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3 years ago

Explain how you determine an equation for a line using a table of input/output values.

1 answer:

6 0

Example :

x y
1 3
2 6
3 9
4 12

first thing u do is pick any 2 points (x,y) from ur table

(1,3) and (2,6)
now we sub those into the slope formula (y2 - y1) / (x2 - x1) to find the slope
(y2 - y1) / (x2 - x1)
(1,3)....x1 = 1 and y1 = 3
(2,6)...x2 = 2 and y2 = 6
sub
slope = (6 - 3) / (2 - 1) = 3/1 = 3

now we use slope intercept formula y = mx + b
y = mx + b
slope(m) = 3
use any point off ur table...(1,3)...x = 1 and y = 3
now we sub and find b, the y int
3 = 3(1) + b
3 = 3 + b
3 - 3 = b
0 = b

so ur equation is : y = 3x + 0....which can be written as y = 3x...and if u sub any of ur points into this equation, they should make the equation true....if they dont, then it is not correct

and if u need it in standard form..
y = 3x
-3x + y = 0
3x - y = 0 ...this is standard form

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Sid is packing crushed ice into a cone-shaped cThe cone has a height of 5 in. Its base has a diameter of 4 in. What is thevolume

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3 years ago

Read 2 more answers

Find the 13th term of a sequence where the

ValentinkaMS [17]

Answer:

First, you write an explicit formula then plug in the value of n in the formula with the term number you are trying to find, in this case the 13th term.

The steps are in the pictures

the final answer is 121 btw

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3 years ago

The approximate height, h, in meters, traveled by golf balls hit with two different clubs over a horizontal distance of d meters

inna [77]

Answer:

a) 100 metres

b) 10 metres

Step-by-step explanation:

Given: the approximate height, h, in meters, travelled by golf balls hit with two different clubs over a horizontal distance of d meters is given by the functions: $h=-0.002d^2+0.3d\,,\,h=-0.004d^2+0.5d$

To find: distances when the ball is at the same height when either of the clubs are used. Also, to find this height

Solution:

(a)

$-0.002d^2+0.3d=-0.004d^2+0.5d\\d(-0.002d+0.3)=d(-0.004d+0.5)\\-0.002d+0.004d=0.5-0.3\\0.002d=0.2\\d=\frac{0.2}{0.002}=100\,\,metres$

(b)

Put d = 100 in $h=-0.002d^2+0.3d$

$h=-0.002(100)^2+0.3(100)\\=-20+30\\=10\,\,metres$

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3 years ago

Restrictions for 9/2x, 5x/8

brilliants [131]

Answer: hummmm let me think it is 14x 40x

Step-by-step explanation:

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3 years ago

The following results come from two independent random samples taken of two populations.

photoshop1234 [79]

Answer:

$(a)\ \bar x_1 - \bar x_2 = 2.0$

$(b)\ CI =(1.0542,2.9458)$

$(c)\ CI = (0.8730,2.1270)$

Step-by-step explanation:

Given

$n_1 = 60$ $n_2 = 35$

$\bar x_1 = 13.6$ $\bar x_2 = 11.6$

$\sigma_1 = 2.1$ $\sigma_2 = 3$

Solving (a): Point estimate of difference of mean

This is calculated as: $\bar x_1 - \bar x_2$

$\bar x_1 - \bar x_2 = 13.6 - 11.6$

$\bar x_1 - \bar x_2 = 2.0$

Solving (b): 90% confidence interval

We have:

$c = 90\%$

$c = 0.90$

Confidence level is: $1 - \alpha$

$1 - \alpha = c$

$1 - \alpha = 0.90$

$\alpha = 0.10$

Calculate $z_{\alpha/2}$

$z_{\alpha/2} = z_{0.10/2}$

$z_{\alpha/2} = z_{0.05}$

The z score is:

$z_{\alpha/2} = z_{0.05} =1.645$

The endpoints of the confidence level is:

$(\bar x_1 - \bar x_2) \± z_{\alpha/2} * \sqrt{\frac{\sigma_1^2}{n_1}+\frac{\sigma_2^2}{n_2}}$

$2.0 \± 1.645 * \sqrt{\frac{2.1^2}{60}+\frac{3^2}{35}}$

$2.0 \± 1.645 * \sqrt{\frac{4.41}{60}+\frac{9}{35}}$

$2.0 \± 1.645 * \sqrt{0.0735+0.2571}$

$2.0 \± 1.645 * \sqrt{0.3306}$

$2.0 \± 0.9458$

Split

$(2.0 - 0.9458) \to (2.0 + 0.9458)$

$(1.0542) \to (2.9458)$

Hence, the 90% confidence interval is:

$CI =(1.0542,2.9458)$

Solving (c): 95% confidence interval

We have:

$c = 95\%$

$c = 0.95$

Confidence level is: $1 - \alpha$

$1 - \alpha = c$

$1 - \alpha = 0.95$

$\alpha = 0.05$

Calculate $z_{\alpha/2}$

$z_{\alpha/2} = z_{0.05/2}$

$z_{\alpha/2} = z_{0.025}$

The z score is:

$z_{\alpha/2} = z_{0.025} =1.96$

The endpoints of the confidence level is:

$(\bar x_1 - \bar x_2) \± z_{\alpha/2} * \sqrt{\frac{\sigma_1^2}{n_1}+\frac{\sigma_2^2}{n_2}}$

$2.0 \± 1.96 * \sqrt{\frac{2.1^2}{60}+\frac{3^2}{35}}$

$2.0 \± 1.96* \sqrt{\frac{4.41}{60}+\frac{9}{35}}$

$2.0 \± 1.96 * \sqrt{0.0735+0.2571}$

$2.0 \± 1.96* \sqrt{0.3306}$

$2.0 \± 1.1270$

Split

$(2.0 - 1.1270) \to (2.0 + 1.1270)$

$(0.8730) \to (2.1270)$

Hence, the 95% confidence interval is:

$CI = (0.8730,2.1270)$

8 0

3 years ago