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anyanavicka [17]
2 years ago
9

Our friend purchased a medium pizza for $10.31 with a 30% off coupon. What is the price of a medium pizza without a coupon? Do n

ot include units (dollars) in your answer.
Mathematics
1 answer:
Komok [63]2 years ago
3 0

Answer:

the price of a medium pizza without a coupon is 14.73

Step-by-step explanation:

The computation of the price of a medium pizza without a coupon is given below:

Given that

The Price of medium pizza be 10.31 after considering the 30% off coupon

That means it is 70% value

So the 100% value would be

= 10.31 × 1 ÷ 0.70

= 14.73

hence, the price of a medium pizza without a coupon is 14.73

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-28/((-12+9)-(9+12/3)+1)
Mazyrski [523]
The answer is -1.86




3 0
3 years ago
100 points + brainiest
Scrat [10]

Answer:

  • 35 grams of 11.43% solution

Step-by-step explanation:

The end solution has same volume as initial volumes of mixture.

  • 10 + 25 = 35 g

The concentration of solution is calculated as below

  • 10*0.15 + 25*0.1 = 4 g
  • 4 / 35*100% = 11.43% (rounded)

6 0
2 years ago
Find the surface area of the triangular prism. Will mark brainliest!
klio [65]

Answer:

720 sq ft

Step-by-step explanation:

Formula:

2 B + P x H

3 0
3 years ago
Read 2 more answers
An article suggests that a poisson process can be used to represent the occurrence of structural loads over time. suppose the me
kirill115 [55]

Answer:

a) \lambda_1 = 2*2 = 4

And let X our random variable who represent the "occurrence of structural loads over time" we know that:

X(2) \sim Poi (4)

And the expected value is E(X) = \lambda =4

So we expect 4 number of loads in the 2 year period.

b) P(X(2) >6) = 1-P(X(2)\leq 6)= 1-[P(X(2) =0)+P(X(2) =1)+P(X(2) =2)+...+P(X(2) =6)]

P(X(2) >6) = 1- [e^{-4}+ \frac{e^{-4}4^1}{1!}+ \frac{e^{-4}4^2}{2!} +\frac{e^{-4}4^3}{3!} +\frac{e^{-4}4^4}{4!}+\frac{e^{-4}4^5}{5!}+\frac{e^{-4}4^6}{6!}]

And we got: P(X(2) >6) =1-0.889=0.111

c)  e^{-2t} \leq 2

We can apply natural log in both sides and we got:

-2t \leq ln(0.2)

If we multiply by -1 both sides of the inequality we have:

2t \geq -ln(0.2)

And if we divide both sides by 2 we got:

t \geq \frac{-ln(0.2)}{2}

t \geq 0.8047

And then we can conclude that the time period with any load would be 0.8047 years.

Step-by-step explanation:

Previous concepts

The exponential distribution is "the probability distribution of the time between events in a Poisson process (a process in which events occur continuously and independently at a constant average rate). It is a particular case of the gamma distribution". The probability density function is given by:

P(X=x)=\lambda e^{-\lambda x}

The exponential distribution is "the probability distribution of the time between events in a Poisson process (a process in which events occur continuously and independently at a constant average rate). It is a particular case of the gamma distribution"

Solution to the problem

Let X our random variable who represent the "occurrence of structural loads over time"

For this case we have the value for the mean given \mu = 0.5 and we can solve for the parameter \lambda like this:

\frac{1}{\lambda} = 0.5

\lambda =2

So then X(t) \sim Poi (\lambda t)

X follows a Poisson process

Part a

For this case since we are interested in the number of loads in a 2 year period the new rate would be given by:

\lambda_1 = 2*2 = 4

And let X our random variable who represent the "occurrence of structural loads over time" we know that:

X(2) \sim Poi (4)

And the expected value is E(X) = \lambda =4

So we expect 4 number of loads in the 2 year period.

Part b

For this case we want the following probability:

P(X(2) >6)

And we can use the complement rule like this

P(X(2) >6) = 1-P(X(2)\leq 6)= 1-[P(X(2) =0)+P(X(2) =1)+P(X(2) =2)+...+P(X(2) =6)]

And we can solve this like this using the masss function:

P(X(2) >6) = 1- [e^{-4}+ \frac{e^{-4}4^1}{1!}+ \frac{e^{-4}4^2}{2!} +\frac{e^{-4}4^3}{3!} +\frac{e^{-4}4^4}{4!}+\frac{e^{-4}4^5}{5!}+\frac{e^{-4}4^6}{6!}]

And we got: P(X(2) >6) =1-0.889=0.111

Part c

For this case we know that the arrival time follows an exponential distribution and let T the random variable:

T \sim Exp(\lambda=2)

The probability of no arrival during a period of duration t is given by:

f(T) = e^{-\lambda t}

And we want to find a value of t who satisfy this:

e^{-2t} \leq 2

We can apply natural log in both sides and we got:

-2t \leq ln(0.2)

If we multiply by -1 both sides of the inequality we have:

2t \geq -ln(0.2)

And if we divide both sides by 2 we got:

t \geq \frac{-ln(0.2)}{2}

t \geq 0.8047

And then we can conclude that the time period with any load would be 0.8047 years.

3 0
3 years ago
Use radical notation to rewrite each expression and simplify<br> 9 3/2
scoundrel [369]

Answer:

  9^{\frac{3}{2}}=\sqrt{9^3}=27

Step-by-step explanation:

The fractional exponent m/n is often translated to radical form as ...

  x^{\frac{m}{n}}=\sqrt[n]{x^m}

In this case, I find it easier to evaluate as ...

  x^{\frac{m}{n}}=(\sqrt[n]{x})^m=\boxed{(\sqrt{9})^3=3^3=27}

7 0
3 years ago
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