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miss Akunina [59]
3 years ago
7

Which equation is represented by the graph below?y=in x​

Mathematics
2 answers:
Margaret [11]3 years ago
8 0
Can you add a picture if you can in order for me to solve this question.
nadya68 [22]3 years ago
5 0

Answer:

The answer is c

Step-by-step explanation:

trust me

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Evaluate 2x2 - 1 when x = 3.
tigry1 [53]
X should equal 11
2(3)2-1
6(2)-1
12-1
11
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3 years ago
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Have 5 pieces of candy and 29 people. How many pieces are the candy divided into for everyone to get a piece.
Ivanshal [37]

Answer:

everyone gets 5 pieces and the rest gets thrown out

Step-by-step explanation:

29/5=5.8

but obviously ur not gonna break the candy into .8 so everyone just gets 5 so its even

8 0
3 years ago
When Carson runs the 400 meter dash, his finishing times are normally distributed with a mean of 63 seconds and a standard devia
Gnom [1K]

Answer: in 95% of races, his finishing time will be between 62 and 64 seconds.

Step-by-step explanation:

The empirical rule states that for a normal distribution, nearly all of the data will fall within three standard deviations of the mean . The empirical rule is further illustrated below

68% of data falls within the first standard deviation from the mean.

95% fall within two standard deviations.

99.7% fall within three standard deviations.

From the information given, the mean is 63 seconds and the standard deviation is 5 seconds.

2 standard deviations = 2 × 0.5 = 1

63 - 1 = 62 seconds

63 + 1 = 64 seconds

Therefore, in 95% of races, his finishing time will be between 62 and 64 seconds.

6 0
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
What is located closest to the orgin
jeka94
The zero is closed to the origin
6 0
3 years ago
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