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3 years ago

How many relatives does it have in 8 and 9 generations ???

Mathematics

2 answers:

Elina [12.6K]3 years ago

8 0

its C .34 and 55

i got it right on apex

OleMash [197]3 years ago

7 0

The Fibonacci sequence works like this. Take the first two generations, add them together and that gives you the third. Add the second and third to get the fourth. Add the fourth and fifth to get the sixth.

So you need to add the sixth and seventh to get the eighth.
13 + 21= 34

Add the seventh and eighth to get the ninth.
21 + 34= 55

Answer is C- 34, 55

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A billing company that collects bills for doctors' offices in the area is concerned that the percentage of bills being paid by

omeli [17]

Answer:

1) A. H0: p = 0.30

HA: p not equal to 0.30

2) A. The Independence Assumption is met.

C. The Randomization Condition is met.

D. The Success/Failure Condition is met.

3) Test statistic z = 2.089

P-value = 0.0367

4) C. Reject H0. There is sufficient evidence to suggest that the percentage of bills paid by medical insurance has changed.

Step-by-step explanation:

1) This is a hypothesis test for a proportion.

The claim is that there is a significant change in the percent of bills being paid by medical insurance.

As we are looking for evidence of a difference, no matter if it is higher or lower than the null hypothesis proportion, the alternative hypothesis is defined by a unequal sign.

Then, the null and alternative hypothesis are:

$H_0: \pi=0.3\\\\H_a:\pi\neq 0.3$

2) Cheking the conditions:

The independence assumption and the randomization condition are met as the bills were selected randomly from the population.

The 10% condition can not be checked, as we do not know the size of the population.

The success/failure condition is met as the products np and n(1-p) are bigger than 10 (the number of successes and failures are both bigger than 10).

3) The significance level is assumed to be 0.05.

The sample has a size n=9260.

The sample proportion is p=0.31.

The standard error of the proportion is:

$\sigma_p=\sqrt{\dfrac{\pi(1-\pi)}{n}}=\sqrt{\dfrac{0.3*0.7}{9260}}\\\\\\ \sigma_p=\sqrt{0.000023}=0.005$

Then, we can calculate the z-statistic as:

$z=\dfrac{p-\pi-0.5/n}{\sigma_p}=\dfrac{0.31-0.3-0.5/9260}{0.005}=\dfrac{0.01}{0.005}=2.089$

This test is a two-tailed test, so the P-value for this test is calculated as:

$\text{P-value}=2\cdot P(z>2.089)=0.0367$

As the P-value (0.0184) is smaller than the significance level (0.05), the effect is significant.

The null hypothesis is rejected.

There is enough evidence to support the claim that there is a significant change in the percent of bills being paid by medical insurance.

5 0

2 years ago

Which the slope-intercept form of the equation of each line the slope and y-intercept?

Delvig [45]

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6 0

3 years ago

10 1/4 divided by 151 3/16

Svetlanka [38]

10 1/4 / 151 3/16 = 4/59

Have a great day :)

8 0

3 years ago

Read 2 more answers

As a bowling instructor, you calculate your students' averages during tournaments. In 5 games, one bowler had the following scor

erastovalidia [21]

Answer:

154.2

Step-by-step explanation:

143 plus

156 plus

172 plus

133 plus

167 = 771

divide by 5 equals 154.2

4 0

3 years ago

The following describes a sample. The information given includes the five number summary, the sample size, and the largest and s

lys-0071 [83]

Answer:

$L= Q_1 - 1.5*IQR = 210 -1.5*40 =150$

$U= Q_3 + 1.5*IQR = 250 +1.5*40 =310$

And if we analyze the info provided we have 500 values

Tails: 160,165,167,171,175,...,268,269,269,269,270,270

So as we can see on the tails any values is <150 or >310 so then for this case we cannot consider as outliers any of the values on the tails of the distribution.

Step-by-step explanation:

For this case we have the 5 number summary

(160,210,220,250,270)

So then we have:

minimum = 160 , Q1 = 210, Q2= Median=220, Q3 = 250, Max=270

If we find the interquartile range we got:

$IQR = Q_3 -Q_1 = 250-210 =40$

For this case we need to find the lower and upper limit with the following formulas:

$L= Q_1 - 1.5*IQR = 210 -1.5*40 =150$

$U= Q_3 + 1.5*IQR = 250 +1.5*40 =310$

And if we analyze the info provided we have 500 values

Tails: 160,165,167,171,175,...,268,269,269,269,270,270

So as we can see on the tails any values is <150 or >310 so then for this case we cannot consider as outliers any of the values on the tails of the distribution.

4 0

3 years ago