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4 years ago

The sides of an open lot are measured in feet. Which unit could you use to describe the

Mathematics

2 answers:

artcher [175]4 years ago

8 0

Answer:

square feet

Step-by-step explanation:

Area has units of squared length.

Svet_ta [14]4 years ago

7 0

Answer:

c. Square feet

Step-by-step explanation:

it wouldn't be cubic feet, because that is used to measure volume. It wouldn't be feet, because that is for the sides, and yards is to measure length or width, not area. so the answer would be C. Square feet

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ASAP: I need as soon as possible with this question I have taken a picture of. If you could tell me how to do it and the answer,

VladimirAG [237]

Answer:

The answer to your question is h = 125.85 ft = 126 ft

Step-by-step explanation:

Process

1.- Determine two equations to solve the problem

$tan 14 = \frac{h}{x}$ (1)

$tan 47 = \frac{h}{x - 386}$ (2)

from (1) h = xtan14

substitute in (2) tan 47 = $\frac{xtan14}{x - 386}$

solve for x tan47(x - 386) = xtan14

1.07x - 413.9 = 0.25x

1.07x - 0.25x = 413.9

0.82x = 413.9

x = 413.9/0.82

x = 504.76 ft

2.- Calculate h

h = 504.76 tan 14

h = 125.85 ft = 126 ft

4 0

4 years ago

Why isn't 9 divisible by 5622

Jobisdone [24]

Answer:

It is. Only you dont get a whole number.

Step-by-step explanation:

7 0

4 years ago

The coordinates of point T are (0,6). The midpoint of ST is (3, -4). Find the coordinates of point S

bezimeni [28]

Answer:

S = (-8, 0 )

Step-by-step explanation:

7 0

3 years ago

A random sample of 10 parking meters in a beach community showed the following incomes for a day. Assume the incomes are normall

Vlad1618 [11]

Answer: (2.54,6.86)

Step-by-step explanation:

Given : A random sample of 10 parking meters in a beach community showed the following incomes for a day.

We assume the incomes are normally distributed.

Mean income : $\mu=\dfrac{\sum^{10}_{i=1}x_i}{n}=\dfrac{47}{10}=4.7$

Standard deviation : $\sigma=\sqrt{\dfrac{\sum^{10}_{i=1}{(x_i-\mu)^2}}{n}}$

$=\sqrt{\dfrac{(1.1)^2+(0.2)^2+(1.9)^2+(1.6)^2+(2.1)^2+(0.5)^2+(2.05)^2+(0.45)^2+(3.3)^2+(1.7)^2}{10}}$

$=\dfrac{30.265}{10}=3.0265$

The confidence interval for the population mean (for sample size <30) is given by :-

$\mu\ \pm t_{n-1, \alpha/2}\times\dfrac{\sigma}{\sqrt{n}}$

Given significance level : $\alpha=1-0.95=0.05$

Critical value : $t_{n-1,\alpha/2}=t_{9,0.025}=2.262$

We assume that the population is normally distributed.

Now, the 95% confidence interval for the true mean will be :-

$4.7\ \pm\ 2.262\times\dfrac{3.0265}{\sqrt{10}} \\\\\approx4.7\pm2.16=(4.7-2.16\ ,\ 4.7+2.16)=(2.54,\ 6.86)$

Hence, 95% confidence interval for the true mean= (2.54,6.86)

7 0

4 years ago

A man travelled 3/8 of his journey by a rail 1/4 by a taxi 1/8 by and the remaining 2km on foot what is the length of his total

Burka [1]

The man travelled in different ways: by rail, by taxi, by ___ and by foot. I placed a blank there because there seems to be a missing word in the given problem above. For sample purposes, let's just assume that is travel by bus.

Since all of these travels are equal to 1 whole journey, you can express each travel as a fraction. When you add them up, the answer would be 1. So,

3/8 + 1/4 + 1/8 + x = 1

The variable x here denotes the fraction of his travel by foot. We are only given the exact distance travelled on foot which is 2 km. We have to find the fraction of the travel by foot to determine the length of the total distance travelled. Solving for x,

x = 1 - 3/8 - 1/4 - 1/8
x = 1/4

That means that the travel by foot comprises 1/4 of the whole journey. Thus,

Let total distance be D.

1.4*D = 2 km
D = 8 km

Therefore, the man travelled a total of 8 kilometers.

4 0

3 years ago