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fenix001 [56]
3 years ago
12

special deck of cards has 20 cards. Nine are green, seven are blue, and four are red. When a card is picked, the color of it is

recorded. An experiment consists of first picking a card and then tossing a coin. A. How many elements are there in the sample space? B. Let A be the event that a red card is picked first, followed by landing a tail on the coin toss. P(A) = Present your answer as a decimal number to 1 decimal place. C. Let B be the event that a green or blue is picked, followed by landing a tail on the coin toss. Are the events A and B mutually exclusive? D. Let C be the event that a green or red is picked, followed by landing a tail on the coin toss. Are the events A and C mutually exclusive?
Mathematics
1 answer:
Sever21 [200]3 years ago
7 0

Answer:

Step-by-step explanation:

Given that special deck of cards has 20 cards. Nine are green, seven are blue, and four are red. When a card is picked, the color of it is recorded. An experiment consists of first picking a card and then tossing a coin

A) Sample space will have Green, Head,  or Green, Tail .... Red, head, red, tail

No of elements in sample space = no of colours x no of outcomes in coin toss

= 4x2 = 8

B) A= getting (RT)

P(A) = Prob of getting red card and tail on coin

= P (R) *P(T)

=\frac{4}{20} *\frac{1}{2} \\=\frac{1}{10}

C) B be the event that a green or blue is picked, followed by landing a tail on the coin toss

B = getting green card and tail

Getting green card tail is mutally exclusive with red card and tail as there is no common element between green and blue.

D) C= red or green card is picked followed by tail.

Here A and C have a common element as getting red and tail.  So not mutually exclusive

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The zeroes ( where the graph cuts the x axis) ar (-2,0) AND (2,0)

tHE FACTOrIAL FORM IS (x - 2)(x + 2)

Its B

3 0
3 years ago
What number is 60 percent of 29
nevsk [136]
60% is 0.60 so you do 0.60 x 29 which = to 17.4
so your answer is 17.40 or 17.4 is 60% of 29
7 0
3 years ago
We are standing on the top of a 320 foot tall building and launch a small object upward. The object's vertical altitude, measure
STALIN [3.7K]

Answer:

The highest altitude that the object reaches is 576 feet.

Step-by-step explanation:

The maximum altitude reached by the object can be found by using the first and second derivatives of the given function. (First and Second Derivative Tests). Let be h(t) = -16\cdot t^{2} + 128\cdot t + 320, the first and second derivatives are, respectively:

First Derivative

h'(t) = -32\cdot t +128

Second Derivative

h''(t) = -32

Then, the First and Second Derivative Test can be performed as follows. Let equalize the first derivative to zero and solve the resultant expression:

-32\cdot t +128 = 0

t = \frac{128}{32}\,s

t = 4\,s (Critical value)

The second derivative of the second-order polynomial presented above is a constant function and a negative number, which means that critical values leads to an absolute maximum, that is, the highest altitude reached by the object. Then, let is evaluate the function at the critical value:

h(4\,s) = -16\cdot (4\,s)^{2}+128\cdot (4\,s) +320

h(4\,s) = 576\,ft

The highest altitude that the object reaches is 576 feet.

6 0
3 years ago
Suppose you choose a team of two people from a group of n > 1 people, and your opponent does the same (your choices are allow
jonny [76]

Answer:

The number of possible choices of my team and the opponents team is

 \left\begin{array}{ccc}n-1\\E\\n=1\end{array}\right     i^{3}

Step-by-step explanation:

selecting the first team from n people we have \left(\begin{array}{ccc}n\\1\\\end{array}\right)  = n possibility and choosing second team from the rest of n-1 people we have \left(\begin{array}{ccc}n-1\\1\\\end{array}\right) = n-1

As { A, B} = {B , A}

Therefore, the total possibility is \frac{n(n-1)}{2}

Since our choices are allowed to overlap, the second team is \frac{n(n-1)}{2}

Possibility of choosing both teams will be

\frac{n(n-1)}{2}  *  \frac{n(n-1)}{2}  \\\\= [\frac{n(n-1)}{2}] ^{2}

We now have the formula

1³ + 2³ + ........... + n³ =[\frac{n(n+1)}{2}] ^{2}

1³ + 2³ + ............ + (n-1)³ = [x^{2} \frac{n(n-1)}{2}] ^{2}

=\left[\begin{array}{ccc}n-1\\E\\i=1\end{array}\right] =   [\frac{n(n-1)}{2}]^{3}

4 0
3 years ago
An urn contains 8 red chips, 10 green chips, and 2 white chips. A chip is drawn and replaced, and then a second chip is drawn.
Harlamova29_29 [7]

Answer:

(A) 0.04

(B) 0.25

(C) 0.40

Step-by-step explanation:

Let R = drawing a red chips, G = drawing green chips and W = drawing white chips.

Given:

R = 8, G = 10 and W = 2.

Total number of chips = 8 + 10 + 2 = 20

P(R) = \frac{8}{20}=\frac{2}{5}\\P(G)=  \frac{10}{20}=\frac{1}{2}\\P(W)=  \frac{2}{20}=\frac{1}{10}

As the chips are replaced after drawing the probability of selecting the second chip is independent of the probability of selecting the first chip.

(A)

Compute the probability of selecting a white chip on the first and a red on the second as follows:

P(1^{st}\ white\ chip, 2^{nd}\ red\ chip)=P(W)\times P(R)\\=\frac{1}{10}\times \frac{2}{5}\\ =\frac{1}{25} \\=0.04

Thus, the probability of selecting a white chip on the first and a red on the second is 0.04.

(B)

Compute the probability of selecting 2 green chips:

P(2\ Green\ chips)=P(G)\times P(G)\\=\frac{1}{2} \times\frac{1}{2}\\ =\frac{1}{4}\\ =0.25

Thus, the probability of selecting 2 green chips is 0.25.

(C)

Compute the conditional probability of selecting a red chip given the first chip drawn was white as follows:

P(2^{nd}\ red\ chip|1^{st}\ white\ chip)=\frac{P(2^{nd}\ red\ chip\ \cap 1^{st}\ white\ chip)}{P (1^{st}\ white\ chip)} \\=\frac{P(2^{nd}\ red\ chip)P(1^{st}\ white\ chip)}{P (1^{st}\ white\ chip)} \\= P(R)\\=\frac{2}{5}\\=0.40

Thus, the probability of selecting a red chip given the first chip drawn was white is 0.40.

6 0
3 years ago
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