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3 years ago

If the diameter of a circle is half of a yard, then its radius is how many inches?

1 answer:

Ber [7]3 years ago

6 0

First of all we need to realize that half of a yard is as much as 18 inches. If a diameter of the circle is 18 inches, then its radius is 18 ÷ 2 = 9 inches.

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Find the area under the standard normal distribution curve for each.

yawa3891 [41]

From the standard normal tables, obtain
P(z<1.95) = 0.9744
P(z<0.37) = 0.6443
P(z<1.32) = 0.9066
P(z<1.82) = 0.9656
P(z<1.05) = 0.8531
P(z<2.05) = 0.9798
P(z<0.03) = 0.5120
P(z<0.53) = 0.7019

Answers:
Range of z Area
-------------- ---------------------------------------
a. 0 - 1.95 0.9744
b. 0 - 0.37 0.6443
c. 1.32 - 1.82 0.9656 - 0.9066 = 0.0590
d. 1.05 - 2.05 0.9798 - 0.8531 = 0.1267
e. 0.03 - 0.53 0.7019 - 0.5120 = 0.1899

8 0

3 years ago

When an inclined plane is used,

Triss [41]

Answer:

the mechanical advantage increases as the height increases.

Step-by-step explanation:

3 0

3 years ago

one piece of ash was ejected from the volcano with the velocity of 368ft/sec. The height H, in feet of our ash projectile is giv

insens350 [35]

This graph is not a minus graph so it just keeps going on and on to infinite.
I’ve attached a pic of the graph

5 0

3 years ago

quadrilateral ABCD is dilated by a scale factor of 2 centered around (2,2). Which statement is true about the dilation?

Setler [38]

Answer:

The vertices of the image will have the coordinates

A(-5, 1) → A'(2x-2, 2y-2) → A'(-12, 0)

B(-4, 3) → B'(2x-2, 2y-2) → B'(-10, 4)

C(1, 2) → C'(2x-2, 2y-2) → C'(0, 2)

D(-3, 0) → D'(2x-2, 2y-2) → D'(-8, -2)

Step-by-step explanation:

<em>Note: You missed to mention the vertices of a quadrilateral ABCD. So, I am assuming the following vertices. It would anyways clear your concept.</em>

<em />

Let us suppose

A(-5, 1)

B(-4, 3)

C(1, 2)

D(-3, 0)

We know that the rule of the dilation with the center of dilation at (2,2) by a scale factor of 2 is:

(x, y) → (2x-2, 2y-2)

Therefore, the vertices of the image will have the coordinates

A(-5, 1) → A'(2x-2, 2y-2) → A'(-12, 0)

B(-4, 3) → B'(2x-2, 2y-2) → B'(-10, 4)

C(1, 2) → C'(2x-2, 2y-2) → C'(0, 2)

D(-3, 0) → D'(2x-2, 2y-2) → D'(-8, -2)

6 0

2 years ago

(HELP ASAP)

wariber [46]

Answer:

45 km^2 (C)

Step-by-step explanation:

Isolate the smaller square -

We only need two side lengths, take 3km and 3km.

3×3 = 9km^2, that's the area of the smaller square.

Isolate the larger square -

Take 6km and the other 6km

6×6 = 36km^2

Add the square's areas together to get the whole area -

36 + 9 = 45km^2

All you need to do to solve these problems is to isolate the shapes and find their area's, and finally add them together.

8 0

3 years ago