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Romashka-Z-Leto [24]
3 years ago
7

Toby unexpectedly went to the hospital to get stitches for a bad cut on his leg. He doesn't have much money in his bank account.

How should Toby pay for the hospital visit?
debit card
debit card

credit card
credit card

check
check

cash
Mathematics
2 answers:
avanturin [10]3 years ago
8 0

Answer:

debit card

Step-by-step explanation:

ollegr [7]3 years ago
7 0
A.) debit card , don’t use credit !
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Draw the standard normal distribution curve
kirza4 [7]

Standard Normal Distribution. As discussed in the introductory section, normal distributions do not necessarily have the same means and standard deviations. A normal distribution with a mean of 0 and a standard deviation of 1 is called a standard normal distribution.


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3 years ago
Select the question that is different from the other three.
dlinn [17]
yup and your going to jail :)
3 0
3 years ago
Need help Will give you the brainliest
marshall27 [118]

Answer:

The missing area for both sides is 48ft²

the missing dimension is 12ft

Step-by-step explanation:

Add all the areas up

12+36+12+36=96

192-96=96

96÷2=48

48÷4=12

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3 years ago
A number cube is rolled once. Answer these questions below?
Yakvenalex [24]

Answer:

1/2

Step-by-step explanation:

1) the sample space is 1 through 6

2) numbers less than 4 on the number cube 1, 2, and 3 3#s

total amount of #s is 6

probability = numbers less than 4/total amount of #s:

3/6

1/2

6 0
3 years ago
Show that ( 2xy4 + 1/ (x + y2) ) dx + ( 4x2 y3 + 2y/ (x + y2) ) dy = 0 is exact, and find the solution. Find c if y(1) = 2.
fredd [130]

\dfrac{\partial\left(2xy^4+\frac1{x+y^2}\right)}{\partial y}=8xy^3-\dfrac{2y}{(x+y^2)^2}

\dfrac{\partial\left(4x^2y^3+\frac{2y}{x+y^2}\right)}{\partial x}=8xy^3-\dfrac{2y}{(x+y^2)^2}

so the ODE is indeed exact and there is a solution of the form F(x,y)=C. We have

\dfrac{\partial F}{\partial x}=2xy^4+\dfrac1{x+y^2}\implies F(x,y)=x^2y^4+\ln(x+y^2)+f(y)

\dfrac{\partial F}{\partial y}=4x^2y^3+\dfrac{2y}{x+y^2}=4x^2y^3+\dfrac{2y}{x+y^2}+f'(y)

f'(y)=0\implies f(y)=C

\implies F(x,y)=x^2y^3+\ln(x+y^2)=C

With y(1)=2, we have

8+\ln9=C

so

\boxed{x^2y^3+\ln(x+y^2)=8+\ln9}

8 0
3 years ago
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