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Feliz [49]
3 years ago
8

Find the next two terms of the geometric sequence 567, 189, 63 … .

Mathematics
2 answers:
Archy [21]3 years ago
7 0

Answer:

As we can observe,

F(1) = 567

F(2) = 189 = 567/3 = F(1)/3

F(3) = 63 = 189/3 = F(2)/3

=> F(4) = F(3)/3 = 63/3 = 21

Hope this helps!

:)

german3 years ago
7 0

Answer:

21 then 7

Step-by-step explanation:

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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
Plz I need help Plz I need help Plz I need help Plz I need help Plz I need help!!!!!!!!!!!!!!!!!!!!!!
Dafna11 [192]

Answer:

LEAST DIFFICULT: B

MOST DIFFICULT: I

Step-by-step explanation:

LEAST DIFFICULT:

The easiest systems of equations to solve are ones in which x or y is equal to just a number , because in this case, you can just sub in that number in the other equations. Though not the only one, B is an example of this.

MOST DIFFICULT:

More difficult systems of equations to solve are the ones with different amounts of x and y as it requires more computation. Set I is an example of this

8 0
3 years ago
7 x 5 x35 = 701223123
9966 [12]

Answer:

don't know how to answer that lol

Step-by-step explanation:

6 0
3 years ago
How to find (c,d,e)<br>Please do a step by step working.Thank you.
Natasha2012 [34]
(a)
The inverse is when you swap the variables and solve for y.
g(t) = 2t - 1 (Note: g(t) represents y)
rewrite as: y = 2t - 1
swap the variables: t = 2y - 1
solve for y: t + 1 = 2y
                   \frac{t + 1}{2} = y
Answer for (a): g^{-1}(t) =  \frac{t + 1}{2}

(b)
Same steps as part (a) above:
h(t) = 4t + 3
rewrite as: y = 4t + 3
swap the variables: t = 4y + 3
solve for y: y =\frac{t - 3}{4}

Answer for (b): h^{-1}(t) = \frac{t - 3}{4}

(c)
g^{-1} ( h^{-1}(t)) =  g^{-1} (\frac{t - 3}{4})
replace all t's in the g^{-1}(t) equation with \frac{t - 3}{4}
 g^{-1} (\frac{t - 3}{4}) = \frac{ \frac{t-3}{4} + 1}{2}
= \frac{ \frac{t-3}{4} +  \frac{4}{4}}{2} = \frac{ \frac{t - 3 + 4}{4}}{2} = \frac{ \frac{t + 1}{4}}{2} =  \frac{t + 1}{8}
Answer for (c): g^{-1} ( h^{-1}(t)) = \frac{t + 1}{8}

 (d)
h(g(t)) = h(2t - 1) = 4(2t - 1) + 3 = 8t - 4 + 3 = 8t - 1
Answer for (d): h(g(t)) = 8t - 1

(e)
h(g(t)) = 8t - 1
   y = 8 t - 1
   t = 8y - 1
  t + 1 = 8y
\frac{t + 1}{8} = y
Answer for (e): inverse of h(g(t)) = \frac{t + 1}{8}
 
























8 0
3 years ago
It might be D but i dont know
FrozenT [24]

Answer:

I'm sorry I don't have answer but.

it's not C!

8 0
2 years ago
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