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natulia [17]
3 years ago
9

PLEASE HELP ME AND EXPLAIN AND I WILL MARK YOU BRAINLIES!!

Mathematics
1 answer:
stira [4]3 years ago
8 0
Team B because mode is how often a number appears in a list and 60 is greater than 48 or 30, which are the modes of Team A and Team C
Hope this helps!
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Matt earned $28,500 last year. He paid $8,265 for rent. What percent of his earnings did Matt pay for rent ?
Katena32 [7]

Answer: 29%

Step-by-step explanation:

You need to divide the rent, 8265, by the amount of money he earns.

8265/28500 is .29

Multiply that by 100 to get it in percent form

5 0
3 years ago
Questions is in the picture. no links please illl give brainly and 50 points only if answered properly
ohaa [14]

Answer:

subtract

a) 13/20

b) 1/2

c) 5/24

d) 7/18

e) 7/20

f) 3/40

Add or subtract

a) 4/9

b) 1/2

c) 11/20

d) 1/4

e) 19/12 <u>or </u>1 \frac{7}{12}

f) 4/21

3 0
3 years ago
Read 2 more answers
Find the y-intercept of f(x) = x^3 + 3x^2 - X-3<br> HURRY
Grace [21]

Answer:

y-intercepts: (-3,0) , (-1,0) , (1,0)

Step-by-step explanation:

3 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
4 years ago
A scatterplot is shown on the graph below. Which of these could be the line of best fit? x = 100 y = 100 y = x – 100 y = x + 100
kkurt [141]

Answer:

55 when y dived and subtracting

Step-by-step explanation:

djiiwwuwb skis wujw que wjsjs what s

6 0
3 years ago
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