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3 years ago

Rewrite the following linear equations in slope-intercept form. write your answer with no spaces. v+4=-2(x-1)

Mathematics

2 answers:

lukranit [14]3 years ago

6 0

V + 4 = -2(x - 1)

v + 4 = -2x + 2

v = -2x - 2. [ Final answer]

Hope this helps!

Vedmedyk [2.9K]3 years ago

3 0

V+4 = -2(x-1)
⇒ v+4=-2x+2
⇒v+4-4=-2x+2-4
⇒v=-2x-2

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Patrick can paint a house by himself in t+4 hours. Karen can paint the same house by herself in t+9 hours. If they work together

vladimir2022 [97]

Patrick will be able to paint the home on his own after a ten-hour effort.

This is further explained below.

<h3>How much time will it take for Patrick to paint the house by himself?</h3>

Generally, the equation for time is mathematically given as

1/(t+4) + 1/(t+9) = 1/t

Therefore

[1/(t+4)] (t+4)(t+9)(t) + [1/(t+9)](t+4)(t+9)(t) = (1/t)(t+4)(t+9)(t)

(t² + 9t) + (t² + 4t) = t² + 13t + 36

Place the equation on the left side of the page and make it equal to 0. Use the additive inverse property as follows:

t² + t² - t² + 9t + 4t - 13t - 36 = 0

2t² - t² + 13t - 13t - 36 = 0

t² - 36 = 0

Calculate the product of the difference of two squares using the formula t2 - 36 = 0.

Both the total and the difference between the two terms constitute the factors.

(t + 6) (t - 6) = 0

t + 6 = 0

t = -6

t - 6 = 0

t = 6

Choose the positive value for time, t = 6 hours.

It will take 10 hours for Patrick to paint the house alone.

Patrick will be able to paint the home on his own after a ten-hour effort.

Read more about time

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1 year ago

The ideal (daytime) noise-level for hospitals is 45 decibels with a standard deviation of 10 db; which is to say, this may not b

WINSTONCH [101]

Answer:

(a) A 99% confidence interval for the actual mean noise level in hospitals is (44.02 db, 49.98 db).

(b) We can be 90% confident that the actual mean noise level in hospitals is 47 db with a margin of error of 1.89 db.

(c) Unless our sample (of 81 hospitals) is among the most unusual 2% of samples, the actual mean noise level in hospitals is between 44.41 db and 49.59 db.

Step-by-step explanation:

The problem is incomplete. The questions are:

(a) A 99% confidence interval for the actual mean noise level in hospitals is (44.02 db, 49.98 db).

For a 99% CI, the value of z is z=2.58

Then, the confidence interval for the mean is:

$M-z\sigma/\sqrt{n}\leq\mu\leq M-z\sigma/\sqrt{n}\\\\47-2.58*10/\sqrt{75} \leq\mu\leq47+2.58*10/\sqrt{75}\\\\47-2.98\leq\mu\leq47+2.98\\\\44.02\leq\mu\leq 49.98$

(b) We can be 90% confident that the actual mean noise level in hospitals is 47 db with a margin of error of 1.89 db.

For a 90% CI, the value of z is z=1.64.

Then, we can calculate the margin of error as:

$E=z*\sigma/\sqrt{n}=1.64*10/\sqrt{75}=1.89$

(c) Unless our sample (of 81 hospitals) is among the most unusual 2% of samples, the actual mean noise level in hospitals is between 44.41 db and 49.59 db.

The 2% tails data corresponds, in the standard normal distirbution, to the values of z whose absolute value is higher than 2.33.

The values of db for these critical values are:

$X_1=M+z_1*\sigma/\sqrt{n}=47+(-2.33)*10/\sqrt{81}=47-2.59=44.41\\\\\\ X_2=M+z_2*\sigma/\sqrt{n}=47+(2.33)*10/\sqrt{81}=47+2.59=49.59$