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egoroff_w [7]
3 years ago
12

What is 4/5÷9/10=_?4/5 please explain this is a homework question

Mathematics
2 answers:
Strike441 [17]3 years ago
8 0
I hope this helps you

kifflom [539]3 years ago
8 0
The awnser is false I hope this helps u
You might be interested in
Jane wants to estimate the proportion of students on her campus who eat cauliflower. after surveying 39 ​students, she finds 4 w
stellarik [79]

Let x be the number of students who eat cauliflower.

Therefore, x=4.

Let n be the total number of students surveyed.

Therefore, n=39

Thus, \hat p=\frac{4}{39} =0.10256

Now, for 90% confidence level, from the table we know that Z=1.645.

The formula for the interval range of proportion of students is :

p= \hat p\pm Z\sqrt{\frac{\hat p(1-\hat p)}{n}}

Plugging in the values we get:

p=0.10256\pm 1.645\sqrt{\frac{0.10256(1-0.10256)}{39}}=0.10256\pm 0.04858=0.15114, 0.05398

Thus, Jane is 90% confident that the population proportion p, for students who eat cauliflower in her campus is between 5.398% and 15.114% (after converting the answer we got to percentage).

7 0
4 years ago
If sinx = p and cosx = 4, work out the following forms :<br><br><br>​
Kay [80]

Answer:

$\frac{p^2 - 16} {4p^2 + 16} $

Step-by-step explanation:

I will work with radians.

$\frac {\cos^2 \left(\frac{\pi}{2}-x \right)+\sin(-x)-\sin^2 \left(\frac{\pi}{2}-x \right)+\cos \left(\frac{\pi}{2}-x \right)} {[\sin(\pi -x)+\cos(-x)] \cdot [\sin(2\pi +x)\cos(2\pi-x)]}$

First, I will deal with the numerator

$\cos^2 \left(\frac{\pi}{2}-x \right)+\sin(-x)-\sin^2 \left(\frac{\pi}{2}-x \right)+\cos \left(\frac{\pi}{2}-x \right)$

Consider the following trigonometric identities:

$\boxed{\cos\left(\frac{\pi}{2}-x \right)=\sin(x)}$

$\boxed{\sin\left(\frac{\pi}{2}-x \right)=\cos(x)}$

\boxed{\sin(-x)=-\sin(x)}

\boxed{\cos(-x)=\cos(x)}

Therefore, the numerator will be

$\sin^2(x)-\sin(x)-\cos^2(x)+\sin(x) \implies \sin^2(x)- \cos^2(x)$

Once

\sin(x)=p

\cos(x)=4

$\sin^2(x)-\cos^2(x) \implies p^2-4^2 \implies \boxed{p^2-16}$

Now let's deal with the numerator

[\sin(\pi -x)+\cos(-x)] \cdot [\sin(2\pi +x)\cos(2\pi-x)]

Using the sum and difference identities:

\boxed{\sin(a \pm b)=\sin(a) \cos(b) \pm \cos(a)\sin(b)}

\boxed{\cos(a \pm b)=\cos(a) \cos(b) \mp \sin(a)\sin(b)}

\sin(\pi -x) = \sin(x)

\sin(2\pi +x)=\sin(x)

\cos(2\pi-x)=\cos(x)

Therefore,

[\sin(\pi -x)+\cos(-x)] \cdot [\sin(2\pi +x)\cos(2\pi-x)] \implies [\sin(x)+\cos(x)] \cdot [\sin(x)\cos(x)]

\implies [p+4] \cdot [p \cdot 4]=4p^2+16p

The final expression will be

$\frac{p^2 - 16} {4p^2 + 16} $

8 0
3 years ago
Name the image of C(6, –4) under a rotation 90 counter-clockwise about the origin.
Rufina [12.5K]

Answer:

d.) (4,6)

Step-by-step explanation:

(-y,x) is the formula for a 90 counter-clockwise rotation.

So, it's (4,6)

6 0
3 years ago
12^4 + 1^4 equals what
belka [17]

Answer:

20737

Step-by-step explanation:

12^4=12*12*12*12= 20736

1^4= 1*1*1*1=1

20736+1=20737

6 0
3 years ago
A rhombus diagonals bisects opposite <br>a)angles <br>b)line​
hichkok12 [17]

Answer:

The answer to this question is angles

3 0
3 years ago
Read 2 more answers
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