Emma is building a scale model of a jet airplane. The real airplane has a wingspan of 200 feet and a length of 240 feet.

A force of 500.0 is represented graphically with its tail at the origin and the tip pointed in a direction 30.0° above the posit

trapecia [35]

Answer:

$F^{'}=(250\sqrt{3},250 })$

Step-by-step explanation:

We have F´ =500 and $\alpha$ =30º, so x and y components:

F´ = $(F_{x} , F_{y})$ this is

$F_{x} = F^{'} *Cos\alpha$

$F_{y} = F^{'} *Sin\alpha$

$F_{x} = F`*Cos\alpha =500*Cos30=500*\frac{\sqrt{3} }{2} =250\sqrt{3}$

$F_{y}=F*Sin\alpha =500*Sin30=500*\frac{1}{2}=250$ ;

Finally

F' = $(250\sqrt{3} , 250)$

3 0

2 years ago

Two angles are complementary. the measure of ∠abc is x° and the measure of ∠dbc is (3x + 10)°. what is the value of x? enter you

Marta_Voda [28]

This problem can be expressed through the equation:
x+3x+10= 90

To solve it we subtract 10 on both sides
→x+3x=80
Add x and 3x
→4x=80
And divide each side by 4
→x=80/4
→x=20

ANSWER:
20

8 0

2 years ago

robert plant 7 trees behind westwood school. he planted 6 times as many trees in front of the school. how many trees did he plan

Sati [7]

49 in total or 42 trees in the front

4 0

2 years ago

Read 2 more answers

A random sample of 35 undergraduate students who completed two years of college were asked to take a basic mathematics test. The

lisov135 [29]

Answer:

$(75.1 -72.1) -1.66 \sqrt{\frac{12.8^2}{35} +\frac{14.6^2}{50}}= -1.96$

$(75.1 -72.1) +1.66 \sqrt{\frac{12.8^2}{35} +\frac{14.6^2}{50}}= 7.96$

And the confidence interval would be given by:

$-1.96 \leq \mu_1 -\mu_2 \leq 7.96$

And since the confidence interval contains the value 0 we have enough evidence to conclude that we don't have significant differences between the two means at 10% of significance.

Step-by-step explanation:

For this case we have the following info given :

$\bar X_1= 75.1$ represent the sample mean for the scores of the undergraduate students

$s_1 = 12.8$ represent the standard deviation for the undergraduate students

$n_1 =35$ the sample size for the undergraduate

$\bar X_2= 72.1$ represent the sample mean for the scores of the high school students

$s_2 = 14.6$ represent the standard deviation for the high school students

$n_2 =50$ the sample size for the high school

The confidence interval for the true difference of means is given by:

$(\bar X_1 -\bar X_2) \pm t_{\alpha/2} \sqrt{\frac{s^2_1}{n_1} +\frac{s^2_2}{n_2}}$

The degrees of freedom are given by:

$df=n_1 +n_2 -2= 35+50-2=83$

The confidence level is 90% and the significance level is $\alpha=0.1$ and $\alpha/2 =0.05$ then the critical value would be:

$t_{\alpha/2}= 1.99$

And replacing the info we got:

$(75.1 -72.1) -1.66 \sqrt{\frac{12.8^2}{35} +\frac{14.6^2}{50}}= -1.96$

$(75.1 -72.1) +1.66 \sqrt{\frac{12.8^2}{35} +\frac{14.6^2}{50}}= 7.96$

And the confidence interval would be given by:

$-1.96 \leq \mu_1 -\mu_2 \leq 7.96$

And since the confidence interval contains the value 0 we have enough evidence to conclude that we don't have significant differences between the two means at 10% of significance.

8 0

1 year ago

If trapezoid JKLM with vertices) (3, 4), K (6,4), L (8, 1) and M (1,1) is rotated 270 degrees counterclockwise, what are the

Stels [109]

Answer:

$J' = (4,-3)$