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3 years ago

Which sequence is geometric and has 14 as its fifth term and 12 as the common ratio?

1 answer:

7 0

Geometric sequence is
an=a1(r)^(n-1)
an=nth erm
a1=first term
r=common ratio

5th term=14
a5=a1(12)^(5-1)=15

a1(12)^4=15
a1(20736)=15
divide both sides by 20736
a1=15/20736=5/6912
geometric sequence is
an=(5/6912)(12)^(n-1)

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PLEASE HELP!! ASAP!!!!!

Dovator [93]

The linear function with the same y-intercept with the graphed function is: table A.

<h3>What is a Linear Function?</h3>

The equation that models a linear function is, y = mx + b, where m is the slope and b is the y-intercept.

Slope of the graphed function = rise/run = - 2/1 = -2

Using one of the points on the line (x, y) = (5, 0) and the slope, m = -2, find the y-intercept (b) by substituting the values into y = mx + b:

0 = -2(5) + b

0 = -10 + b

10 = b

b = 10

The slope (m) of the graphed function is -2, and the y-intercept (b) is: 10.

Slope (m) of table A = change in y/change in x = (14 - 8)/(3 - 1) = 3

Substitute a point (x, y) = (1, 8) and slope (m) = 3 into y = mx + b to find the y-intercept (b):

8 = 3(1) + b

8 - 3 = b

5 = b

b = 5

Therefore the table with the same y-intercept as the graphed function is table A.

Learn more about linear function on:

brainly.com/question/4025726

#SPJ1

5 0

2 years ago

A baseball team played 147 regular season games. The ratio of the number of games they won to the number of games they lost was

bulgar [2K]

Answer:42 losses

Step-by-step explanation:

142 x 2/7

7 0

2 years ago

F(x) = (128/127)(1/2)x, x = 1,2,3,...7. determine the requested values: round your answers to three decimal places (e.g. 98.765)

Marrrta [24]

A.
$\mathbb P(X\le 1)=\mathbb P(X=1)=\dfrac{128}{127}\left(\dfrac12\right)^1=\dfrac{64}{127}$

b.
$\mathbb P(X>1)=1-\mathbb P(X\le1)=1-\dfrac{64}{127}=\dfrac{63}{127}$

c.
$\mathbb E(X)=\displaystyle\sum_{x=1}^7 x\,f_X(x)=\frac{64}{127}\sum_{x=1}^7 x\left(\frac12\right)^{x-1}$

Suppose $f(y)=\displaystyle\sum_{x=0}^7 y^x$ . Then $f'(y)=\displaystyle\sum_{x=1}^7 xy^{x-1}$ . So if we can find a closed form for $f(y)$ , in terms of $y$ , we can find $\mathbb E(X)$ by evaluating the derivative of $f(y)$ at $y=\dfrac12$ .

$f(y)=\displaystyle\sum_{x=0}^7 y^x=y^0+y^1+y^2+\cdots+y^6+y^7$
$y\,f(y)=y^1+y^2+y^3+\cdots+y^7+y^8$
$f(y)-y\,f(y)=y^0-y^8$
$(1-y)f(y)=1-y^8$
$f(y)=\dfrac{1-y^8}{1-y}$
$\implies f'(y)=\dfrac{7y^8-8y^7+1}{(1-y)^2}$
$\implies\mathbb E(X)=\dfrac{64}{127}f'\left(\dfrac12\right)=\dfrac{64}{127}\times\dfrac{247}{64}=\dfrac{247}{127}$

d.
$\mathbb V(X)=\mathbb E(X^2)-\mathbb E(X)^2$

We find $\mathbb E(X^2)$ in a similar manner as in (c).

$\mathbb E(X^2)=\displaystyle\sum_{x=1}^7 x^2\,f_X(x)=\frac{32}{127}\sum_{x=1}^7x^2\left(\frac12\right)^{x-2}$

Now,

$f(y)=\displaystyle\sum_{x=0}^7y^x$
$\implies f'(y)=\displaystyle\sum_{x=1}^7xy^{x-1}$
$\implies f''(y)=\displaystyle\sum_{x=2}^7x(x-1)y^{x-2}$

We know that

$f''(y)=-\dfrac{42y^8-96y^7+56y^6-2}{(1-y)^3}$
$\implies f''\left(\dfrac12\right)=\dfrac{219}{16}$

We also have

$f''(y)=\displaystyle\sum_{x=2}^7x(x-1)y^{x-2}$
$f''(y)=\displaystyle\sum_{x=2}^7x^2y^{x-2}-\sum_{x=2}^7xy^{x-2}$
$f''(y)=\displaystyle\frac1{y^2}\left(\sum_{x=2}^7x^2y^x-\sum_{x=2}^7xy^x\right)$
$f''(y)=\displaystyle\frac1{y^2}\left(\bigg(\sum_{x=1}^7x^2y^x-y\bigg)-\bigg(\sum_{x=1}^7xy^x-y\bigg)\right)$
$f''(y)=\displaystyle\frac1{y^2}\left(\sum_{x=1}^7x^2y^x-\sum_{x=1}^7xy^x\right)$

so that when $y=\dfrac12$ , we get

$\dfrac{219}{16}=4\left(\dfrac{127}{128}\mathbb E(X^2)-\dfrac{127}{128}\mathbb E(X)\right)\implies\mathbb E(X^2)=\dfrac{685}{127}$

Then

$\mathbb V(X)=\dfrac{685}{127}-\left(\dfrac{247}{127}\right)^2=\dfrac{25,986}{16,129}$

6 0

4 years ago

Find the sum and product of each of these pairs of num- bers. express your answers as a base 3 expansion.

alisha [4.7K]

Adding in base 3 is very much like adding in base 10, but instead of carrying digits when their sum exceeds 9, we have to watch for when the sum exceeds 2:

.   112
+ 210
- - - - -
.   322 -> 1022

.   12021
+    2122
- - - - - - -
.   14143 -> 14150 -> 14220 -> 21220

.   20001
+    1111
- - - - - - -
.   21112

.   120021
+      2002
- - - - - - - -
.    122023 -> 122030 -> 122100

5 0

3 years ago

Can someone tell me if I’m doing this right? I think I have to multiply 5.735 but I don’t know

devlian [24]

Answer:

you need a calculator for the answer

4 0

2 years ago