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defon
4 years ago
7

48+49+50+51+...112+113

Mathematics
1 answer:
Stels [109]4 years ago
4 0

Answer: 423

Step-by-step explanation:

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Kisachek [45]

Answer

X= 20 degree angle

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3 years ago
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Verify the identity cot(x-pi/2)=-tan x
vampirchik [111]
To verify the identity we need the following identities:

i) \displaystyle{ \cot(x)= \frac{\cos x}{\sin x}

ii) \displaystyle{ \sin (x-y)=\ sinx\cdot\ cosy -\ siny\cdot\ cosx

iii) \displaystyle{ \cos (x-y)=\ cosx\cdot \ cosy +\ sinx\cdot\ siny.

Also, we have know that \displaystyle{ \sin \frac{ \pi }{2}=1 and \displaystyle{ \cos \frac{ \pi }{2}=0.


Thus, \displaystyle{ \cot(x-\frac{\pi}{2})= \frac{\cos (x-\frac{\pi}{2})}{\sin (x-\frac{\pi}{2})}

By (ii) and (iii) we have:

\displaystyle{ \frac{\cos (x-\frac{\pi}{2})}{\sin (x-\frac{\pi}{2})}= \frac{\ cosx\cdot \ cos\frac{\pi}{2} +\ sinx\cdot\ sin\frac{\pi}{2}}{\ sinx\cdot\ cos\frac{\pi}{2} -\ sin\frac{\pi}{2}\cdot\ cosx} = \frac{\ sinx}{-\cos x}

by simplifying \displaystyle{ \sin \frac{ \pi }{2}=1 and \displaystyle{ \cos \frac{ \pi }{2}=0.

Now, \displaystyle{ \frac{\ sinx}{-\cos x} is clearly -tanx.
4 0
3 years ago
The following data give the estimated prices of a 6-ounce can or a 7.06-ounce pouch of water-packed tuna for 14 different brands
Vikentia [17]

Answer:

a) Range = 1.39

b) the sample variance is 0.1597

c) the sample standard deviation is 0.3996

Step-by-step explanation:

Given the data in the question;

a) Find the range.

To determine the range, we simple subtract the smallest value from the largest value. i.e

Range = largest value - smallest value

from data set; our smallest is 0.53 while our largest value is 1.92

so

Range = 1.92 - 0.53 = 1.39

b) Find the sample variance.

To determine our variance, we use the following formula;

∑(X_{i} - x_{bar})² / n - 1 = [ ∑X_{i}²/n-1 ] - [ \frac{n}{n-1}(x_{bar})²]

where x_{bar} =  ∑X_{i}/n

n is sample size = 14 so lets calculate ∑X_{i}

∑X_{i} = 0.99 + 1.92 + 1.23 + 0.85 + 0.65 + 0.53 + 1.41  + 1.12 + 0.63 + 0.67 + 0.69 + 0.60 + 0.60 + 0.66

∑X_{i}  =  12.55    

∑X_{i}² = 0.99² + 1.92² + 1.23² + 0.85² + 0.65² + 0.53² + 1.41²  + 1.12² + 0.63² + 0.67² + 0.69² + 0.60² + 0.60² + 0.66²

∑X_{i}² = 13.3253

so

our  x_{bar} =  ∑X_{i}/n = 12.55 / 14 = 0.8964

so our Variance  will be;

= [ ∑X_{i}²/n-1 ] - [ \frac{n}{n-1}(x_{bar})²]

= [ 13.3253 / 14-1 ] - [ \frac{14}{14-1} (0.8964)²]

= 1.025 - 0.8653

= 0.1597

Therefore, the sample variance is 0.1597

c) Find the sample standard deviation.

we know that standard deviation is the square root of variance;

standard deviation = √Variance

standard deviation = √0.1597

standard deviation = 0.3996

Therefore, the sample standard deviation is 0.3996

7 0
3 years ago
How do i write the product of q and 11 as an algebraic expression
san4es73 [151]

Answer:

since the product meaning multiplication, it basically means 11 (q) which is 11q.

3 0
3 years ago
Distribute -9 (-4+8) please and thank you
qaws [65]

Answer:

36-72

Step-by-step explanation:

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5 0
3 years ago
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