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zysi [14]
3 years ago
5

Find the area under the standard normal curve to the left of z = 1.5

Mathematics
2 answers:
-BARSIC- [3]3 years ago
5 0

Answer:

The answer is 0.9332 .

DaniilM [7]3 years ago
3 0

Answer:

0.9332

Step-by-step explanation:

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A right rectangular pyramid is sliced vertically (down) by a plane passing
Anton [14]

A right rectangular pyramid when sliced vertically, the  shape of the cross-section is known as Triangle.

<h3>What is A triangle?</h3>

This is known to be a kind of shape that is said to be in  a closed form and it is also known to be a 2-dimensional shape that has 3 sides, 3 angles, and also 3 vertices.

Note that when the when the right rectangular pyramid is sliced vertically (down) by a plane passing through the  of the pyramid, the new shape of the cross-section is a triangle.

See full question below

A right rectangular pyramid is sliced vertically (down) by a plane passing through the  of the pyramid. What is the shape of the cross-section?

A. Rectangle

B. Pyramid

C. Triangle

D. Trapezoid

See full question below

Learn more about triangle from

brainly.com/question/17335144

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3 0
2 years ago
Someone please help me? Answe is NOT 4/7 , tried that already
pishuonlain [190]
Yes the answer is not 4/7 its 7/4
5 0
3 years ago
Help please thank you
Maslowich

Answer:

it does not say anthing it says its restikted

Step-by-step explanation:

8 0
3 years ago
Read 2 more answers
What is the ratio of 20 praise and 2000 rupees<br>​
AnnZ [28]
1:100 because 20 divide by 20=1 and 2000 divide by 20 =100
8 0
3 years ago
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Consider the function on the interval (0, 2π). f(x) = sin x + cos x (a) Find the open intervals on which the function is increas
Annette [7]

Answer:Increasing in x∈(0,π/4)∪(5π/4,2π) decreasing in(π/4,5π/4)

Step-by-step explanation:

given f(x) = sin(x) + cos(x)

f(x) can be rewritten as \sqrt{2} [\frac{sin(x)}{\sqrt{2} }+\frac{cos(x)}{\sqrt{2} }  ]..................(a)\\\\\ \frac{1}{\sqrt{2} } = cos(45) = sin(45)\\\\

Using these result in equation a we get

f(x) = \sqrt{2} [ cos(45)sin(x)+sin(45)cos(x)]\\\\= \sqrt{2} [sin(45+x)]..........(b)

Now we know that for derivative with respect to dependent variable is positive for an increasing function

Differentiating b on both sides with respect to x we get

f '(x) = f '(x)=\sqrt{2}  \frac{dsin(45+x)}{dx}\\ \\f'(x)=\sqrt{2} cos(45+x)\\\\f'(x)>0=>\sqrt{2} cos(45+x)>0

where x∈(0,2π)

we know that cox(x) > 0 for x∈[0,π/2]∪[3π/2,2π]

Thus for cos(π/4+x)>0 we should have

1) π/4 + x < π/2  => x<π/4  => x∈[0,π/4]

2) π/4 + x > 3π/2  => x > 5π/4  => x∈[5π/4,2π]

from conditions 1 and 2 we have  x∈(0,π/4)∪(5π/4,2π)

Thus the function is decreasing in x∈(π/4,5π/4)

5 0
2 years ago
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