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3 years ago

If someone answers this correctly Ill be there friend

Mathematics

2 answers:

TEA [102]3 years ago

7 0

Answer:

The answer is C

Step-by-step explanation:

Because it's going over 3 & up 1. Since it's rise over run it would be 1/3.

Alex17521 [72]3 years ago

4 0

Answer should be c..

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Find f(5) for f(x)= 1/9(3)^x<br> A. 27<br> B. 3<br> C. 9<br> D. 81

zhuklara [117]

Answer:

A 27

Step-by-step explanation:

f(5) means x = 5, so substitute 5 into 1/9(3)^x

f(5) = 1/9(3^5) = 1/9(243) = 27

7 0

3 years ago

Write equations for the horizontal and vertical lines passing through the point (6,4).

liberstina [14]

Check the picture below.

4 0

2 years ago

Why Do i SUCK AT MATH!!!!!!!!!!!!!!!!! :(

blagie [28]

Answer:

Math is not a very easy subject and not everyone can master it.All that is needed are focus,dedication and confidence.You should try practicing in your weak areas or seek help to get a better understanding.

Step-by-step explanation:

I really hope this helps :)

8 0

3 years ago

From a sample of 37 graduate students, the mean number of months of work experience prior to entering an MBA program was 35.15.

kiruha [24]

Answer: (29.35, 40.95)

Step-by-step explanation:

Formula to find the confidence interval for population mean :-

$\overline{x}\pm z^*\dfrac{\sigma}{\sqrt{n}}$

, where $\overline{x}$ = sample mean.

z*= critical z-value

n= sample size.

$\sigma$ = Population standard deviation.

As per given , we have

$\overline{x}=35.15$

$\sigma=18$

n= 37

Using z-table, the critical z-value for 95% confidence = z* = 1.96

Then, the confidence interval for the population mean will be :

$35.15\pm (1.96)\dfrac{18}{\sqrt{37}}$

$=35.15\pm (1.96)\dfrac{1.7}{6.08276}$

$=35.15\pm (1.96)(2.95918)$

$\approx35.15\pm5.80$

$=(35.15-5.80,\ 35.15+5.80)=(29.35,\ 40.95)$

Hence, a 95% confidence interval for the population mean = (29.35, 40.95)

8 0

3 years ago

How do I get (tan^2(x)-sin^2(x))/tan(x) equal to (sin^2(x))/cot(x)

shepuryov [24]

$LHS\\ \\ =\frac { \tan ^{ 2 }{ x-\sin ^{ 2 }{ x } } }{ \tan { x } } \\ \\ =\frac { 1 }{ \tan { x } } \left( \tan ^{ 2 }{ x-\sin ^{ 2 }{ x } } \right)$

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

$\\ \\ =\frac { \cos { x } }{ \sin { x } } \cdot \frac { \sin ^{ 2 }{ x\left( 1-\cos ^{ 2 }{ x } \right) } }{ \cos ^{ 2 }{ x } } \\ \\ =\frac { \cos { x } }{ \sin { x } } \cdot \frac { \sin ^{ 2 }{ x\cdot \sin ^{ 2 }{ x } } }{ \cos ^{ 2 }{ x } } \\ \\ =\frac { \cos { x } \sin ^{ 4 }{ x } }{ \sin { x\cos ^{ 2 }{ x } } } \\ \\ =\frac { \sin ^{ 3 }{ x } }{ \cos { x } }$

$\\ \\ =\sin ^{ 2 }{ x } \cdot \frac { \sin { x } }{ \cos { x } } \\ \\ =\sin ^{ 2 }{ x } \cdot \frac { 1 }{ \frac { \cos { x } }{ \sin { x } } } \\ \\ =\sin ^{ 2 }{ x } \cdot \frac { 1 }{ \cot { x } } \\ \\ =\frac { \sin ^{ 2 }{ x } }{ \cot { x } } \\ \\ =RHS$

8 0

3 years ago