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3 years ago

Please help ASAP!

Mathematics

1 answer:

Misha Larkins [42]3 years ago

6 0

The probability of pulling a boys name first is 8/13. Then the probably of pulling a girls second is 2/5.

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A submarine dives at a constant speed of 10 m/s at a diving angle measured from the vertical

valina [46]

Answer:

155.3 m

Step-by-step explanation:

We have:

θ: is the angle measured from the vertical = 75 °

v = is the speed of the submarine = 10 m/s

t = 1 min

The deep of the front end of the submarine at the end of a 1-minute drive is given by the following ratio:

$cos (\theta) = \frac{x}{d}$

Where:

d: is the distance traveled by the submarine in 1 minute

x: is the deep to find

The distance, d, is:

$d = v*t = 10 \frac{m}{s}* 1 min * \frac{60 s}{1 min} = 600 m$

Now, the deep is:

$cos (75) = \frac{x}{600}$

$x = cos(75)*600 = 155.3 m$

Therefore, the deep of the front end of the submarine at the end of a 1-minute drive is 155.3 m.

I hope it helps you!

3 0

3 years ago

Please help!! I need ASAP 10 points <3

Readme [11.4K]

5m+8 that’s it I’m pretty sure

5 0

3 years ago

In order to estimate the difference between the average hourly wages of employees of two branches of a department store, the fol

Julli [10]

Answer:

$0.071,1.928$

Step-by-step explanation:

Downtown Store North Mall Store

Sample size n 25 20

Sample mean $\bar{x}$ $9 $8

Sample standard deviation s $2 $1

$n_1=25\\n_2=20$

$\bar{x_1}=9\\ \bar{x_2}=8$

$s_1=2\\s_2=1$

$x_1-x_2=9-8=1$

Standard error of difference of means = $\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}$

Standard error of difference of means = $\sqrt{\frac{2^2}{25}+\frac{1^2}{20}}$

Standard error of difference of means = $0.458$

Degree of freedom = $\frac{\sqrt{(\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}})^2}{\frac{(\frac{s_1^2}{n_1})^2}{n_1-1}+\frac{(\frac{s_2^2}{n_2})^2}{n_2-1}}$

Degree of freedom = $\frac{\sqrt{(\frac{2^2}{25}+\frac{1^2}{20}})^2}{\frac{(\frac{2^2}{25})^2}{25-1}+\frac{(\frac{1^2}{20})^2}{20-1}}$

Degree of freedom =36

So, z value at 95% confidence interval and 36 degree of freedom = 2.0280

Confidence interval = $(x_1-x_2)-z \times SE(x_1-x_2),(x_1-x_2)+z \times SE(x_1-x_2)$

Confidence interval = $1-(2.0280)\times 0.458,1+(2.0280)\times 0.458$

Confidence interval = $0.071,1.928$

Hence Option A is true

Confidence interval is $0.071,1.928$

4 0

2 years ago

Help with math ASAP!!!

melomori [17]

Answer:

its b

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

wo point charges lie on the $x$ axis. A charge of +6.24 $\mu C$ is at the origin, and a charge of -9.55 $\mu C$ is at $x$ = 12.0

Andreyy89

Answer:

$E_n=34,467,075.42\ N/C$

Step-by-step explanation:

Electric Field

The electric field produced by a point charge Q at a distance d is given by

$\displaystyle E=K\cdot \frac{Q}{d^2}$

Where

$K = 9\cdot 10^9\ Nw.m^2/c^2$

The net electric field is the vector addition of the individual electric fields produced by each charge. The direction is given by the rule: If the charge is positive, the electric field points outward, if negative, it points inward.

Let's calculate the electric fields of each charge at the given point. The first charge $q_1=+6.24\mu C=6.24\cdot 10^{-6}C$ is at the origin. We'll calculate its electric field at the point x=-3.85 cm. The distance between the charge and the point is d=3.85 cm = 0.0385 m, and the electric field points to the left:

$\displaystyle E_1=9\cdot 10^9\cdot \frac{6.24\cdot 10^{-6}}{0.0385^2}$

$E_1=37,888,345.42\ N/C$

Similarly, for $q_2=-9.55\mu C=-9.55\cdot 10^{-6}C$ , the distance to the point is 12 cm + 3.85 cm = 15.85 cm = 0.1585 m. The electric field points to the right:

$\displaystyle E_2=9\cdot 10^9\cdot \frac{9.55\cdot 10^{-6}}{0.1585^2}$

$E_2=3,421,270\ N/C$

Since E1 and E2 are opposite, the net field is the subtraction of both

$E_n=37,888,345.42\ N/C-3,421,270\ N/C$

$\boxed{E_n=34,467,075.42\ N/C}$

6 0

3 years ago