Find the exact value by using a half-angle identity. (6 points)

A casino riverboat travels up the Mississippi Rive at 12 mph and then back to its original port at a rate of 15 mph. The round t

o-na [289]

Speed up to Mississippi Rive = 12 mph

Speed back to original port = 15 mph

Total time taken by the riverboat = 7.5 hours

Total distance covered = ?

We know that the distance covered each side would be the same.

So Distance up to Mississippi Rive = Distance back to original port

⇒ ${\text} Speed up to Mississippi Rive * Time taken to reach Mississippi Rive = Speed back to the original port * Time back to the original port$ .......... (i)

Let Time taken to reach Mississippi Rive = x .............. (ii)

⇒ Time taken to reach back to original port = 7.5 - x .......... (iii)

Substituting (ii) and (iii) in (i)

⇒ ${\text} 12 * x = 15 * (7.5 - x)$

⇒ ${\text} 12x = 112.5 - 15x$

⇒ ${\text} 27x = 112.5$

⇒ ${\text} x = [tex]\frac{112.5}{27}$ [/tex]

Using value of x to determine distance traveled:

Distance Traveled up to Mississippi Rive = ${\text}Speed up to Mississippi Rive * Time taken to reach Mississippi Rive$

⇒ Distance Traveled up to Mississippi Rive = ${\text}12 * \frac{112.5}{27}$

⇒ Distance Traveled up to Mississippi Rive = 50 miles

Hence, the riverboat traveled total of 50 + 50 = 100 miles

7 0

3 years ago

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Consider a sampling distribution with p equals 0.15p=0.15 and samples of size n each. Using the appropriate formulas, find the

Grace [21]

Answer:

$a.\ \mu_p=750\ \ , \sigma_p=0.005\\\\b.\ \mu_p=150\ \ , \sigma_p=0.0113\\\\c.\ \mu_p=75\ \ , \sigma_p=0.0160$

Step-by-step explanation:

a. Given p=0.15.

-The mean of a sampling proportion of n=5000 is calculated as:

$\mu_p=np\\\\=0.15\times 5000\\\\=750$

-The standard deviation is calculated using the formula:

$\sigma_p=\sqrt{\frac{p(1-p)}{n}}\\\\=\sqrt{\frac{0.15(1-0.15)}{5000}}\\\\=0.0050$

Hence, the sample mean is μ=750 and standard deviation is σ=0.0050

b. Given that p=0.15 and n=1000

#The mean of a sampling proportion of n=1000 is calculated as:

$\mu_p=np\\\\=1000\times 0.15\\\\\\=150$

#-The standard deviation is calculated as follows:

$\sigma_p=\sqrt{\frac{p(1-p)}{n}}\\\\\\=\sqrt{\frac{0.15\times 0.85}{1000}}\\\\\\=0.0113$

Hence, the sample mean is μ=150 and standard deviation is σ=0.0113

c. For p=0.15 and n=500

#The mean is calculated as follows:

$\mu_p=np\\\\\\=0.15\times 500\\\\=75$

#The standard deviation of the sample proportion is calculated as:

$\sigma_p=\sqrt{\frac{p(1-p)}{n}}\\\\\\=\sqrt{\frac{0.15\times 0.85}{500}}\\\\\\=0.0160$

Hence, the sample mean is μ=75 and standard deviation is σ=0.0160

4 0

3 years ago

How to write 11001195 in words

avanturin [10]

Answer:

Eleven million one thousand one hundred ninety-five

6 0

3 years ago

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Pls pls pls pls help

artcher [175]

Answer:

see below

Step-by-step explanation:

A. Reflection across the y-axis replaces each x coordinate with its opposite.

B. Rotation 90° CW does the transformation (x, y) ⇒ (y, -x)

C. This is a translation left 2 units.

D. Rotation 90° CCW does the transformation (x, y) ⇒ (-y, x)

E. This is a translation left 7 and down 2.

__

In the attached graphs, we have identified a point that is a corresponding point on each figure. This is so you can see how the various transformations move it and the rest of the figure in relation to it. 90° arcs are shown so you can see the rotations more easily.

The figures are labeled and color coded in accordance with the problem statement.

8 0

3 years ago

F(x)=-9x^2-2x and g(x)=-3x^2+6x-9, find (f-g)(x) and (f-g)(-4)

Alona [7]

Answer:

(f - g)(x)= - 6 {x}^{2} - 8x + 9

(f - g)(-4!)= - 55

Step-by-step explanation:

$f(x) = - 9 {x}^{2} - 2x, \: \: g(x) = - 3 {x}^{2} + 6x - 9 \\ (f - g)(x) = f(x) - g(x) \\ = - 9 {x}^{2} - 2x - (- 3 {x}^{2} + 6x - 9) \\ = - 9 {x}^{2} - 2x + 3 {x}^{2} - 6x + 9 \\ \purple{ \boxed{ \bold{(f - g)(x)= - 6 {x}^{2} - 8x + 9}}} \\ (f - g)( - 4)= - 6 {( -4 )}^{2} - 8( - 4) + 9 \\ = - 6 \times 16 + 32 + 9 \\ = - 96 + 41 \\ \red{ \boxed{ \bold{(f - g)( - 4)= - 55}}}$

6 0

3 years ago