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aliya0001 [1]
3 years ago
5

Let f(x) = x – 3 and g(x) = x + 11. Find f(x) • g(x).. . A] x2 – 8x – 33. . B] x2 – 8x + 14. . C] x2 + 8x + 14. . D] x2 + 8x – 3

3
Mathematics
2 answers:
GenaCL600 [577]3 years ago
7 0
F(x) • g(x) 
<span>= ( x - 3 )( x + 11 ) </span>
<span>= x² + 8x - 33 </span>
<span>Answer : D </span>
Katarina [22]3 years ago
3 0
F(x) = x-3 and g(x) = x+11 

f(x) *g(x) = (x-3)(x+11) = x^2 +11x -3x -33 = x^2 +8x -33 

so choice D. is right 

hope helped 
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So, we'll say Mia gets $4.50 a day, since a standard weekend is two days, therefore adding up to $9 a weekend. So if Mia were to do her chores for 4 weekends, she'd have a total of $36 but not $40. So if she were to work an extra weekend, so 5 weekends, she'd have a total of $45. So you can say 5 weekends in order to earn more than $40.
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2 years ago
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mina [271]

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b

Step-by-step explanation:

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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
Please help I keep getting this Wrong
Airida [17]

Answer:

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4 0
3 years ago
The angle of depression from a helicopter to a landing pad is 37 degrees. if the helicopter is 1250 feet from the ground. what i
djyliett [7]

The horizontal distance from the helicopter to the landing pad is 1658.81 feet

<em><u>Solution:</u></em>

The figure is attached below

Triangle ABC is a rightangled triangle

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BC is the horizontal distance from the helicopter to the landing pad

BC = ?

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tan \theta = \frac{opposite}{adjacent}

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Thus the horizontal distance from the helicopter to the landing pad is 1658.81 feet

5 0
3 years ago
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