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anastassius [24]
3 years ago
11

To ride a horse at the state fair costs 2$ per minute. Pablo buys a 16 minute ride. Next time her plans to ride 25% longer. How

much will he pay?
Help please i will reward brainly
Mathematics
1 answer:
Mariana [72]3 years ago
6 0

Answer:

40 dollars

Step-by-step explanation:

25% of 16 minutes is 4 mins. 16+4=20 mins.

so 20*2=40 dollars

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The sum of two numbers is 44. One number is 3 times as large as the other. What are the numbers?
jok3333 [9.3K]
Your numbers are 11 and 32

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2 years ago
There are nine water bottles in a fridge. Three full boxes are added. Two more boxes are added, both of which have one fewer bot
Degger [83]

The number of bottles in a full box is 12.

<u><em>Explanation</em></u>

Suppose, the number of bottles in a full box is  x

So, the number of bottles in <u>three</u> full boxes =3x

Now, two more boxes are added, both of which have <u>one fewer bottle than the other three</u>. So, the number of bottles in these two boxes = 2(x-1)

As there were 9 bottles initially and now there are 67 bottles in the fridge, <em><u>so the equation will be</u></em>....

9+3x+2(x-1)=67\\ \\ 9+3x+2x-2=67\\ \\ 5x+7=67\\ \\ 5x=67-7\\ \\ 5x=60\\ \\ x=\frac{60}{5}=12

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3 years ago
Lie detectors have a 15% chance of concluding that a person is lying even when they are telling the truth. a bank conducts inter
Otrada [13]
Part A:

Given that lie <span>detectors have a 15% chance of concluding that a person is lying even when they are telling the truth. Thus, lie detectors have a 85% chance of concluding that a person is telling the truth when they are indeed telling the truth.

The case that the lie detector correctly determined that a selected person is saying the truth has a probability of 0.85
Thus p = 0.85

Thus, the probability that </span>the lie detector will conclude that all 15 are telling the truth if <span>all 15 applicants tell the truth is given by:

</span>P(X)={ ^nC_xp^xq^{n-x}} \\  \\ \Rightarrow P(15)={ ^{15}C_{15}(0.85)^{15}(0.15)^0} \\  \\ =1\times0.0874\times1=0.0874
<span>

</span>Part B:

Given that lie detectors have a 15% chance of concluding that a person is lying even when they are telling the truth. Thus, lie detectors have a 85% chance of concluding that a person is telling the truth when they are indeed telling the truth.

The case that the lie detector wrongly determined that a selected person is lying when the person is actually saying the truth has a probability of 0.25
Thus p = 0.15

Thus, the probability that the lie detector will conclude that at least 1 is lying if all 15 applicants tell the truth is given by:

P(X)={ ^nC_xp^xq^{n-x}} \\ \\ \Rightarrow P(X\geq1)=1-P(0) \\  \\ =1-{ ^{15}C_0(0.15)^0(0.85)^{15}} \\ \\ =1-1\times1\times0.0874=1-0.0874 \\  \\ =0.9126


Part C:

Given that lie detectors have a 15% chance of concluding that a person is lying even when they are telling the truth. Thus, lie detectors have a 85% chance of concluding that a person is telling the truth when they are indeed telling the truth.

The case that the lie detector wrongly determined that a selected person is lying when the person is actually saying the truth has a probability of 0.15
Thus p = 0.15

The mean is given by:

\mu=npq \\  \\ =15\times0.15\times0.85 \\  \\ =1.9125


Part D:

Given that lie detectors have a 15% chance of concluding that a person is lying even when they are telling the truth. Thus, lie detectors have a 85% chance of concluding that a person is telling the truth when they are indeed telling the truth.

The case that the lie detector wrongly determined that a selected person is lying when the person is actually saying the truth has a probability of 0.15
Thus p = 0.15

The <span>probability that the number of truthful applicants classified as liars is greater than the mean is given by:

</span>P(X\ \textgreater \ \mu)=P(X\ \textgreater \ 1.9125) \\  \\ 1-[P(0)+P(1)]
<span>
</span>P(1)={ ^{15}C_1(0.15)^1(0.85)^{14}} \\  \\ =15\times0.15\times0.1028=0.2312<span>
</span>
8 0
3 years ago
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