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suter [353]
3 years ago
7

a rectangular room is 2 times as long as it is wide and the perimeter is 24 meters find the dimension of the room

Mathematics
1 answer:
GuDViN [60]3 years ago
3 0
24= 2l+2w
l=2w

Substitute for l
24= 2(2w)+2w
24= 6w
4=w

Plug in the w value
l=2(4)
l=8

Final answer: Length= 8 m, Width= 4 m
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Write in miles per hour. Round to the nearest tenth. 211 ft/s
IgorC [24]

Answer:

143.864

Step-by-step explanation:

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3 years ago
Y<br> 47<br> 27<br> 2+<br> -2 -2<br> 1<br> 2<br> What is the slope of the line
Kryger [21]

Answer:

answer =3/4

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3 0
3 years ago
A Survey of 85 company employees shows that the mean length of the Christmas vacation was 4.5 days, with a standard deviation of
GenaCL600 [577]

Answer:

The 95% confidence interval for the population's mean length of vacation, in days, is (4.24, 4.76).

The 92% confidence interval for the population's mean length of vacation, in days, is (4.27, 4.73).

Step-by-step explanation:

We have the standard deviations for the sample, which means that the t-distribution is used to solve this question.

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 85 - 1 = 84

95% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 84 degrees of freedom(y-axis) and a confidence level of 1 - \frac{1 - 0.95}{2} = 0.975. So we have T = 1.989.

The margin of error is:

M = T\frac{s}{\sqrt{n}} = 1.989\frac{1.2}{\sqrt{85}} = 0.26

In which s is the standard deviation of the sample and n is the size of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 4.5 - 0.26 = 4.24 days

The upper end of the interval is the sample mean added to M. So it is 4.5 + 0.26 = 4.76 days

The 95% confidence interval for the population's mean length of vacation, in days, is (4.24, 4.76).

92% confidence interval:

Following the sample logic, the critical value is 1.772. So

M = T\frac{s}{\sqrt{n}} = 1.772\frac{1.2}{\sqrt{85}} = 0.23

The lower end of the interval is the sample mean subtracted by M. So it is 4.5 - 0.23 = 4.27 days

The upper end of the interval is the sample mean added to M. So it is 4.5 + 0.23 = 4.73 days

The 92% confidence interval for the population's mean length of vacation, in days, is (4.27, 4.73).

8 0
3 years ago
BRAINLIEST ASAP! PLEASE HELP ME :)
Afina-wow [57]

Answer:

\large \boxed{\textbf{30.54 yr}}

Step-by-step explanation:

For species A,

P =2000e^{0.05t}

For species B,

P =5000e^{0.02t}

When their populations are equal,

\begin{array}{rcl}2000e^{0.05t} & = & 5000e^{0.02t}\\e^{0.05t} & = & 2.5e^{0.02t}\\0.05t & = & \ln2.5 + 0.02t\\0.05t & = & 0.9163 + 0.02t\\0.03t & = & 0.9163\\t& = & \dfrac{0.9163}{0.03}\\\\& = & \mathbf{30.54}\\\end{array}\\\text{The populations of species A and B will be the same in $\large \boxed{\textbf{30.54 yr}}$}

The diagram below shows the population of Species A overtaking that of Species B in about 30.5 yr.

3 0
3 years ago
What are the opposites of these numbers
MakcuM [25]
Would it be
1) -4.25
2) -5 1/4
3) -1/2
4) -2 1/3
5) 3.85
6) 6.1
8 0
3 years ago
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