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2 years ago

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For dinner last night, Vinnie ordered a large supreme pizza with extra anchovies. The regular price of the pizza is $24 but Vinn

ie had a 20% off coupon. When Vinnie went to pay the bill, an 10% sales tax was added onto the discounted price.
How much did Vinnie pay for the pizza?

1 answer:

MaRussiya [10]2 years ago

8 0

Answer:

$20.94

Step-by-step explanation:

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A machine can make 56 parts in 4 hours.It can also make 70 parts in 5 hours. Write an equation that relates the number of parts

ira [324]

Answer:

126 parts in 9 and a half hours

Step-by-step explanation:

56 parts in 4 hours

that means 56 parts in 240 mins

9 hours and 30 minutes in minutes is 570 minutes

56m parts:240 mins

540 mins

to get to 540 from 240, you multiply by 2.25.

you can work this out by doing 540 divided by 240

now, multiply 56 by 2.25.

the answer is 126 parts

5 0

3 years ago

Find the solution to the equations.<br> 2х-у=-3<br> х+у=0<br><br> 0,0<br><br> -1,1<br><br> 1,-1

kobusy [5.1K]

It’s 0,0. The correct answer!

3 0

3 years ago

Read 2 more answers

Help me guys please need to finish this ASAP

mezya [45]

Answer:

B Wednesday

Step-by-step explanation: Don't cheat on nwea btw

6 0

2 years ago

Read 2 more answers

A researcher compares differences in positivity between participants in a low-, middle-, or upper-middle-class family. If she ob

777dan777 [17]

Answer:

(2, 12)

Step-by-step explanation:

Analysis of variance (ANOVA) "is used to analyze the differences among group means in a sample".

The sum of squares "is the sum of the square of variation, where variation is defined as the spread between each individual value and the grand mean"

If we assume that we have $3$ groups and on each group from $j=1,\dots,n_j$ we have $n_j$ individuals on each group we can define the following formulas of variation:

$SS_{total}=\sum_{j=1}^p \sum_{i=1}^{n_j} (x_{ij}-\bar x)^2$

$SS_{between=Treatment}=SS_{model}=\sum_{j=1}^p n_j (\bar x_{j}-\bar x)^2$

$SS_{within}=SS_{error}=\sum_{j=1}^p \sum_{i=1}^{n_j} (x_{ij}-\bar x_j)^2$

And we have this property

$SST=SS_{between}+SS_{within}$

The degrees of freedom for the numerator on this case is given by $df_{num}=k-1=3-1=2$ where k =3 represent the number of groups.

The degrees of freedom for the denominator on this case is given by $df_{den}=df_{between}=N-K=15-3=12$ .

And the total degrees of freedom would be $df=N-1=15 -1 =14$

And the correct answer would be 2 degrees of freedom for the numerator and 12 for the denominator

(2, 12)

3 0

3 years ago

Find the general solution to each of the following ODEs. Then, decide whether or not the set of solutions form a vector space. E

Ipatiy [6.2K]

Answer:

(A) $y=ke^{2t}$ with $k\in\mathbb{R}$ .

(B) $y=ke^{2t}/2-1/2$ with $k\in\mathbb{R}$

(C) $y=k_1e^{2t}+k_2e^{-2t}$ with $k_1,k_2\in\mathbb{R}$

(D) $y=k_1e^{2t}+k_2e^{-2t}+e^{3t}/5$ with $k_1,k_2\in\mathbb{R}$ ,

Step-by-step explanation

(A) We can see this as separation of variables or just a linear ODE of first grade, then $0=y'-2y=\frac{dy}{dt}-2y\Rightarrow \frac{dy}{dt}=2y \Rightarrow \frac{1}{2y}dy=dt \ \Rightarrow \int \frac{1}{2y}dy=\int dt \Rightarrow \ln |y|^{1/2}=t+C \Rightarrow |y|^{1/2}=e^{\ln |y|^{1/2}}=e^{t+C}=e^{C}e^t} \Rightarrow y=ke^{2t}$ . With this answer we see that the set of solutions of the ODE form a vector space over, where vectors are of the form $e^{2t}$ with $t$ real.

(B) Proceeding and the previous item, we obtain $1=y'-2y=\frac{dy}{dt}-2y\Rightarrow \frac{dy}{dt}=2y+1 \Rightarrow \frac{1}{2y+1}dy=dt \ \Rightarrow \int \frac{1}{2y+1}dy=\int dt \Rightarrow 1/2\ln |2y+1|=t+C \Rightarrow |2y+1|^{1/2}=e^{\ln |2y+1|^{1/2}}=e^{t+C}=e^{C}e^t \Rightarrow y=ke^{2t}/2-1/2$ . Which is not a vector space with the usual operations (this is because $-1/2$ ), in other words, if you sum two solutions you don't obtain a solution.

(C) This is a linear ODE of second grade, then if we set $y=e^{mt} \Rightarrow y''=m^2e^{mt}$ and we obtain the characteristic equation $0=y''-4y=m^2e^{mt}-4e^{mt}=(m^2-4)e^{mt}\Rightarrow m^{2}-4=0\Rightarrow m=\pm 2$ and then the general solution is $y=k_1e^{2t}+k_2e^{-2t}$ with $k_1,k_2\in\mathbb{R}$ , and as in the first items the set of solutions form a vector space.

(D) Using C, let be $y=me^{3t}$ we obtain that it must satisfies $3^2m-4m=1\Rightarrow m=1/5$ and then the general solution is $y=k_1e^{2t}+k_2e^{-2t}+e^{3t}/5$ with $k_1,k_2\in\mathbb{R}$ , and as in (B) the set of solutions does not form a vector space (same reason! as in (B)).

4 0

3 years ago