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Effectus [21]
3 years ago
10

Assume that 1200 births are randomly selected and exactly 595 of the births are girls. Use subjective judgment to determine whet

her the given outcome is unlikely, and also determine whether it is unusual in the sense that the result is far from what is typically expected. Determine whether exactly 595 girls out of 1200 randomly selected births is unlikely.
A. It is not unlikely because girls are slightly less common than boys.
B- t is unlikely because it is not 600 as expected.
C. It is unlikely because the probability of this particular outcome is very small, considering all of the other possible outcomes.
D. It is not unlikely because 595 is about half of 1200.

Determine whether exactly 595 girls out of 1200 randomly selected births is unusual.
A. It is unusual because girls are more common than boys.
B. It is unusual because it is not 600 as expected.
C. It is not unusual because 595 girls out of 1200 births is the only possible outcome.
D. It is not unusual because 595 is about the number of girls expected.
Mathematics
1 answer:
emmainna [20.7K]3 years ago
6 0

Answer:

Lets assume that 1200 births are randomly selected and exactly 595 of the births are girls.

In percentage this is : \frac{595}{1200}\times100=49.58\%

This is approx 50%. We can say that almost random 50% are girls.

So, It is unlikely because the probability of this particular outcome is very​ small, considering all of the other possible outcomes.

Part B: It is not unusual because 595 is about the number of girls expected.

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Ratling [72]
Point? I'm pretty sure or the origin point
5 0
3 years ago
The velocity of an automobile starting from rest is given by the equation below, where v is measured in feet per second and t is
maria [59]

Answer:

a. At t = 5 s

a(5)=\frac{1785}{\left(6(5)+17\right)^2}=\frac{1785}{2209}\approx0.808 \frac{ft}{s^2}

b. At t = 10 s

a(10)=\frac{1785}{\left(6(10)+17\right)^2}=\frac{255}{847}\approx0.301 \frac{ft}{s^2}

c. At t = 20 s

a(20)=\frac{1785}{\left(6(20)+17\right)^2}=\frac{1785}{18769}\approx0.095 \frac{ft}{s^2}

Step-by-step explanation:

We know that the velocity function is given by

                                                   v(t)=\frac{105t}{6t+17}

Acceleration is the rate of change of velocity so we take the derivative of the velocity function with respect to time.

a(t)=\frac{dv}{dt}=\frac{d}{dt} (\frac{105t}{6t+17})

\mathrm{Take\:the\:constant\:out}:\quad \left(a\cdot f\right)'=a\cdot f\:'\\\\105\frac{d}{dt}\left(\frac{t}{6t+17}\right)\\\\\mathrm{Apply\:the\:Quotient\:Rule}:\quad \frac{d}{{dx}}\left( {\frac{{f\left( x \right)}}{{g\left( x \right)}}} \right) = \frac{{\frac{d}{{dx}}f\left( x \right)g\left( x \right) - f\left( x \right)\frac{d}{{dx}}g\left( x \right)}}{{g^2 \left( x \right)}}

105\cdot \frac{\frac{d}{dt}\left(t\right)\left(6t+17\right)-\frac{d}{dt}\left(6t+17\right)t}{\left(6t+17\right)^2}\\\\105\cdot \frac{1\cdot \left(6t+17\right)-6t}{\left(6t+17\right)^2}\\\\a(t)=\frac{1785}{\left(6t+17\right)^2}

a. At t = 5 s

a(5)=\frac{1785}{\left(6(5)+17\right)^2}=\frac{1785}{2209}\approx0.808 \frac{ft}{s^2}

b. At t = 10 s

a(10)=\frac{1785}{\left(6(10)+17\right)^2}=\frac{255}{847}\approx0.301 \frac{ft}{s^2}

c. At t = 20 s

a(20)=\frac{1785}{\left(6(20)+17\right)^2}=\frac{1785}{18769}\approx0.095 \frac{ft}{s^2}

3 0
4 years ago
PLEASE HELP ASAPPPPP THANK YOU
Arada [10]

Answer:

i wish i could help u!!!!!!

Step-by-step explanation:

uuuhhhhh i hate that i kinda suck at math

anyway, i'll give this my best shot

The midsegment of a triangle is a line segment that connects the midpoints of two sides ofrigidity to structures. In the support for the image shown shown, the crossbar GJK is a midsegment of GFK

You can use a compass and straightedge to construct the midsegments of a triangle.

Hope this helps! It's the best i can do!

8 0
3 years ago
Use Newton's method to approximate a root of the equation 3x^(7) + 2x^(4) + 2 = 0 as follows. Let x1 = 3 be the initial approxim
Maurinko [17]

Answer:

x_{2}\approx 2.59, x_{4}\approx 4.3

Step-by-step explanation:

1) Given that the formula for the Newton Method is:

x_{n+1}=x_{n}-\frac{f(x_{n})}{f(x_{n+1})}

2) The Function and its 1st derivative

f(x)=3x^{7}+2x^{4}+2\\f'(x)=21x^{6}+8x^{3}

3) Let's work with the first approximation. Plugging in in the function:

f(x)=3x^{7}+2x^{4}+2\Rightarrow f(3)=3(3)^{7}+2(3)^{3}+2\Rightarrow f(3)=6617\\f'(x)=21x^{6}+8x^{3}\Rightarrow f'(3)=21(3)^{6}+8(3)^{4}\Rightarrow f'(3)=15957\\\\x_{n+1}=x_{n}-\frac{f(x_{n})}{f'(x_{n})}\Rightarrow x_{2}=x_{1}-\frac{f(x_{1})}{f'(x_{1})}\\x_{2}=3-\frac{6617}{15957}\Rightarrow x_{2}\approx 2.59\\\\

4) To approximate the other root x

f(x)=3x^{7}+2x^{4}+2\Rightarrow f(5)=3(5)^{7}+2(5)^{3}+2\Rightarrow f(5)=234627\\f'(x)=21x^{6}+8x^{3}\Rightarrow f'(5)=21(5)^{6}+8(5)^{4}\Rightarrow f'(5)=333125\\\\x_{n+1}=x_{n}-\frac{f(x_{n})}{f'(x_{n})}\Rightarrow x_{4}=x_{3}-\frac{f(x_{1})}{f'(x_{1})}\\x_{4}=5-\frac{234627}{333125}\Rightarrow x_{4}\approx 4.3

5 0
3 years ago
A 5 cm×11 cm\greenD{5\,\text{cm} \times 11\,\text{cm}}5cm×11cmstart color #1fab54, 5, start text, c, m, end text, times, 11, sta
sashaice [31]

Answer:

42.5 square.cm

Step-by-step explanation:

Find the diagram attached

Area of the shaded portion = area of square - area of circle

Area of square = L²

Area of square = 11 × 11

Area of square = 121 square centimetres

Area of circle = pir²

r is the radius of the circle = 5cm

Area of the circle = 3.14×5²

Area of the circle = 3.14×25

Area of the circle = 78.5 square. cm

Area of the shaded region = 121 - 78.5

Area of the shaded region = 42.5 square.cm

8 0
3 years ago
Read 2 more answers
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