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4 years ago

11

Your friend brings 15 chocolate cupcakes and 15 vanilla cupcakes to school. Students will take turns picking a pair of cupcakes

at random. What is the probability that the first student will pick 2 chocolate cupcakes?

2 answers:

Zolol [24]4 years ago

7 0

Probability of 1st chocolate = 15/30 = 1/2
Probability of 2nd after picking chocolate = 14/29

Multiply the 2 probabilities together
1/2 * 14/29 = 14/58 = 7/29

Gekata [30.6K]4 years ago

6 0

I believe it is 1/3, but probability is not my strong suit.

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(3x + 8y)(3x - 8y)

Step-by-step explanation:

this is the difference of 2 perfect squares.

9x2 - 64y2

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4 years ago

These are Questions/answers for #1, 2, 3, 17, 18, 19 & 20 if you can't see what it says in the picture (Please answer all qu

kkurt [141]

Answer:

See below for answers and explanations (along with a graph for #3)

Step-by-step explanation:

<u>Problem #1</u>

Applying scalar multiplication, $-4w=-4\langle-96,-180\rangle=\langle384,720\rangle$ .

Its magnitude would be $||-4w||=\sqrt{384^2+720^2}=816$ .

Its direction would be $\displaystyle\theta=tan^{-1}\biggr(\frac{720}{384}\biggr)\approx61.927^\circ\approx62^\circ$ .

Thus, B) 816; 62° is the correct answer

<u>Problem #2</u>

Find the time it takes for the ball to cover 13ft:

$x=(24\cos48^\circ)t\\13=(24\cos48^\circ)t\\t\approx0.8095$

Find the height of the ball at the time it takes for the ball to cover 13ft:

$y=6.1+(24\sin48^\circ)t-16t^2\\y=6.1+(24\sin48^\circ)(0.8095)-16(0.8095)^2\\y\approx10.053$

Thus, A) 10.053 is the correct answer

<u>Problem #3</u>

We have $u=\langle0,-8\rangle$ and $v=\langle6,0\rangle$ as our vectors. Thus, $u+v=\langle0+6,-8+0\rangle=\langle6,-8\rangle$ . Attached below is the correct graph. You can also solve the problem visually by using the parallelogram method where the resultant vector is the diagonal of the parallelogram.

<u>Problem 4 (#7)</u>

<u /> $\displaystyle t \cdot v=(7)(-10)+(-3)(-8)=-70+24=-46$

Thus, C) -46 is the correct answer

<u>Problem 5 (#8)</u>

Find $r$ and $\theta$ :

$r=\sqrt{x^2+y^2}=\sqrt{2^2+(-8)^2}=\sqrt{4+64}=\sqrt{68}=2\sqrt{17}\approx8.246$

$\displaystyle\theta=tan^{-1}\biggr(\frac{y}{x}\biggr)=tan^{-1}\biggr(\frac{-8}{2}\biggr)\approx-75.964^\circ$

Find the true direction angle accounting for Quadrant IV:

$\theta=360^\circ-75.964^\circ=284.036^\circ$

Write the complex number in polar/trigonometric form:

$z=8.246(\cos284.036^\circ+i\sin284.036^\circ)$

Thus, C) 8.246(cos 284.036° + i sin 284.036°) is the correct answer

<u>Problem 6 (#12)</u>

Eliminate the parameter and find the rectangular equation:

$x=3t\\\frac{x}{3}=t\\ \\y=t^2+5\\y=(\frac{x}{3})^2+5\\y=\frac{x^2}{9}+5\\9y=x^2+45\\0=x^2-9y+45\\x^2-9y+45=0$

Thus, D) x^2-9y+45=0 is the correct answer

<u>Problem 7 (#13)</u>

Find the magnitude of the vector:

$||v||=\sqrt{(-77)^2+36^2}=85$

Find the true direction of the vector accounting for Quadrant II:

$\displaystyle \theta=tan^{-1}\biggr(\frac{36}{-77}\biggr)\approx-25^\circ=180^\circ-25^\circ=155^\circ$

Write the vector in trigonometric form:

$w=85\cos155^\circ i+85\sin155^\circ j$

Thus, D) w=85cos155°i+85sin155°j is the corrwect answer

<u>Problem 8 (#15)</u>

$\frac{z_1+z_2}{2}=\frac{(3-7i)+(-9-19i)}{2}=\frac{-6-26i}{2}=-3-13i=(-3,-13)$

Thus, C) (-3,-13) is the correct answer

<u>Problem 9 (#16)</u>

Treat the golf ball and wind as vectors:

$u=\langle1.3\cos140^\circ,1.3\sin140^\circ\rangle$ <-- Golf Ball

$v=\langle1.2\cos50^\circ,1.2\sin50^\circ\rangle$ <-- Wind

Add the vectors:

$u+v=\langle1.3\cos140^\circ+1.2\cos50^\circ,1.3\sin140^\circ+1.2\sin50^\circ\rangle\approx\langle-0.225,1.755\rangle$

Find the magnitude of the resultant vector:

$||u+v||=\sqrt{(-0.225)^2+1.755^2}\approx1.769$

Find the true direction of the resultant vector accounting for Quadrant II:

$\displaystyle \theta=\tan^{-1}\biggr(\frac{1.755}{-0.225}\biggr)\approx-82.694^\circ\approx-83^\circ=180^\circ-83^\circ=97^\circ$

Thus, B) 1.769 m/s; 97° is the correct answer

<u>Problem 10 (#17)</u>

Identify the vectors and add them:

$u+v+w=\langle50\cos20^\circ,50\sin20^\circ\rangle+\langle13\cos90^\circ,13\sin90^\circ\rangle+\langle35\cos280^\circ,35\sin280^\circ\rangle=\langle50\cos20^\circ+13\cos90^\circ+35\cos280^\circ,50\sin20^\circ+13\sin90^\circ+35\sin280^\circ\rangle=\langle53.062,-4.367\rangle$

Find the magnitude of the resultant vector:

$||u+v+w||=\sqrt{53.062^2+(-4.367)^2}\approx53.241$

Find the true direction of the resultant vector accounting for Quadrant IV:

$\displaystyle \theta=\tan^{-1}\biggr(\frac{-4.367}{53.062}\biggr)\approx-4.705^\circ\approx-5^\circ=360^\circ-5^\circ=355^\circ$

Thus, A) 53.241, 355° is the correct answer

<u>Problem 11 (#18)</u>

We observe that $z_1=-8-6i$ and $z_2=4-4i$ , hence, $z_1+z_2=(-8-6i)+(4-4i)=-4-10i$

Thus, Q is the correct answer

<u>Problem 12 (#19)</u>

Find the dot product of the vectors:

$F_1\cdot F_2=(12000*14500)+(7000*-5000)=174000000+(-35000000)=139000000$

Find the magnitude of each vector:

$||F_1||=\sqrt{12000^2+7000^2}=1000\sqrt{193}\\||F_2||=\sqrt{14500^2+(-5000)^2}=500\sqrt{941}$

Find the angle between the two vectors:

$\displaystyle \theta=\cos^{-1}\biggr(\frac{F_1\cdot F_2}{||F_1||||F_2||}\biggr)\\ \theta=\cos^{-1}\biggr(\frac{139000000}{(1000\sqrt{193})(500\sqrt{941})}\biggr)\\\theta\approx49.282^\circ\approx49^\circ$

Thus, C) 49° is the correct answer

<u>Problem 13 (#20)</u>

Using scalar multiplication, $7v-2w=7\langle8,-3\rangle-2\langle-12,4\rangle=\langle56,-21\rangle-\langle-24,8\rangle=\langle56-(-24),-21-8\rangle=\langle80,-29\rangle=80i-29j$

Thus, A) -80i - 29j is the correct answer

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2 years ago

A parallelogram is shown below:

Oliga [24]

Answer:

area of whole triangle is 3

area of triangle cut in half is 1.5

Step-by-step explanation:

Area: (.5)(8)(3/4)=3

Area of each triangle. It would be split down the middle. from a to c

3/2=1.5

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3 years ago