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2 years ago

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If 2 cards are selected from a standard deck of 52 cards.The first card is placed back in the deck before the 2nd card is drawn.

What is the probability that a heart and a club will be drawn ?

1 answer:

Mkey [24]2 years ago

7 0

There are 13 cards of each suit.

Picking a heart would be 13/52 which reduces to 1/4

Then replacing the card and picking a club would be the same : 1/4

Picking a heart then a club would be 1/4 x 1/4 = 1/16

Answer: 1/16

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Consider the points A(5, 3t+2, 2), B(1, 3t, 2), and C(1, 4t, 3). Find the angle ∠ABC given that the dot product of the vectors B

Vilka [71]

Answer:

66.42°

Step-by-step explanation:

<u>Given:</u>

A(5, 3t+2, 2)

B(1, 3t, 2)

C(1, 4t, 3)

BA • BC = 4

Step 1: Find t.

First we have to find vectors BA and BC. We do that by subtracting the coordinates of the initial point from the coordinates of the terminal point.

In vector BA B is the initial point and A is the terminal point.

BA = OA - OB = (5-1, 3t+2-3t, 2-2) = (4, 2, 0)

BC = OC - OB = (1-1, 4t-3t, 3-2) = (0, t, 1)

Now we can find t because we know that BA • BC = 4

BA • BC = 4

To find dot product we calculate the sum of the produts of the corresponding components.

BA • BC = (4)(0) + (2)(t) + (0)(1)

4 = (4)(0) + (2)(t) + (0)(1)

4 = 0 + 2t + 0

4 = 2t

2 = t

t = 2

Now we know that:

BA = (4, 2, 0)

BC = (0, 2, 1)

Step 2: Find the angle ∠ABC.

Dot product: a • b = |a| |b| cos(angle)

BA • BC = 4

|BA| |BC| cos(angle) = 4

To get magnitudes we square each compoment of the vector and sum them together. Then square root.

$|BA| = \sqrt{4^2 + 2^2 + 0^2} = \sqrt{20} = 2\sqrt{5}$

$|BC| = \sqrt{0^2 + 2^2 + 1^2} = \sqrt{5}$

$2\sqrt{5}\sqrt{5}\cos{(m\angle{ABC})} = 4$

$10\cos{(m\angle{ABC})} = 4$

$\cos(m\angle{ABC}) = \frac{4}{10}=\frac{2}{5}$

$m\angle{ABC} = cos^{-1}{(\frac{2}{5})}$

$m\angle{ABC} = 66.4218^{\circ}$

Rounded to two decimal places:

$m\angle{ABC} = 66.42^\circ$

3 0

1 year ago

"find the cost per ounce of a sunscreen made from 140 oz of lotion that cost $3.45 per ounce and 90 oz of lotion that cost $14.0

Olegator [25]

Sunscreen approx. 40.6 cents an oz
Lotion 6.4 cents an oz

5 0

3 years ago

Do you guys know which coordinates im supposed to graph on here?

nadezda [96]

Answer:

SOLVE FOR Y

graph y= 2x + 2

sp graph the points

(0,2) and then (1,4)

connect the points and extend the line.

Step-by-step explanation:

add 2x to both sides

-2x+y=2

+2x +2x

y= 2+2x

reverse

y= 2x+2

3 0

3 years ago

Simplify 4√16 + 8√16

Mumz [18]

$4\sqrt{16} + 8\sqrt{16}$ :

Since $\sqrt{1} =1$ , we get:

$4\cdot \:1\cdot \:6+8\cdot \:1\cdot \:6=24+48=72$

-----------------------------------------------------------------------------------

$7\sqrt{4} *3\sqrt{8}$

We have $\sqrt{4}=2$

simplify $\sqrt{8} = \sqrt{2^3} = \sqrt{2^2*2}$

then the radical rule is used: $\quad \sqrt[n]{ab}=\sqrt[n]{a}\sqrt[n]{b}$

⇒ $\sqrt{2^2*2} = \sqrt{2^2} \sqrt{2}= \sqrt{4}\sqrt{2} =2\sqrt{2}$

Now we have $\sqrt{4}=2$ and $\sqrt{8}=2 \sqrt{2}$

$7\sqrt{4} *3\sqrt{8} = 7*2*3*2\sqrt{2} = 84 \sqrt{2}$

-----------------------------------------------------------------------------------

$2\sqrt{4}+5 \sqrt{9}$

$2\sqrt{4} =2*2= 4$ and $5\sqrt{9} =5*3= 15$

$2\sqrt{4}+5 \sqrt{9} = 4+15 = 19$

-----------------------------------------------------------------------------------

$4\sqrt{5} -2\sqrt{5} = 2\sqrt{5}$

since $2\sqrt{5}$ is the half of $4\sqrt{5}$

-----------------------------------------------------------------------------------

$-4\sqrt{2}\:+\:5\sqrt{2} =1 \sqrt{2} = \sqrt{2}\\$ <em>(-4+5=1)</em>

4 0

2 years ago

41% of U.S. adults have very little confidence in newspapers. You randomly select 10 U.S. adults. Find the probability that the

lys-0071 [83]

Answer:

a) 0.2087 = 20.82% probability that the number of U.S. adults who have very little confidence in newspapers is exactly five.

b) 0.1834 = 18.34% probability that the number of U.S. adults who have very little confidence in newspapers is at least six.

c) 0.3575 = 35.75% probability that the number of U.S. adults who have very little confidence in newspapers is less than four.

Step-by-step explanation:

For each adult, there are only two possible outcomes. Either they have very little confidence in newspapers, or they do not. The answers of each adult are independent, which means that the binomial probability distribution is used to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

41% of U.S. adults have very little confidence in newspapers.

This means that $p = 0.41$

You randomly select 10 U.S. adults.

This means that $n = 10$

(a) exactly five

This is $P(X = 5)$ . So

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 5) = C_{10,5}.(0.41)^{5}.(0.59)^{5} = 0.2087$

0.2087 = 20.82% probability that the number of U.S. adults who have very little confidence in newspapers is exactly five.

(b) at least six

This is:

$P(X \geq 6) = P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10)$

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 6) = C_{10,6}.(0.41)^{6}.(0.59)^{4} = 0.1209$

$P(X = 7) = C_{10,7}.(0.41)^{7}.(0.59)^{3} = 0.0480$

$P(X = 8) = C_{10,8}.(0.41)^{8}.(0.59)^{2} = 0.0125$

$P(X = 9) = C_{10,9}.(0.41)^{9}.(0.59)^{1} = 0.0019$

$P(X = 10) = C_{10,10}.(0.41)^{10}.(0.59)^{0} = 0.0001$

Then

$P(X \geq 6) = P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10) = 0.1209 + 0.0480 + 0.0125 + 0.0019 + 0.0001 = 0.1834$

0.1834 = 18.34% probability that the number of U.S. adults who have very little confidence in newspapers is at least six.

(c) less than four.

This is:

$P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)$

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{10,0}.(0.41)^{0}.(0.59)^{10} = 0.0051$

$P(X = 1) = C_{10,1}.(0.41)^{1}.(0.59)^{9} = 0.0355$

$P(X = 2) = C_{10,2}.(0.41)^{2}.(0.59)^{8} = 0.1111$

$P(X = 3) = C_{10,3}.(0.41)^{3}.(0.59)^{7} = 0.2058$

So

$P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.0051 + 0.0355 + 0.1111 + 0.2058 = 0.3575$

0.3575 = 35.75% probability that the number of U.S. adults who have very little confidence in newspapers is less than four.

5 0

2 years ago