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3 years ago

15

Plz hurry

1 answer:

Ivan3 years ago

5 0

Answer:

Data for store #1 shows greater variability and

The mean for store #1 is greater than the mean for store #2

Step-by-step explanation:

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Silas just bought a used car for $5000. It loses value at a rate of 30% per year. The value of the car in n years is modeled by

Fantom [35]

Answer:

$1200.50

plug 5 for n

5000(.7)^4

5000*.2401

=1200.5

8 0

3 years ago

A spinner has three unusual sections: red, yellow, and blue.. The table shows the results of Nolan’s spins. Find the probability

Anika [276]

Answers:

Red = 1/3
Yellow = 7/15
Blue = 1/5

====================================================

Explanation:

Add up all the frequencies in the table to see that there are 120+168+72 = 360 spins total.

Of those 360 spins, 120 of them land on red. The probability of landing on red is 120/360 = (120*1)/(120*3) = 1/3
Since 168 land on yellow, this means the probability of landing on yellow is 168/360 = (24*7)/(24*15) = 7/15
Lastly, we have 72 occurrences of blue out of 360 total spins. The probability of landing on blue is 72/360 = (72*1)/(72*5) = 1/5

4 0

3 years ago

What is the solution to the equation-3(h+5)+2 = 4(h+6)- 9?<br> h= 4<br> h= -2<br> h = 2<br> h = 4

ruslelena [56]

Expanding
-3h-15+2=4h+24-9
-3h-4h=24-9-2+15
-7h=28
h=4

8 0

3 years ago

Determine formula of the nth term 2, 6, 12 20 30,42

nalin [4]

Check the forward differences of the sequence.

If $\{a_n\} = \{2,6,12,20,30,42,\ldots\}$ , then let $\{b_n\}$ be the sequence of first-order differences of $\{a_n\}$ . That is, for n ≥ 1,

$b_n = a_{n+1} - a_n$

so that $\{b_n\} = \{4, 6, 8, 10, 12, \ldots\}$ .

Let $\{c_n\}$ be the sequence of differences of $\{b_n\}$ ,

$c_n = b_{n+1} - b_n$

and we see that this is a constant sequence, $\{c_n\} = \{2, 2, 2, 2, \ldots\}$ . In other words, $\{b_n\}$ is an arithmetic sequence with common difference between terms of 2. That is,

$2 = b_{n+1} - b_n \implies b_{n+1} = b_n + 2$

and we can solve for $b_n$ in terms of $b_1=4$ :

$b_{n+1} = b_n + 2$

$b_{n+1} = (b_{n-1}+2) + 2 = b_{n-1} + 2\times2$

$b_{n+1} = (b_{n-2}+2) + 2\times2 = b_{n-2} + 3\times2$

and so on down to

$b_{n+1} = b_1 + 2n \implies b_{n+1} = 2n + 4 \implies b_n = 2(n-1)+4 = 2(n + 1)$

We solve for $a_n$ in the same way.

$2(n+1) = a_{n+1} - a_n \implies a_{n+1} = a_n + 2(n + 1)$

Then

$a_{n+1} = (a_{n-1} + 2n) + 2(n+1) \\ ~~~~~~~= a_{n-1} + 2 ((n+1) + n)$

$a_{n+1} = (a_{n-2} + 2(n-1)) + 2((n+1)+n) \\ ~~~~~~~ = a_{n-2} + 2 ((n+1) + n + (n-1))$

$a_{n+1} = (a_{n-3} + 2(n-2)) + 2((n+1)+n+(n-1)) \\ ~~~~~~~= a_{n-3} + 2 ((n+1) + n + (n-1) + (n-2))$

and so on down to

$a_{n+1} = a_1 + 2 \displaystyle \sum_{k=2}^{n+1} k = 2 + 2 \times \frac{n(n+3)}2$

$\implies a_{n+1} = n^2 + 3n + 2 \implies \boxed{a_n = n^2 + n}$

6 0

2 years ago

Can someone plzzz heelppp meeee!!!

mel-nik [20]

Answer:

.

Step-by-step explanation:

5 0

3 years ago