All categories

3 years ago

How can I solve this system of equations: x-y=-3 and -5x-2y=1

1 answer:

8 0

<span> x-y=-3 so y = x + 3
-5x-2y=1
substitute </span> y = x + 3 into -5x-2y=1
-5x - 2y = 1
-5x - 2(x + 3) = 1
-5x - 2x - 6 = 1
-7x= 7
x = -1

y = x + 3 = -1 + 3 = 2

answer
(-1, 2)

You might be interested in

If it is 9:36 pm what time would it be a hour and a half later

svlad2 [7]

Answer:

an hour would make it 10, then another 30 minutes added to the 30 would make it 11, so 11:06 pm

Step-by-step explanation:

4 0

3 years ago

How do you write –51. 6 as a mixed nuber

Ganezh [65]

Answer:

-51 6/10 or -51 60/100

Step-by-step explanation:

4 0

2 years ago

For each level of precision, find the required sample size to estimate the mean starting salary for a new CPA with 95 percent co

Rzqust [24]

Answer:

(a) Margin of error ( E) = $2,000 , n = 54

(b) Margin of error ( E) = $1,000 , n = 216

(c) Margin of error ( E) = $500 , n= 864

Step-by-step explanation:

Given -

Standard deviation $\sigma$ = $7,500

$\alpha$ = 1 - confidence interval = 1 - .95 = .05

$Z_{\frac{\alpha}{2}}$ = $Z_{\frac{.05}{2}}$ = 1.96

let sample size is n

(a) Margin of error ( E) = $2,000

Margin of error ( E) = $Z_{\frac{\alpha}{2}}\frac{\sigma}{\sqrt{n}}$

E = $Z_{\frac{.05}{2}}\frac{7500}{\sqrt{n}}$

Squaring both side

$E^{2} = 1.96^{2}\times\frac{7500^{2}}{n}$

$n =\frac{1.96^{2}}{2000^{2}} \times 7500^{2}$

n = 54.0225

n = 54 ( approximately)

(b) Margin of error ( E) = $1,000

E = $Z_{\frac{\alpha}{2}}\frac{\sigma}{\sqrt{n}}$

1000 = $Z_{\frac{.05}{2}}\frac{7500}{\sqrt{n}}$

Squaring both side

$1000^{2} = 1.96^{2}\times\frac{7500^{2}}{n}$

$n =\frac{1.96^{2}}{1000^{2}} \times 7500^{2}$

n = 216

E = $Z_{\frac{\alpha}{2}}\frac{\sigma}{\sqrt{n}}$

500 = $Z_{\frac{.05}{2}}\frac{7500}{\sqrt{n}}$

Squaring both side

$500^{2} = 1.96^{2}\times\frac{7500^{2}}{n}$

$n =\frac{1.96^{2}}{500^{2}} \times 7500^{2}$

n = 864

7 0

3 years ago

Can someone plz explain how to do this?

marissa [1.9K]

Answer:

the answer is f.

Step-by-step explanation:

it is the only sensible answer. every 5 years it increases by ~40%.

6 0

3 years ago

If 2k, 5k-1 and 6k+2 are the first 3 terms of an arithmetic sequence, find k and the 8th term

Klio2033 [76]

Answer:

see explanation

Step-by-step explanation:

The common difference d of an arithmetic sequence is

d = $a_{2}$ - $a_{1}$ = $a_{3}$ - $a_{2}$

Substitute in values and solve for k, that is

5k - 1 - 2k = 6k + 2 - (5k - 1)

3k - 1 = 6k + 2 - 5k + 1

3k - 1 = k + 3 ( subtract k from both sides )

2k - 1 = 3 ( add 1 to both sides )

2k = 4 ⇒ k = 2

--------------------------------------------------------

The n th term of an arithmetic sequence is

$a_{n}$ = $a_{1}$ + (n - 1)d

$a_{1}$ = 2k = 2 × 2 = 4 and

d = 5k - 1 - 2k = 3k - 1 = (3 × 2) - 1 = 5

Hence

$a_{8}$ = 4 + (7 × 5) = 4 + 35 = 39

4 0

3 years ago