Lines m and p are perpendicular if the slope of line is 1, 25th what is the slope of line p

A real estate agency offers three apartments for rent in New Mexico City. The expected total revenue per month from monthly rent

lutik1710 [3]

Monthly maintenance costs of 20% of the rent for the first two apartments and 25% of the rent for the third apartment, which is a total amount of $345 (0.2)x + (0.2)y + (0.25)z = 345 I'm not 1005 sure tho

7 0

2 years ago

Which of the following are dimensionally consistent? (Choose all that apply.)(a) a=v / t+xv2 / 2(b) x=3vt(c) xa2=x2v / t4(d) x=v

Bumek [7]

Complete Question

The complete question is shown on the first uploaded image

Answer:

A

is dimensionally consistent

B

is not dimensionally consistent

C

is dimensionally consistent

D

is not dimensionally consistent

E

is not dimensionally consistent

F

is dimensionally consistent

G

is dimensionally consistent

H

is not dimensionally consistent

Step-by-step explanation:

From the question we are told that

The equation are

$A) \ \ a^3 = \frac{x^2 v}{t^5}$

$B) \ \ x = t$

$C \ \ \ v = \frac{x^2}{at^3}$

$D \ \ \ xa^2 = \frac{x^2v}{t^4}$

$E \ \ \ x = vt+ \frac{vt^2}{2}$

$F \ \ \ x = 3vt$

$G \ \ \ v = 5at$

$H \ \ \ a = \frac{v}{t} + \frac{xv^2}{2}$

Generally in dimension

x - length is represented as L

t - time is represented as T

m = mass is represented as M

Considering A

$a^3 = (\frac{L}{T^2} )^3 = L^3\cdot T^{-6}$

and $\frac{x^2v}{t^5 } = \frac{L^2 L T^{-1}}{T^5} = L^3 \cdot T^{-6}$

Hence

$a^3 = \frac{x^2 v}{t^5}$ is dimensionally consistent

Considering B

$x = L$

and

$t = T$

Hence

$x = t$ is not dimensionally consistent

Considering C

$v = LT^{-1}$

and

$\frac{x^2 }{at^3} = \frac{L^2}{LT^{-2} T^{3}} = LT^{-1}$

Hence

$v = \frac{x^2}{at^3}$ is dimensionally consistent

Considering D

$xa^2 = L(LT^{-2})^2 = L^3T^{-4}$

and

$\frac{x^2v}{t^4} = \frac{L^2(LT^{-1})}{ T^5} = L^3 T^{-5}$

Hence

$xa^2 = \frac{x^2v}{t^4}$ is not dimensionally consistent

Considering E

$x = L$

;

$vt = LT^{-1} T = L$

and

$\frac{vt^2}{2} = LT^{-1}T^{2} = LT$

Hence

$E \ \ \ x = vt+ \frac{vt^2}{2}$ is not dimensionally consistent

Considering F

$x = L$

and

$3vt = LT^{-1}T = L$ Note in dimensional analysis numbers are

not considered

Hence

$F \ \ \ x = 3vt$ is dimensionally consistent

Considering G

$v = LT^{-1}$

and

$at = LT^{-2}T = LT^{-1}$

Hence

$G \ \ \ v = 5at$ is dimensionally consistent

Considering H

$a = LT^{-2}$

,

$\frac{v}{t} = \frac{LT^{-1}}{T} = LT^{-2}$

and

$\frac{xv^2}{2} = L(LT^{-1})^2 = L^3T^{-2}$

Hence

$H \ \ \ a = \frac{v}{t} + \frac{xv^2}{2}$ is not dimensionally consistent

8 0

2 years ago

In measuring reaction time, a psychologist estimates that a standard deviation is .05 seconds. How large a sample of measurement

blondinia [14]

Answer:

97

Step-by-step explanation:

We are asked to find the size of sample to be 95% confident that the error in psychologist estimate of mean reaction time will not exceed 0.01 seconds.

We will use following formula to solve our given problem.

$n\geq (\frac{z_{\alpha/2}\cdot\sigma}{E})^2$ , where,

$\sigma=\text{Standard deviation}=0.05$ ,

$\alpha=\text{Significance level}=1-0.95=0.05$ ,

$z_{\alpha/2}=\text{Critical value}=z_{0.025}=1.96$ .

$E=\text{Margin of error}$

$n=\text{Sample size}$

Substitute given values:

$n\geq (\frac{z_{0.025}\cdot\sigma}{E})^2$

$n\geq (\frac{1.96\cdot0.05}{0.01})^2$

$n\geq (\frac{0.098}{0.01})^2$

$n\geq (9.8)^2$

$n\geq 96.04$

Therefore, the sample size must be 97 in order to be 95% confident that the error in his estimate of mean reaction time will not exceed 0.01 seconds.

5 0

2 years ago

Eileen has a greenhouse and is growing orchids. The table shows the average number of orchids that bloomed over a period of four

Darina [25.2K]

Answer:

First one:

Linearly, because the table shows that the orchids increased by the same amount each month

Step-by-step explanation:

Let y be the no. of Orchids and x be the no. of months

y = 1.4x + 9.6

4 0

2 years ago

Why does a calculator return an error when trying to find inverse sine of 1.055?

anyanavicka [17]

Sin(x) is a function bounded between -1 and 1. A value greater than 1 or smaller than -1 (-1.1, for instance) does not correspond to any real angle.

It then returns error

6 0

2 years ago

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