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3 years ago

The intersection of two planes can be a line true or false

1 answer:

8 0

True.

Let's say that you have two pieces of paper that both represent planes. If you cross them, you'll be able to see that the intersection of these two pieces of paper is a line.

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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%

Step-by-step explanation:

The random variable Y 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 a = 1 and b = 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

Please help! Thanks in advance.

alukav5142 [94]

In this problem, we need to plug in the given x values for $f(x)=0$ and find a and b.

When we plug in 1, we get: $2 \times {1}^{4} - 5 \times {1}^{3} - 14 \times {1}^{2} + a \times 1 + b = 0$

Simplify:
$2 - 5 - 14 + a + b = 0$
$- 17 + a + b = 0$
$a + b = 17$

We got our first statement about the values of the variables. If we find one more we can find those 2 variables.

We have another given root: 4.

Plug it in:
$2 \times {4}^{4} - 5 \times {4}^{3} - 14 \times {4}^{2} + a \times 4 + b = 0$
$512 - 320 - 224 + 4a + b = 0$
$- 32 + 4a + b = 0$
$4a + b = 32$

Now we have our second one. We can combine them:
$a + b = 17 \\ 4a + b = 32$

I use elimination method which is easier here.

Multiply the top equation by -1: $- a - b = - 17 \\ 4a + b = 32$

Add them up:
$3a = 15$

Simplify:
$a = 5$

Now we have a, we can plug in one of those equations to find b:
$5 + b = 17$
$b = 17 - 5$
$b = 12$

So, the answers are $a=5$ and $b=12$ .

6 0

3 years ago

What is 5/12 divided by 3/8

asambeis [7]

Answer:

1 1/9

Step-by-step explanation:

put the problem into the calculator and divide. so simple

5 0

2 years ago

Read 2 more answers

Wats is the answer to this question

sweet-ann [11.9K]

Simplify the radical by breaking the radicand up into a product of known factors in other words take the cube root of 27 then for the exponent 9 you divide by 3 get

3x^3

3 0

2 years ago

A rectangular pyramid has a height of 8 meters. The base is 7 meters in length and 15 meters in width two different triangular c

vagabundo [1.1K]

Answer:

The difference in the areas of the cross section is 32 m²

Step-by-step explanation:

The given parameters are;

The height of the rectangular pyramid = 8 meters

The length of the base of the pyramid = 7 meters

The width of the base of the pyramid = 15 meters

Whereby triangular cross sections are formed through the vertex and perpendicular to the base, and to each other, we have;

The sides of the two triangles consists of the following;

1) Two slant height of the pyramid each

2) The two perpendicular lines joining the midpoints of the opposite sides of the base of the pyramid with length equal to the length of the adjacent side to the sides from which they are drawn which are 15 meters and 7 meters

3) The two lines and the corresponding slant height form triangles cross section which are perpendicular to each other.

the slant height, $h_l$ , is given as follows;

$h_l$ = √(8² + (15/2)²) ≈ 10.966

For the triangular cross section with base = 15 m

The area of the cross section = 1/2 × Base₁₅ × Height = 1/2 × 15 m × 8 m = 60 m²

For the triangular cross section with base = 7 m

The area of the cross section = 1/2 × Base₇ × Height = 1/2 × 7 m × 8 m = 28 m²

The difference in the areas of the cross section = 60 m² - 28 m² = 32 m².

5 0

2 years ago