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Marina CMI [18]
3 years ago
13

Here are the fuel efficiencies (in mpg) of 12 new cars . 28, 14, 48, 22, 14, 18, 52, 36, 32, 20, 12, 55 What is the percentage o

f these cars with a fuel efficiency less than 18 mpg?

Mathematics
2 answers:
SSSSS [86.1K]3 years ago
5 0
25% assuming you don’t count the car with 18 mpg fuel efficiency.

if you add the amount of cars up that have less than 18 mpg and then divide it by the total number (12) you will get a decimal (0.25) to find the percent multiply the decimal by 100
Wewaii [24]3 years ago
4 0
It’s 335.28 added together, so divide it by 13, which you’ll get 25.79
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What is the perimeter of a polygon with verticals at (-1, 3) , (-1, 6) , (2, 10) , (5, 6) , and (5, 3)
Marat540 [252]

Answer:

22 units

Step-by-step explanation:

The perimeter of a polygon is said to be the sum of the length of it's sides.

From the question, we have 5 vertices. This means the polygon is a pentagon. It's given vertices are

A = (−1, 3)

B = (−1, 6) ​

C = (2, 10)

D = (5, 6) ​​

E = (5, 3) ​

To find the distance between two points, we use the formula

d = √[(y2 - y1)² + (x2 - x1)²]

Between A and B, we have

d(ab) = √[(6 - 3)² + (-1 --1)²]

d(ab) = √(3²) + 0

d(ab) = √9 = 3

Between B and C, we have

d(bc) = √[(10 - 6)² + (2 --1)²]

d(bc) = √[4² + 3²]

d(bc) = √(16 + 9) = √25 = 5

Between C and D, we have

d(cd) = √[(6 - 10)² + (5 - 2)²]

d(cd) = √[(-4)² + 3²]

d(cd) = √(16 + 9) = √25 = 5

Between D and E, we have

d(de) = √[(3 - 6)² + (5 - 5)²]

d(de) = √(-3)² + 0

d(de) = √9 = 3

Between E and A, we have

d(ea) = √[(3 - 3)² + (5 --1)²]

d(ea) = √[0 + (6)²]

d(ea) = √36 = 6

The perimeter is given as

d(ab) + d(bc) + d(cd) + d(de) + d(ea) =

3 + 5 + 5 + 3 + 6 = 22 units

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3 years ago
Multiply 5 4/7 x -2 2/5
Lunna [17]
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Sliva [168]
15. 22=4*5+2
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7 0
2 years ago
I need some help on this. Not sure how to solve.
andreyandreev [35.5K]

Answer:

x = 9

Step-by-step explanation:

This shape is a 30-60-90 triangle, which means that the angles of the triangle are each 30°, 60° and 90°.  Given that the measures of the angles of the triangle are known, as well as one side, we can find the measures of all the other sides (example attached).  A 30-60-90 triangle is special because of the relationship of its sides.  The hypotenuse (directly across from the 90° angle) is equal to twice the length (2x) of the shorter leg, in this case labeled 'x' and directly across from the 30° angle.  The longer leg, across from the 60° angle, is equal to multiplying the shorter leg (x) by √3 or x√3.

Since the measure of the hypotenuse is given at 18, we can set the expression of that side equal to 18 and solve for x:

2x = 18  or x = 9


5 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
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