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Alenkinab [10]
3 years ago
7

Find the area under the standard normal curve greater than z=1.25

Mathematics
1 answer:
Svetllana [295]3 years ago
4 0

Answer:

0.1056

Step-by-step explanation:

okay so using a table of standard normal probabilities:

1.) look for 1.2 on the left column and .05 (which will give you .8944)

2.) this will give you the area before it so do

1- 0.8944

3.) You have your answer!

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When a fraction of 12 is taken away from 17, what remain exceed one-third of seventeen by six.
vova2212 [387]
X - the fraction

17- 12x=\frac{1}{3} \times 17+6 \\
17-12x=\frac{17}{3}+6 \\
-12x=\frac{17}{3}+6-17 \\
-12x=\frac{17}{3}-11 \\
-12x=\frac{17}{3}-\frac{33}{3} \\
-12x=-\frac{16}{3} \\
x=-\frac{16}{3} \times (-\frac{1}{12}) \\
x=\frac{16}{36} \\
x=\frac{4}{9}

The fraction is 4/9.
5 0
3 years ago
Show that the curve x = 7 cos(t), y = 5 sin(t) cos(t) has two tangents at (0, 0) and find their equations.
elena-s [515]
Y= 5 y =7 thats the smaller slip and larger
8 0
3 years ago
Help will give brainlyest
bazaltina [42]
X=-9 is that answer
5 0
2 years ago
Read 2 more answers
What sine function represents an amplitude of 4, a period of pi over 2, no horizontal shift, and a vertical shift of −3?
mixer [17]

Answer:

f(x) = 4sin(\frac{\pi}{2}x) - 3, the third one

Step-by-step explanation:

Explaining the sine function:

The sine function is defined by:

S = Asin(p(x - x_{0})) + V

In which A is the amplitude, p = \frac{2\pi}{N} is the period, x_{0} is the horizontal shift and V is the vertical shift.

So, in your problem:

The amplitude is 4, so A = 4.

The period is \frac{\pi}{2}, so p = \frac{\pi}{2}.

There is no horizontal shift, so x_{0} = 0.

The vertical shift is −3, so V = -3.

The sine function that represents these following conditions is

f(x) = 4sin(\frac{\pi}{2}x) - 3, the third one

4 0
3 years ago
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ankoles [38]

Answer:

The expression will be simplified into 26x^4y^3+35xy^6-42xy^5\\

Step-by-step explanation:

It's like addition, when you multiply exponents that are on the same bases as each other.

6 0
3 years ago
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