Name 3 planes that intersect at point O

he diameter of an ice skating rink is 44 feet. In terms of π, what is the area of the rink to the nearest square yard? (9 ft2 =

crimeas [40]

Answer:

The area of the circular ring is 53.73 π sq. yards

Step-by-step explanation:

The diameter of the skating ring = 44 ft

Diameter = 2 x Radius

⇒Radius of the ring = 44 / 2 = 22 ft

Now, 1 ft = 0.333333 yards

⇒22 ft = 22 x 0.333333 yards = 7.3333 yards

Now, Area of the circular Ring = $\pi r^2\\$

or, $A = \pi (7.33)^2$

= 53.73 π sq. yards

or, the area of the circular ring is 53.73 π sq. yards

6 0

2 years ago

PLEASE HELP ASAP 25 PTS + BRAINLIEST TO RIGHT/BEST ANSWER

LUCKY_DIMON [66]

Answer: d

Step-by-step explanation: when you bring 3 into the square root, it becomes isqrt(63). when you apply i, it makes the number sqrt(-63)

pretty sure I did it right this time

7 0

3 years ago

Read 2 more answers

Given the graph below, write the piecewise function.<br><br> How do I solve?

MaRussiya [10]

Answer:

for x < 0 x^2 - 2

for x $\geq$ 0 2x + 1

Step-by-step explanation:

4 0

2 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

2 years ago

C

ElenaW [278]

Hold on let me go do my research I’ll brb .

4 0

3 years ago