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jeka57 [31]
2 years ago
7

It takes a car 2.5 h to go from mile marker 15 to mile marker 130. What is the average speed of the car?

Mathematics
2 answers:
adoni [48]2 years ago
8 0

<u><em>Answer:</em></u>

  • <u><em>42 MPH</em></u>

<u><em>Step-by-step explanation:</em></u>

<u><em>Please give me brainliest :)</em></u>

AysviL [449]2 years ago
6 0

50 MPH I think tell me if im wrong.

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maks197457 [2]
8x + 114x + 3
8x + 114x = 122x
112x + 3

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Answer the question below from I ready
Aleksandr [31]

Answer:

a and b

Step-by-step explanation:

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2 years ago
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Mr Turner buys 12 pencils for $0.57 each . He pays with a $10 bill how much change would he revive
Delvig [45]
12(0.57)= 6.84

10-6.84= 3.16

Mr. Turner’s change is $3.16.


Hope this helps!
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3 years ago
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PLSS HELP ME WITH THIS
kolbaska11 [484]

Answer:

127

Step-by-step explanation:

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8 0
3 years ago
A new test to detect TB has been designed. It is estimated that 88% of people taking this test have the disease. The test detect
Elodia [21]

Answer:

Correct option: (a) 0.1452

Step-by-step explanation:

The new test designed for detecting TB is being analysed.

Denote the events as follows:

<em>D</em> = a person has the disease

<em>X</em> = the test is positive.

The information provided is:

P(D)=0.88\\P(X|D)=0.97\\P(X^{c}|D^{c})=0.99

Compute the probability that a person does not have the disease as follows:

P(D^{c})=1-P(D)=1-0.88=0.12

The probability of a person not having the disease is 0.12.

Compute the probability that a randomly selected person is tested negative but does have the disease as follows:

P(X^{c}\cap D)=P(X^{c}|D)P(D)\\=[1-P(X|D)]\times P(D)\\=[1-0.97]\times 0.88\\=0.03\times 0.88\\=0.0264

Compute the probability that a randomly selected person is tested negative but does not have the disease as follows:

P(X^{c}\cap D^{c})=P(X^{c}|D^{c})P(D^{c})\\=[1-P(X|D)]\times{1- P(D)]\\=0.99\times 0.12\\=0.1188

Compute the probability that a randomly selected person is tested negative  as follows:

P(X^{c})=P(X^{c}\cap D)+P(X^{c}\cap D^{c})

           =0.0264+0.1188\\=0.1452

Thus, the probability of the test indicating that the person does not have the disease is 0.1452.

4 0
3 years ago
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