Find the mean proportion between 1/2 nd 1/8

A path 2.5m wide is running outside a square field whose side is 45m. Determine the area of the path.

aliya0001 [1]

Answer:

The area of the path would be 231.25 squared meters.

Step-by-step explanation:

Consider the path as 45m (the field's area), then add 2.5m to all of the sides. You'll get 47.5m on all sides. Then you do 47.5² to get 2256.25 squared meters. After that, you'll remove the area of the squared field. To do this, do 45m². By doing this, you'll get 2025 squared meters. Lastly, to finish up the question, do 2256.25 - 2025. This would get you 31.25 squared meters, the answer.

4 0

3 years ago

Find the area of a circle with radius equal to 6.<br><br> A =

anyanavicka [17]

Answer:

Step-by-step explanation:

hello :

A= πr² r = 6

so : A= 3.14(36) = 113.04

8 0

4 years ago

Health insurers are beginning to offer telemedicine services online that replace the common office visit. A company provides a v

Veronika [31]

Answer:

$\text {CI} = (60.54, \: 81.46)\\\\$

Therefore, we are 95% confident that actual mean savings for a televisit to the doctor is within the interval of ($60.54 to $81.46)

Step-by-step explanation:

Let us find out the mean savings for a televisit to the doctor from the given data.

Using Excel,

=AVERAGE(number1, number2,....)

The mean is found to be

$\bar{x} = \$71$

Let us find out the standard deviation of savings for a televisit to the doctor from the given data.

Using Excel,

=STDEV(number1, number2,....)

The standard deviation is found to be

$s = \$ 22.35$

The confidence interval is given by

$\text {confidence interval} = \bar{x} \pm MoE\\\\$

Where the margin of error is given by

$$ MoE = t_{\alpha/2}(\frac{s}{\sqrt{n} } ) $ \\\\$

Where n is the sample of 20 online doctor visits, s is the sample standard deviation and $t_{\alpha/2}$ is the t-score corresponding to a 95% confidence level.

The t-score is given by is

Significance level = α = 1 - 0.95 = 0.05/2 = 0.025

Degree of freedom = n - 1 = 20 - 1 = 19

From the t-table at α = 0.025 and DoF = 19

t-score = 2.093

So, the margin of error is

$MoE = t_{\alpha/2}(\frac{s}{\sqrt{n} } ) \\\\MoE = 2.093\cdot \frac{22.35}{\sqrt{20} } \\\\MoE = 2.093\cdot 4.997\\\\MoE = 10.46\\\\$

So the required 95% confidence interval is

$\text {CI} = \bar{x} \pm MoE\\\\\text {CI} = 71 \pm 10.46\\\\\text {CI} = 71 - 10.46, \: 71 + 10.46\\\\\text {CI} = (60.54, \: 81.46)\\\\$

Therefore, we are 95% confident that actual mean savings for a televisit to the doctor is within the interval of ($60.54 to $81.46)

7 0

3 years ago

Helppppp plzzzz ASAP!!!!!!<br> Thank you!!!!!!

Zielflug [23.3K]

Answer:

option 4.

16 square units

Step-by-step explanation:

as we do not have the measures of the sides, but if the points of the vertices with Pythagoras we can calculate the sides.

P = (2 , 4)

S = (4 , 2)

we have to subtract the values of p from s

PS = (4 - 2 , 2 - 4)

PS = (2 , -2)

by pitagoras h ^ 2 = c1 ^ 2 + c2 ^ 2

h: hypotenuse

c1: leg 1

c2: leg 2

PS^2 = 2^2 + -2^2

PS = √ 4 + 4

PS = √8

PS = 2√2

S = (4 , 2)

R = (8 , 6)

SR = (8-4 , 6-2)

SR = (4 , 4)

by pitagoras h ^ 2 = c1 ^ 2 + c2 ^ 2

h: hypotenuse

c1: leg 1

c2: leg 2

SR^2 = 4^2 + 4^2

SR = √ (16 + 16)

SR = √32

SR = 4√2

having the values of 2 of its sides we multiply them and obtain their area

PS * RS = Area

2√2 * 4√2 =

16

3 0

3 years ago

A random sample of 64 observations produced a mean value of 84 and standard deviation of 5.5. The 90% confidence interval for th

ASHA 777 [7]

Answer: The 90% confidence interval for the population mean μ is between 82.85 and 85.15,

Step-by-step explanation:

When population standard deviation is not given ,The confidence interval population proportion is given by ( $\mu$ ):-

$\overline{x}\pm t^*\dfrac{s}{\sqrt{n}}$

, where n= Sample size.

s= Sample standard deviation

$\overline{x}$ = sample mean

t* = Critical t-value (Two-tailed)

As per given , we have

$\overline{x}=84$

n= 64

Degree of freedom : df = n-1=63

s= 5.5

Significance level : $\alpha=1-0.90=0.1$

Two-tailed T-value for df = 63 and $\alpha=1-0.90=0.1$ would be

$t_{\alpha/2,df}=t_{0.05,63}=1.669$ (By t-distribution table)

i.e. t*= 1.669

The 90% confidence interval for the population mean μ would be

$84\pm (1.669)\dfrac{5.5}{\sqrt{64}}$

$=84\pm (1.669)\dfrac{5.5}{8}$

$\approx84\pm 1.15$

$=(84-1.15,\ 84+1.15)=(82.85,\ 85.15)$

∴ The 90% confidence interval for the population mean μ is between 82.85 and 85.15,

6 0

3 years ago