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nexus9112 [7]
3 years ago
15

What is the Area of 20ft & 24ft?

Mathematics
1 answer:
olasank [31]3 years ago
7 0

Answer:

480

Step-by-step explanation:

bebebbebeebbebebebebebebebebbebebeb

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Is this relation a function?
AleksandrR [38]
Yes the relation is a function
4 0
2 years ago
Solve using the Zero Product Property. What is one solution for x? (x – 5) (x + 1) =0
andriy [413]
The answer would be 2.) 5

(5-5) = 0 (5+1) = 6

0x6= 0
4 0
3 years ago
An Internet reaction time test asks subjects to click their mouse button as soon as a light flashes on the screen. The light is
Ganezh [65]

Answer:

(a) 0.20

(b) 31%

(c) 2.52 seconds

Step-by-step explanation:

The random variable <em>Y</em> models the amount of time the subject has to wait for the light to flash.

The density curve represents that of an Uniform distribution with parameters <em>a</em> = 1 and <em>b</em> = 5.

So, Y\sim Unif(1,5)

(a)

The area under the density curve is always 1.

The length is 5 units.

Compute the height as follows:

\text{Area under the density curve}=\text{length}\times \text{height}

                                          1=5\times\text{height}\\\\\text{height}=\frac{1}{5}\\\\\text{height}=0.20

Thus, the height of the density curve is 0.20.

(b)

Compute the value of P (Y > 3.75) as follows:

P(Y>3.75)=\int\limits^{5}_{3.75} {\frac{1}{b-a}} \, dy \\\\=\int\limits^{5}_{3.75} {\frac{1}{5-1}} \, dy\\\\=\frac{1}{4}\times [y]^{5}_{3.75}\\\\=\frac{5-3.75}{4}\\\\=0.3125\\\\\approx 0.31

Thus, the light will flash more than 3.75 seconds after the subject clicks "Start" 31% of the times.

(c)

Compute the 38th percentile as follows:

P(Y

Thus, the 38th percentile is 2.52 seconds.

4 0
2 years ago
A dilation with a scale factor of 2 is applied to the three line segments shown. The resulting images are P′Q′¯¯¯¯¯¯¯, A′B′¯¯¯¯¯
ivanzaharov [21]

Answer:

Step-by-step explanation:

Formula to determine the scale factor,

Scale factor = \frac{\text{Dimension of the image}}{\text{Dimension of the original}}

2 = \frac{P'Q'}{PQ}

2 = \frac{P'Q'}{2}

P'Q' = 4 cm

Similarly, for the length of A'B',

2 = \frac{A'B'}{AB}

2 = \frac{A'B'}{1.5}

A'B' = 3 cm

For the length of M'N',

2 = \frac{M'N'}{MN}

2 = \frac{M'N'}{3}

M'N' = 6 cm

4 0
2 years ago
Can anyone solve this for me?
nydimaria [60]

Answer:

multiply, 12 inches per foot

Step-by-step explanation:

8 0
2 years ago
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