Ask question

Ask question

All categories

3 years ago

9

Find the 20th term in the sequence, where the first term is -2 and the common ratio is-2.

1 answer:

geniusboy [140]3 years ago

6 0

Answer:

Step-by-step explanation:

(1)

a1=-2

r=-2

$a_{20}=(-2)(-2)^{20-1}=-2(-2)^{19}=2^{20}=1048576\\(2)\\a_{1}=1000\\r=\frac{200}{1000} =\frac{1}{5} \\a_{8}=1000(\frac{1}{5} )^{8-1}=1000(.2)^7=0.0128\\(3)\\a_{1}=2\\a_{3}=50\\a_{2}=\sqrt{2*50} =10\\r=\frac{10}{2} =5$

You might be interested in

If point A is located at (-9, 2) on a coordinate plane, and point B is located at (-9, 10), what is the distance between the two

JulsSmile [24]

ANSWER

8 units.

EXPLANATION

The first point, A is located at (-9,2) on the coordinate plane.

The second point, B is also located at (-9,10) on the coordinate plane.

You should be smart and observe that the x-coordinates of both points are the same. This means that, the two points lies on a vertical line.

Hence it will be faster to use the absolute value method rather than the distance formula.

According to this method, you only have to find the absolute value of the difference in the y-values.

$|AB|=|10-2|$

$|AB|= |8| = 8units$

6 0

3 years ago

18.<br> What is the solution of |2x+5|<19?<br> -19 < x < 19<br> -12 -7 -24 < x < 14

elena-14-01-66 [18.8K]

Answer:

thats a tough one

Step-by-step explanation:

4 0

3 years ago

I need help please explain your work for number 7 only

Keith_Richards [23]

Answer:

Transitive property

Step-by-step explanation:

When a=c and b=c, a=b

A would be 16+43

B would be 59

C would be 10+49

5 0

2 years ago

McAllister et al. (2012) compared varsity football and hockey players with varsity athletes from noncontact sports to determine

jonny [76]

Answer:

$t _{critical} = 1.760$

t = 2.2450

d. 0.264

Step-by-step explanation:

The null hypothesis is:

$H_o: \mu_1 - \mu_2 = 0$

Alternative hypothesis;

$H_a : \mu_1 - \mu_2 > 0\\$

The pooled variance t-Test would have been determined if the population variance are the same.

$S_p^2 = \dfrac{(n_1-1)S_1^2+(n_2-1)S^2_2}{(n_1-1)+(n_2-1)}$

$S_p^2 = \dfrac{(8-1)2.507^2+(8-1)2.8282^2}{(8-1)+(8-1)}$

$S_p^2 = 7.14$

The t-test statistics can be computed as:

$t= \dfrac{(x_1-x_2)-(\mu_1 - \mu_2)}{\sqrt{Sp^2 ( \dfrac{1}{n} +\dfrac{1}{n_2})}}$

$t= \dfrac{(9-6)-0}{\sqrt{7.14 ( \dfrac{1}{8} +\dfrac{1}{8})}}$

$t= \dfrac{3}{1.336}$

t = 2.2450

Degree of freedom $df = (n_1 -1) + ( n_2 +1 )$

df = (8-1)+(8-1)

df = 7 + 7

df = 14

At df = 14 and ∝ = 0.05;

$t _{critical} = 1.760$

Decision Rule: To reject the null hypothesis if the t-test is greater than the critical value.

Conclusion: We reject $H_o$ and there is sufficient evidence to conclude that the test scores for contact address s less than Noncontact athletes.

To calculate r²

The percentage of the variance is;

$r^2 = \dfrac{t^2}{t^2 + df}$

$r^2 = \dfrac{2.2450^2}{2.2450^2 + 14}$

$r^2 = \dfrac{5.040025}{5.040025+ 14}$

$r^2 = 0.2647$

7 0

3 years ago

How do you factor -6a+27b-15c

tatyana61 [14]

-3 (2a-9b+5c)

common factor out the lowest common factor which is -3

5 0

3 years ago