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nirvana33 [79]
3 years ago
6

Can anyone help me with this math test I got to get it turned in by midnight tonight so please respond quick!!

Mathematics
2 answers:
Sergio039 [100]3 years ago
8 0

Answer:

The answer is C.

Step-by-step explanation:

You first have to take 50 and divide it by 3 to get 16.6 repeating. From there, you just have to turn it into a fraction which would be between 1 half and 3 fourths.

frutty [35]3 years ago
8 0

Answer:

C

Step-by-step explanation:

50 ÷ 3 = 16.67

½ = 0.5

¾ = 0.75

So, the answer is C ^^

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This is my last question please help !
chubhunter [2.5K]

Answer:

try the third option

Step-by-step explanation:

At least I'm trying

8 0
3 years ago
Px + qy = r<br> 2px - qy = 2r<br> When solving this system of equations, x =
VladimirAG [237]
Px + qy = r
2px - qy = 2r
----------------add
3px = 3r

x = 3r/3p
x = r/p
8 0
4 years ago
Read 2 more answers
The circumference of Circle F is 72 cm. What is the length of Arc DE (the minor arc)?
Tom [10]

You need to first set up the equation:

It would look like 90 divided by 360 multiplied by the given circumference which is 72 cm.

So 90 / 360 x 72

Simply the fraction above, it will give us:

¼ x 72

So the answer would be 18 cm that is the length of DE (minor arc)

4 0
4 years ago
Read 2 more answers
Which of the following is a solution of z^5 = 1 + √3 i?
nordsb [41]

Answer:

Option 2 is right

Step-by-step explanation:

Given that

z^5=1+\sqrt{3} i

We can write this in polar form with modulus and radius

|z^5|= \sqrt{1+3} =2\\tan of Angle t =\sqrt{3} \\

Hence angle = 60 degrees and

|z^5|= 2(cos60+isin60)

Since we have got 5 roots for z, we can write 60, 420, 780, etc. with periods of 360

Using Demoivre theorem we get 5th root would be

5th root of 2 multiplied by 1/5 th of 60, 420, 780,....

z= \sqrt[5]{2} (cos12+isin12)\\z=\sqrt[5]{2} (cos84+isin84)\\\\z=\sqrt[5]{2} (cos156+isin156)\\\\z=\sqrt[5]{2} (cos228+isin228)\\\\z=\sqrt[5]{2} (cos300+isin300)\\

Out of these only 2nd option suits our answer

Hence answer is Option 2.

8 0
3 years ago
A fair coin is flipped twelve times. What is the probability of the coin landing tails up exactly nine times?
seraphim [82]

Answer:

P\left(E\right)=\frac{55}{1024}

Step-by-step explanation:

Given that a fair coin is flipped twelve times.

It means the number of possible sequences of heads and tails would be:

2¹² = 4096

We can determine the number of ways that such a sequence could contain exactly 9 tails is the number of ways of choosing 9 out of 12, using the formula

nCr=\frac{n!}{r!\left(n-r\right)!}

Plug in n = 12 and r = 9

       =\frac{12!}{9!\left(12-9\right)!}

       =\frac{12!}{9!\cdot \:3!}

       =\frac{12\cdot \:11\cdot \:10}{3!}            ∵ \frac{12!}{9!}=12\cdot \:11\cdot \:10

       =\frac{1320}{6}                   ∵ 3!\:=\:3\times 2\times 1=6

       =220

Thus, the probability will be:

P\left(E\right)=\frac{n\left(E\right)}{n\left(S\right)}

         =\frac{220}{4096}

         =\frac{55}{1024}

Thus, the probability of the coin landing tails up exactly nine times will be:

P\left(E\right)=\frac{55}{1024}

4 0
3 years ago
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