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PtichkaEL [24]
3 years ago
10

As Saturn revolves around the sun, it travels at a rate of approximately 6 miles per second. Convert this rate to kilometers per

second. At this rate, how many kilometers will Saturn travel in 20 seconds? In your computations, assume that 1 mile is equal to kilometers. Do no round you answer.
Mathematics
1 answer:
liraira [26]3 years ago
5 0

Answer:

Step-by-step explanation:

Pls help me in homework

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In a recent study, researchers found that 31 out of 150 boys aged 7-13 were overweight or obese. On the basis of this study can
AURORKA [14]

Answer:

1) Null hypothesis:p \leq 0.15  

Alternative hypothesis:p > 0.15  

2) The One-Sample Proportion Test is used to assess whether a population proportion \hat p is significantly different from a hypothesized value p_o.

3) z_{crit}= 1.64

And the rejection zone would be z>1.64

4) Calculate the statistic  

z=\frac{0.207 -0.15}{\sqrt{\frac{0.15(1-0.15)}{150}}}=1.955  

5) Statistical decision  

For this case our calculated value is on the rejection zone, so we have enough evidence to reject the null hypothesis at 5% of significance and we can conclude that the true proportion is higher than 0.15

Step-by-step explanation:

Data given and notation

n=150 represent the random sample taken

X=21 represent the boys overweight

\hat p=\frac{31}{150}=0.207 estimated proportion of boy overweigth

p_o=0.15 is the value that we want to test

\alpha=0.05 represent the significance level

Confidence=95% or 0.95

z would represent the statistic (variable of interest)

p_v represent the p value (variable of interest)  

1) Concepts and formulas to use  

We need to conduct a hypothesis in order to test the true proportion of boys obese is higher than 0.15.:  

Null hypothesis:p \leq 0.15  

Alternative hypothesis:p > 0.15  

When we conduct a proportion test we need to use the z statistic, and the is given by:  

z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}} (1)  

2) The One-Sample Proportion Test is used to assess whether a population proportion \hat p is significantly different from a hypothesized value p_o.

3) Decision rule

For this case we need a value on the normal standard distribution who accumulates 0.05 of the area on the right tail and on this case this value is:

z_{crit}= 1.64

And the rejection zone would be z>1.64

4) Calculate the statistic  

Since we have all the info requires we can replace in formula (1) like this:  

z=\frac{0.207 -0.15}{\sqrt{\frac{0.15(1-0.15)}{150}}}=1.955  

5) Statistical decision  

For this case our calculated value is on the rejection zone, so we have enough evidence to reject the null hypothesis at 5% of significance and we can conclude that the true proportion is higher than 0.15

6 0
3 years ago
If something cost 15$ and he wants to tip 20% how much would it be
ANTONII [103]
The answer to your question would be $18
7 0
3 years ago
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Plz math help me in class !!!!!
Serhud [2]
Sorry but im terrible at math
4 0
3 years ago
Would I do Pythagorean theorem?
kati45 [8]
Yes you do , i learned pythagorean theorem last year in pre-algebra
4 0
3 years ago
Read 2 more answers
How is this solved using trig identities (sum/difference)?
GenaCL600 [577]
FIRST PART
We need to find sin α, cos α, and cos β, tan β
α and β is located on third quadrant, sin α, cos α, and sin β, cos β are negative

Determine ratio of ∠α
Use the help of right triangle figure to find the ratio
tan α = 5/12
side in front of the angle/ side adjacent to the angle = 5/12
Draw the figure, see image attached

Using pythagorean theorem, we find the length of the hypotenuse is 13
sin α = side in front of the angle / hypotenuse
sin α = -12/13

cos α = side adjacent to the angle / hypotenuse
cos α = -5/13

Determine ratio of ∠β
sin β = -1/2
sin β = sin 210° (third quadrant)
β = 210°

cos \beta = -\frac{1}{2}  \sqrt{3}

tan \beta= \frac{1}{3}  \sqrt{3}

SECOND PART
Solve the questions
Find sin (α + β)
sin (α + β) = sin α cos β + cos α sin β
sin( \alpha + \beta )=(- \frac{12}{13} )( -\frac{1}{2}  \sqrt{3})+( -\frac{5}{13} )( -\frac{1}{2} )
sin( \alpha + \beta )=(\frac{12}{26}\sqrt{3})+( \frac{5}{26} )
sin( \alpha + \beta )=(\frac{5+12\sqrt{3}}{26})

Find cos (α - β)
cos (α - β) = cos α cos β + sin α sin β
cos( \alpha + \beta )=(- \frac{5}{13} )( -\frac{1}{2} \sqrt{3})+( -\frac{12}{13} )( -\frac{1}{2} )
cos( \alpha + \beta )=(\frac{5}{26} \sqrt{3})+( \frac{12}{26} )
cos( \alpha + \beta )=(\frac{5\sqrt{3}+12}{26} )

Find tan (α - β)
tan( \alpha - \beta )= \frac{ tan \alpha-tan \beta }{1+tan \alpha  tan \beta }
tan( \alpha - \beta )= \frac{ \frac{5}{12} - \frac{1}{2} \sqrt{3}   }{1+(\frac{5}{12}) ( \frac{1}{2} \sqrt{3})}

Simplify the denominator
tan( \alpha - \beta )= \frac{ \frac{5}{12} - \frac{1}{2} \sqrt{3}   }{1+(\frac{5\sqrt{3}}{24})}
tan( \alpha - \beta )= \frac{ \frac{5}{12} - \frac{1}{2} \sqrt{3} }{ \frac{24+5\sqrt{3}}{24} }

Simplify the numerator
tan( \alpha - \beta )= \frac{ \frac{5}{12} - \frac{6}{12} \sqrt{3} }{ \frac{24+5\sqrt{3}}{24} }
tan( \alpha - \beta )= \frac{ \frac{5-6\sqrt{3}}{12} }{ \frac{24+5\sqrt{3}}{24} }

Simplify the fraction
tan( \alpha - \beta )= (\frac{5-6\sqrt{3}}{12} })({ \frac{24}{24+5\sqrt{3}})
tan( \alpha - \beta )= \frac{10-12\sqrt{3} }{ 24+5\sqrt{3}}

7 0
3 years ago
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