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3 years ago

Ryan buys 61.5 ounces of lemonade for $24.60 find the unit price

Mathematics

1 answer:

Gekata [30.6K]3 years ago

8 0

Answer:

$0.4 for every ounce of lemonade

Step-by-step explanation:

$24.60 ÷ 61.5 ounces of lemonade = 0.4

Hope this is helpful!

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The eighth grade math teachers decided to

Lubov Fominskaja [6]

Answer:

this would be your answer not sure if you need to round 4.085 so pretty much the answer would be 3 because you have enough for about 4 and a half

Step-by-step explanation:

4 0

2 years ago

A couple plans to have children until they get a girl, but they agree they will not have more than three children, even if all

barxatty [35]

Answer:

a)Let X be number of children

X=1,2,3

P(X=1)= 0.47

P(X=2) = 0.2491

P(X=3) = 0.942923

b) 3.7970

c) 1.6232

Step-by-step explanation:

The complete question is:

A couple plans to have children until they get a girl, but they agree they will not have more than three children, even if all are boys. Assume that the probability of having a girl is 47.00%.

a) Create a probability model for the number of children they'll have.

X=1,2,3

P(X=1)=??

P(X=2)= ??

P(X=30=???

(Round to four decimal places as needed

b) Find the expected number of children.

E(X)= ???

c) Find the expected number of boys they'll have.

Expected number of boys= ???

Solution:

Probability of a girl= 0.47

Probability of a boy= 0.53

a) P(X=1)= 0.47

P(X=2) = 0.47× 0.53= 0.2491

P(X=3)= 0.47× 0.53× 0.53 + 0.53× 0.53× 0.53

= 0.942923

b) E(number of children)= 1× P(X=1) + 2 ×P(X=2) + 3 × (PX=3)

= 3.796969

c) Y: number of boys

P(Y=1)= 0.53×0.47= 0.2491

P(Y=2) = 0.53×0.53×0.47=0.46375

P(Y=3)= 0.53× 0.53× 0.53= 0.148875

E(Y)= P(Y=1)×1 + P(Y=2)×2 + P(Y=3)×3

= 0.148875×3 +0.46375×2+0.2491 ×1

= 1.6232

5 0

2 years ago

Lurinda ordered some boxes of greating cards online the cost of the cards is $6.50n + $3 where n is the number of boxes orderd a

Neporo4naja [7]

Answer:

$55

Step-by-step explanation:

For this question you would have to multiply

6.50*8 which equals $52

Then add 3 dollars for the total 52+3=55

6 0

3 years ago

15.30 find the inverse laplace transform of: 1. (a) f1(s) = 6s 2 8s 3 s(s 2 2s 5) 2. (b) f2(s) = s 2 5s 6 (s 1) 2 (s 4) 3. (c) f

EleoNora [17]

The solution of the inverse Laplace transforms is mathematically given as

$f_{1}(t)=e^{-t}\sin (2 t)$

$f_{2}(t)=\frac{7}{9} e^{-t}+\frac{2}{3} e^{-t}+\frac{2}{9} e^{-4 t}$

$f_{3}(t)=2 e^{-t}-2 e^{-2 t} \cos (2 t)-e^{-2 t} \sin (2 t)$

<h3>What is the inverse Laplace transform?</h3>

Generally, the equation for the function is mathematically given as

$F_{1}(s)=\frac{6 s^{2}+8 s+3}{s\left(s^{2}+2 s+5\right)}$

By Applying the Partial fractions method

$\frac{6 s^{2}+8 s+3}{s\left(s^{2}+2 s+5\right)}=\frac{A}{s}+\frac{B s+C}{s^{2}+2 s+5}$

$6 s^{2}+8 s+3=A\left(s^{2}+2 s+5\right)+(B s+C) s$

$\begin{aligned}&3=5 A \\&A=\frac{3}{5}\end{aligned}$

Considers s^2 coefficient

$\begin{aligned}&6=A+B \\&B=6 \cdot A \\&B=\frac{27}{5}\end{aligned}$

Consider s coeffici ent

$\begin{aligned}&8=2 A+C \\&C=8-2 A \\&C=\frac{34}{5}\end{aligned}$

Putting these values into the previous equation

$&F_{1}(s)=\frac{3}{5 s}+\frac{27 s+34}{5\left(s^{2}+2 s+5\right)} \\\\&F_{1}(s)=\frac{3}{5 s}+\frac{27(s+1)}{5\left((s+1)^{2}+4\right)}+\frac{7 \times 2}{10\left((s+1)^{2}+4\right)}$

By taking Inverse Laplace Transforms

$f_{1}(t)=\frac{3}{5}+\frac{27}{5} e^{-t} \cos (2t) + \frac{7}{10}\\\\$

$f_{1}(t)=e^{-t}\sin (2 t)$

For B

$F_{2}(s)=\frac{s^{2}+5 s+6}{(s+4)(s+1)^{2}}$

By Applying Partial fractions method

$\begin{aligned}&\frac{s^{2}+5 s+6}{(s+4)(s+1)^{2}}=\frac{A}{s+1}+\frac{B}{(s+1)^{2}}+\frac{C}{s+4} \\\\&s^{2}+5 s+6=A(s+1)(s+4)+B(s+4)+C(s+1)^{2}\end{aligned}$

at s=-1

$1-5+6=3 B \\\\B=\frac{2}{3}$

at s=-4

$&16-20+6=9 C \\\\&9 C=2 \\\\&C=\frac{2}{9}$

at s^2 coefficient

1=A+C

A=1-C

A=7/9

inputting Variables into the Previous Equation

$\begin{aligned}&F_{2}(s)=\frac{A}{s+1}+\frac{B}{(s+1)^{2}}+\frac{C}{s+4} \\&F_{2}(s)=\frac{7}{9(s+1)}+\frac{2}{3(s+1)^{2}}+\frac{2}{9(s+4)}\end{aligned}$

By taking Inverse Laplace Transforms

$f_{2}(t)=\frac{7}{9} e^{-t}+\frac{2}{3} e^{-t}+\frac{2}{9} e^{-4 t}$

For C

$F_{3}(s)=\frac{10}{(s+1)\left(s^{2}+4 s+8\right)}$

Using the strategy of Partial Fractions

$\frac{10}{(s+1)\left(s^{2}+4 s+8\right)}=\frac{A}{s+1}+\frac{B s+C}{s^{2}+4 s+8}$

$10=A\left(s^{2}+4 s+8\right)+(B s+C)(s+1)$

S=-1

10=(1-4+8) A

A=10/5

A=2

Consider constants

10=8 A+C

C=10-8 A

C=10-16

C=-6

Considers s^2 coefficient

0=A+B

B=-A

B=-2

inputting Variables into the Previous Equation

$&F_{3}(s)=\frac{2}{s+1}+\frac{-2 s-6}{\left((s+2)^{2}+4\right)} \\\\&F_{3}(s)=\frac{2}{s+1}-\frac{2(s+2)}{\left((s+2)^{2}+4\right)}-\frac{2}{\left((s+2)^{2}+4\right)}$

Inverse Laplace Transforms

$f_{3}(t)=2 e^{-t}-2 e^{-2 t} \cos (2 t)-e^{-2 t} \sin (2 t)$