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ArbitrLikvidat [17]
3 years ago
6

Two boxers bump heads in a boxing match. The referee will check for a concussion on one of the boxers. Consider a null hypothesi

s, , that the boxer does not have a concussion, and an alternative hypothesis, , that the boxer did get a concussion. The referee has two possible decisions: stop the match due to injury (reject the null hypothesis) or allow the match to continue (do not reject the null hypothesis). If the referee makes a Type I error, what happened?
A. He allowed the match to continue, and the boxer did not have a concussion.
B. He stopped the match due to injury, but the boxer did not have a concussion.
C. He stopped the match due to injury, and the boxer did have a concussion.
D. He allowed the match to continue, but the boxer did have a concussion.
Mathematics
1 answer:
Natalka [10]3 years ago
4 0
B,hope this helped!
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Please help and thank you
Masteriza [31]

Answer:A

Step-by-step explanation:

It might be A

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3 years ago
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Which of the following represents 8 square root x5
gavmur [86]

Answer:

5 \sqrt{8}

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A basketball team sells tickets that cost​ $10, $20,​ or, for VIP​ seats,​ $30. The team has sold 3296 tickets overall. It has s
MissTica

Answer:

1022 of $10 tickets, 1326 of $20 tickets, 948 of VIP $30 tickets

Step-by-step explanation:

We can set-up a system of equations to find the number of each kind of ticket. We know there are the cheap tickets c, the medium tickets m and the VIP tickets v. Since 3296 tickets were purchased, then c+m+v=3296.

We also know that 304 more medium priced tickets have been sold then cheap tickets. We write 304+c=m.

Lastly, we know that a total of $65,180 was sold. We write 10c+20m+30v=65,180.

We will solve by substituting one equation into the other. We substitute m=304+c into c+m+v=3296. Simplify and isolate the variable v.

c+m+v=3296

c+(304+c)+v=3296

c+304+c+v=3296

2c+304-304+v=3296-304

2c-2c+v=2992-2c

v=2992-2c

We will now substitute this into the remaining equation with our first substitution.

10c+20m+30v=65,180

10c+20(304+c)+30(2992-2c)=65,180

10c+6080+20c+89760-60c=65180

-30c+95840=65180

-30c+95840-95840=65180-95840

-30c=-30660

c=1022

This means 1,022 tickets purchased were $10 tickets. We substitute this equation back into m=304+c to find m.

m=304+1022

m=1326

This means 1,326 were $20 tickets.

Lastly we know 948 VIP tickets were bought because 1022+1326+948=3296

5 0
4 years ago
What is the expression for 6 times the product of 8 and 8
Kryger [21]

Answer:

(8 x 8) x 6 or 6(8 x 8)

Step-by-step explanation:

5 0
3 years ago
Find the general solution of x'1 = 3x1 - x2 + et, x'2 = x1.
Ann [662]
Note that if {x_2}'=x_1, then {x_2}''={x_1}', and so we can collapse the system of ODEs into a linear ODE:

{x_2}''=3{x_2}'-x_2+e^t
{x_2}''-3{x_2}'+x_2=e^t

which is a pretty standard linear ODE with constant coefficients. We have characteristic equation

r^2-3r+1=\left(r-\dfrac{3+\sqrt5}2\right)\left(r+\dfrac{3+\sqrt5}2\right)=0

so that the characteristic solution is

{x_2}_C=C_1e^{(3+\sqrt5)/2\,t}+C_2e^{-(3+\sqrt5)/2\,t}

Now let's suppose the particular solution is {x_2}_p=ae^t. Then

{x_2}_p={{x_2}_p}'={{x_2}_p}''=ae^t

and so

ae^t-3ae^t+ae^t=-ae^t=e^t\implies a=-1

Thus the general solution for x_2 is

x_2=C_1e^{(3+\sqrt5)/2\,t}+C_2e^{-(3+\sqrt5)/2\,t}-e^t

and you can find the solution x_1 by simply differentiating x_2.
7 0
4 years ago
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