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3 years ago

Henry is also eating pizza. He ate 7/8 of a whole pizza 7/8 of a whole pizza that had 12 pieces. How many pieces did he eat

Mathematics

1 answer:

MakcuM [25]3 years ago

6 0

Answer:

10 1/2

Step-by-step explanation:

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73.50 divided by 21. Ishsdbdh

valina [46]

3.5 is the answer!!!

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2 years ago

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The lengths of pregnancies are normally distributed with a mean of 268 days and a standard deviation of 15 days. a. Find the pro

Bogdan [553]

Answer:

Step-by-step explanation:

Let X be the length of pregnancy.

X is N(268, 15)

a) Prob of pregnancy lasting 307 days or longer

= P(X>307) = $P(Z>\frac{307-268}{15} )\\=P(Z>2.6)\\$

=0.5-0.4953

=0.0047

b) Lowest 2% is 2nd percentile

Z=-2.55

X score = 268-2.55(15)

= 229.75 days

If length of days >230 days then it is not premature.

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3 years ago

Verify that this trigonometric equation is an identity?

Vlada [557]

$\cot x\sec^4 x=\cot x+2\tan x+\tan^3x\\\\L=\dfrac{\cos x}{\sin x}\cdot\dfrac{1}{\cos^4x}=\dfrac{1}{\sin x}\cdot\dfrac{1}{\cos^3x}=\dfrac{1}{\sin x\cos^3x}\\\\R=\dfrac{\cos x}{\sin x}+2\cdot\dfrac{\sin x}{\cos x}+\dfrac{\sin^3x}{\cos^3x}\\\\=\dfrac{\cos x\cos^3x}{\sin x\cos^3x}+\dfrac{2\sin x\cos^2x}{\cos x\sin x\cos^2x}+\dfrac{\sin^3x\sin x}{\cos^3x\sin x}\\\\=\dfrac{\cos^4x+2\sin^2x\cos^2x+\sin^4x}{\sin x\cos^3x}\\\\=\dfrac{(\cos^2x)^2+2\sin^2x\cos^2x+(\sin^2x)^2}{\sin x\cos^3x}$

$=\dfrac{(\cos^2x+\sin^2x)^2}{\sin x\cos^3x}=\dfrac{1}{\sin x\cos^3x}=L\\\\Used:\\\tan(a)=\dfrac{\sin(a)}{\cos(a)}\\\cot(a)=\dfrac{\cos(a)}{\sin(a)}\\\sec(a)=\dfrac{1}{\cos(a)}\\\sin^2a+\cos^2a=1\\(a+b)^2=a^2+2ab+b^2$

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3 years ago

Pls help me answer this question

omeli [17]

The answer is -269/30 ;)

3 0

3 years ago

Please help with this —

Igoryamba

3 times 8 is 24 and adding 7 is 31 so 31/1 divided by 19/1

U must flip the 19/1 to 1/19 and make it multiplication sign

31/1 times 1/19 is 31/19 or 1.63157

5 0

3 years ago

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