:

$P(A\ Student\ prefers\ another\ kind)=\frac{No.\ of\ students\ who\ prefer\ another\ kind}{Total\ no.\ of\ students\ surveyed}$

$=\frac{31}{216}\\=0.144$

Thus, the probability distribution table is displayed below:

6 0

3 years ago

F(x) = 9 + 4x f(0) = f(-1) = Find the value of x for which f(x) =6 x=

puteri [66]

Answer: x=-3/4

Step-by-step explanation:

Since we know f(x)=6, we can set it equal to the equation.

6=9+4x [subtract 9 on both sides]

-3=4x [divide both sides by 4]

x=-3/4

8 0

3 years ago

Experimentally verify sum of three angles of an triangle is 180⁰.

svetlana [45]

Draw line a through points A and B. Draw line b through point C and parallel to line a.

Since lines a and b are parallel, <)BAC = <)B'CA and <)ABC = <)BCA'.

It is obvious that <)B'CA + <)ACB + <)BCA' = 180 degrees.

Thus <)ABC + <)BCA + <)CAB = 180 degrees.

Lemma

If ABCD is a quadrilateral and <)CAB = <)DCA then AB and DC are parallel.

Proof

Assume to the contrary that AB and DC are not parallel.

Draw a line trough A and B and draw a line trough D and C.

These lines are not parallel so they cross at one point. Call this point E.

Notice that <)AEC is greater than 0.

Since <)CAB = <)DCA, <)CAE + <)ACE = 180 degrees.

Hence <)AEC + <)CAE + <)ACE is greater than 180 degrees.

Contradiction. This completes the proof.

Definition

Two Triangles ABC and A'B'C' are congruent if and only if

|AB| = |A'B'|, |AC| = |A'C'|, |BC| = |B'C'| and,

<)ABC = <)A'B'C', <)BCA = <)B'C'A', <)CAB = <)C'A'B'.

4 0

3 years ago