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kaheart [24]
2 years ago
5

What type of error occurs if you fail to reject h0 when, in fact, it is not true?.

Mathematics
1 answer:
blagie [28]2 years ago
5 0

The type of error that occurs  if you fail to reject h0 when, in fact, it is not true is a type 2 error.

<h3>What is a type 2 error?</h3>

Type 2 error is when the null hypothesis is not rejected even though it is not false. Type 1 error is when the null hypothesis is rejected when it is true. A type 2 error leads to a false negative which is also known as an error of omission.

To learn more about type 2 error, please check: brainly.com/question/20914617

#SPJ1

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Accidents on highways are one of the main causes of death or injury in developing countries and the weather conditions have an i
pashok25 [27]

Answer:

0.105 = 10.5% probability that an accident results in a death.

Step-by-step explanation:

What is the probability that an accident results in a death?

5% of 60%(sunny)

25% of 20%(foggy)

12.5% of 20%(rainy)

So

p = 0.05*0.6 + 0.25*0.2 + 0.125*0.2 = 0.105

0.105 = 10.5% probability that an accident results in a death.

6 0
3 years ago
Write 3x^2-18x-6 in vertex form
Vedmedyk [2.9K]
The standard form of a quadratic equation is \displaystyle{ y=ax^2+bx+c, while the vertex form is:

                      y=a(x-h)^2+k, where (h, k) is the vertex of the parabola.

What we want is to write \displaystyle{ y=3x^2-18x-6 as y=a(x-h)^2+k

First, we note that all the three terms have a factor of 3, so we factorize it and write:

\displaystyle{ y=3(x^2-6x-2).


Second, we notice that x^2-6x are the terms produced by (x-3)^2=x^2-6x+9, without the 9. So we can write:

x^2-6x=(x-3)^2-9, and substituting in \displaystyle{ y=3(x^2-6x-2) we have:

\displaystyle{ y=3(x^2-6x-2)=3[(x-3)^2-9-2]=3[(x-3)^2-11].

Finally, distributing 3 over the two terms in the brackets we have:

y=3[x-3]^2-33.


Answer: y=3(x-3)^2-33
6 0
3 years ago
Match the following items by evaluating the expression for x = -7. x-2 x-1 x0 x1 x2
frosja888 [35]

Answer:

{x}^{ - 2}  =  \frac{1}{ {x}^{2} }  \:  \: –> {7}^{ - 2}  =  \frac{1}{ {7}^{2} }  =  \frac{ 1}{49}  \\  \\  {x}^{ - 1}  =  \frac{1}{ {x}^{1} }   \: –>  {7}^{ - 1}  =  \frac{1}{ {7}^{1} } =  \frac{1}{7}  \\  \\  {x}^{0} –> {7}^{0}   = 1 \\  \\  {x}^{1}  –> {7}^{1}  = 7 \\  \\  {x}^{2}  –> {7}^{2}  = 49

7 0
2 years ago
Can someone help me with this
Anon25 [30]

Answer:

1] x=4.21

2] 52.5

3] 3.82

4] 38.73

5] 32.005

6] 31.21

Step-by-step explanation:

8 0
3 years ago
Can you please explain and answer this question to me, (don't mind the blue dots, they are the answers of what I think, please l
Fofino [41]
I started with c, but you could’ve started with any box.

So I took the numbers in the c box (the black box in my picture) and divided to see which were multiples of two. They all were.

Then I went to a. All of a’s numbers were divisible by 7.

Then b. Two of them are divisible by 2, so that’s not the answer. None of them were divisible by 7, so there’s your answer!

Then d. Two are divisible by 7, so your rules is divisible by 2.

I hope this helps!

6 0
3 years ago
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