Ifm AGD = 72°, then what is m DGB? G/1 B 3/2 E 1 6 Answer for: m DGB

If f(x) = x2 + 2x - 4 and g(x) = 3x + 1 Find f(g(x)) = [? ]x² +[? ]x+ [?]

zubka84 [21]

Your anwser is quit school ):73

6 0

3 years ago

Uhh I got this answer wrong my brain wasn’t working so I guessed and picked wet and salty

Dahasolnce [82]

Answer:

dry and hot

Step-by-step explanation:

It looks very similar to a cactus which also lives in this in a dry and hot area

4 0

3 years ago

Read 2 more answers

-3 +3(n+8)= 2(1 + 6n) - 8

lilavasa [31]

Answer:

3 =n

Step-by-step explanation:

-3 +3(n+8)= 2(1 + 6n) - 8

Distribute

-3 +3n +24 = 2 +12n -8

Combine like terms

21+3n = 12n -6

Subtract 3n from each side

21+3n-3n = 12n-3n -6

21 = 9n -6

Add 6 to each side

21+6 = 9n-6+6

27 = 9n

Divide each side by 9

27/9 = 9n/9

27/9 =n

Divide the top and bottom by 3

3 =n

4 0

3 years ago

Read 2 more answers

The length of a rectangle is 5 inches more than it's width, x. The area of a rectangle can be represented by the equation.x^2+5x

Digiron [165]

Answer:

the width=15; the length=20

Step-by-step explanation:

1. to solve the given equation: x²+5x-300=0;

x₁= -20; x₂=15, where 'x' can be only positive value. It means x=15;

2. according to the condtition the length is more, it means 15+5=20.

3. the width is x=15; the length is 20.

8 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