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Studentka2010 [4]
2 years ago
12

A,B and C are the vertices of a triangle,

Mathematics
2 answers:
Harlamova29_29 [7]2 years ago
3 0
SSS (side, side, side)
irinina [24]2 years ago
3 0

Answer:

SSS (side, side, side)

hope it helps:))!!!

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Answer my question please help
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3 years ago
A triangle has an area of 72 square inches. If the base of the triangle has a length of 18 inches, what is the height of the tri
Alecsey [184]
It could be 8 since 8x18= 144 and 144 divided by 2 is 72
8 0
3 years ago
Ricardo bought 6 pounds of peanuts at the local wholesaler for $12. His friend Saniya bought 4 pounds of peanuts at the supermar
guapka [62]

Answer:

ricardo

he payed 2 per pound while she payed 2.50

7 0
3 years ago
Read 2 more answers
Find the circumference of the circle shown below if the diameter equals 20 inches. Use 3.14 for π
Tanzania [10]

Answer:

62.8 inches

Step-by-step explanation:

c = πd

c = 3.14 * 20

c = 62.8 inches

4 0
3 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
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