Brian ate 1/5 of a pie. Then Colin ate 1/5 of the pie, and David ate 1/5 of the pie. How much of the pie was left?

Sam sees a rocket traveling straight up at an angle of elevation of 24 degrees. If Sam is located 5 miles from the rocket Launch

julsineya [31]

Given :

Angle of elevation, Ф = 24°.

Distance of Sam from launchpad, D = 5 miles.

To Find :

How high is the rocket.

Solution :

Let, height is h.

We know, tan Ф is given by :

$tan \ \phi=\dfrac{h}{d}\\\\tan \ 24^{\circ}=\dfrac{h}{5}\\\\h = 5\times tan \ 24^{\circ}\\\\h = 2.226 \ miles$

Hence, this is the required solution.

7 0

3 years ago

Please help!!

Lelu [443]

I answer for one shoves is 11pound and two is 22pound sorry if it is wrong

8 0

3 years ago

On a test, the marks for ten students are: 2 students, 85%; 4 students, 78%; 3

anzhelika [568]

Answer:

77.2%

Step-by-step explanation:

The average formula is:

$Average = \frac{Sum/of/all/numbers}{How/many/numbers/there/are}$

Let's first find the sum of all the numbers (remember sum means to add):

(2)(85) + (4)(78) + (3)(66) + 92 = ?

170 + 312 + 198 + 92 = 772

There are marks for 10 students so that is what goes at the bottom (10 students, 10 test scores)

Plug both things into the equation:

$Average = \frac{772}{10} \\= 77.2$

Therefore, the average mark on the test was 77.2%

<em>Hope this helps!!</em>

6 0

3 years ago

In an analysis of healthcare data, ages have been rounded to the nearest multiple of 5 years. The difference between the true ag

Mashcka [7]

Answer:

The correct answer is D. 0.77

Step-by-step explanation:

$X_i \text{ is the difference between true and reported age. where i = 1,2,3......48 }\\\text{The rounded age are uniformly distributed. So, }X_i = 0\\\\and,\thinspace O_{X_i}^2=\frac{\text{(Upper Bound - Lower Bound})^2}{5\times 2.5\text{ ( Rounded value )}}\\\\O_{X_i}^2=\frac{(2.5 + 2.5)^2}{12}\approx 2.083\\\\O_X_i=1.443$

$\bar{X_{48}}=\frac{X_1+X_2+.......+X_{48}}{48}\\\\\bar{X_{48}}=\frac{\frac{48}{X_i}}{48}=0\\\\O_{\bar{X}_{48}}^2=\frac{48\cdot O_{X_i}^2}{48^2}=\frac{O_{X_i}^2}{48}\\\\\text{So, }\bar{X_i}\text{is approximately normal with mean 0 and standard deviation = }\\\\\frac{1.443}{\sqrt{48}}\approx 0.2083$

$=Pr[\frac{-1}{4}\leq \bar{X}_{48}\leq \frac{1}{4}]\\\\=Pr[\frac{-0.25}{0.2083}\leq Z\leq \frac{0.25}{0.2083}]\\\\=Pr[-1.2\leq Z\leq 1.2]\\=Pr(Z\leq 1.2)-Pr(Z\leq -1.2)\\=Pr(Z\leq 1.2)-(1-Pr(Z\leq 1.2))\\=2\times 0.8849 -1\text{ ( Z value for 1.2 is 0.8849 )}\\\approx 0.77$

Hence, the correct answer is 0.77

7 0

3 years ago

Do the ratios 4ft./6ft. and 12sec/18sec form a proportion? Why or why not?

Ilya [14]

Answer:

Yes! There are 3 different explanations I have. Pick your favorite.

Step-by-step explanation:

So we are asked to see if the following equation holds:

$\frac{4 \text{ ft }}{6 \text{ ft}}=\frac{12 \text{ sec }}{18 \text{sec}}$

The units cancel out ft/ft=1 and sec/sec=1.

So we are really just trying to see if 4/6 is equal to 12/18

Or you could cross multiply and see if the products are the same on both sides:

$\frac{4}{6}=\frac{12}{18}$

Cross multiply:

$4(18)=6(12)$

$72=72$

Since you have the same thing on both sides then the ratios given were proportional.

OR!

Put 4/6 and 12/18 in your calculator. They both come out to have the same decimal expansion of .66666666666666666(repeating) so the ratios gives are proportional.

OR!

Reduce 12/18 and reduce 4/6 and see if the reduced fractions are same.

12/18=2/3 (I divided top and bottom by 6)

4/6=2/3 (I divided top and bottom by 2)

They are equal to the same reduced fraction so they are proportional.

8 0

3 years ago