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Rashid [163]
3 years ago
14

Solve for 2. Round to the nearest tenth, if necessary.

Mathematics
1 answer:
Zepler [3.9K]3 years ago
4 0

Answer:

x = 46.9

Step-by-step explanation:

Since this is a right triangle, we can use trig functions

sin theta = opp / hyp

sin  54 = x / 58

58 sin 54 =x

x=46.92298

To the nearest tenth

x = 46.9

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You have a biased coin for which p(h)=pp(h)=p. you toss the coin 2020 times. what is the probability that you observe 88 heads a
Nat2105 [25]
The question describes a binomial probability with p(h) = p, then p(t) = 1 - p and number of trials (n) = 20

The probability of a binomial distribution is given by

P(x)=\, ^nC_xp^x(1-p)^{n-x}

Part A:

The probability of observing 8 heads and 12 tails is given by:

P(8)=\, ^{20}C_8p^8(1-p)^{20-8}=\, ^{20}C_8p^8(1-p)^{12}



Part B:

<span>You observe more than 8 heads and more than 8 tails, when you observe 9 heads and 11 tails, 10 heads and 10 tails, and 11 heads and 9 tails.

Therefore, the probability of </span><span>observing more than 8 heads and more than 8 tails</span> is given by:

P(9)+P(10)+P(11) \\  \\ =\, ^{20}C_9p^9(1-p)^{20-9}+\, ^{20}C_{10}p^{10}(1-p)^{20-10}+\, ^{20}C_{11}p^{11}(1-p)^{20-11} \\  \\ =\, ^{20}C_9p^9(1-p)^{11}+\, ^{20}C_{10}p^{10}(1-p)^{10}+\, ^{20}C_{11}p^{11}(1-p)^{9}
7 0
3 years ago
(3.2 * 10000000) -3
andreev551 [17]

The exponents distribute over factors:

(ab)^c = a^c \times b^c

so, you have

(3.2 \times 10^6)^{-3} = 3.2^{-3} \times (10^6)^{-3}

Let's focus on each factor: applying the definition of negative exponents, you have

3.2^{-3} = \cfrac{1}{3.2^3} \approx 0.03

While for the power of 10, you can use the following rule

(a^b)^c = a^{bc} to write

(10^6)^{-3} = 10^{6\cdot (-3)} = 10^{-18}

So, the expression evaluates to

0.03\times 10^{-18} = 3\times 10^{-20}

3 0
4 years ago
5. mr. and mrs. pinkerton have four boys. every time people notice this they comments to
Mars2501 [29]

The simulation that can best represent the chances of the Pinkertons having a boy or a girl is Flipping a coin.

<h3>What best shows the chances of having either a boy or girl?</h3>

The question asks for the simulation that can represent the likelihood of having a boy or a girl.

Because there are two outcomes, they each have a 50% probability.

A good simulation to show this would therefore be flipping a coin because this has two outcomes as well which are heads or tails.

Find out more on using a coin for probability at brainly.com/question/1334780.

#SPJ1

6 0
2 years ago
2. Find the area of the regular polygon.
frozen [14]

Answer:

a. 80 units

b. 482.843 units²

Step-by-step explanation:

<em>First</em><em> </em><em>things</em><em> </em><em>first</em><em>,</em><em> </em><em>we</em><em> </em><em>need</em><em> </em><em>to</em><em> </em><em>solve</em><em> </em><em>for</em><em> </em><em>the</em><em> </em><em>side</em><em> </em><em>of</em><em> </em><em>the</em><em> </em><em>octagon</em><em>. </em><em>We</em><em> </em><em>can</em><em> </em><em>do</em><em> </em><em>so</em><em> </em><em>by</em><em> </em><em>using</em><em> </em><em><u>Pythagorean</u></em><em><u> </u></em><em><u>Theorem</u></em><em><u>:</u></em><em> </em><em>a</em><em>²</em><em> </em><em>+</em><em> </em><em>b</em><em>²</em><em> </em><em>=</em><em> </em><em>c</em><em>²</em><em> </em><em>where</em><em> </em><em>c</em><em> </em><em>is</em><em> </em><em>the</em><em> </em><em><u>hypotenuse</u></em><em> </em><em>and</em><em> </em><em>a</em><em> </em><em>and</em><em> </em><em>b</em><em> </em><em>are</em><em> </em><em>the</em><em> </em><em>other</em><em> </em><em>two</em><em> </em><em>sides</em><em>. </em>

12² + b² = 13²

144 + b² = 169

b² = 25

b = 5

<em>Now</em><em> </em><em>that</em><em> </em><em>we</em><em> </em><em>know</em><em> </em><em>half</em><em> </em><em>of</em><em> </em><em>the</em><em> </em><em>side</em><em> </em><em>of</em><em> </em><em>the</em><em> </em><em>octagon</em><em> </em><em>is</em><em> </em><em>5</em><em>,</em><em> </em><em>we</em><em> </em><em>know</em><em> </em><em>that</em><em> </em><em>one</em><em> </em><em>side</em><em> </em><em>of</em><em> </em><em>the</em><em> </em><em>octagon</em><em> </em><em>is</em><em> </em><em>1</em><em>0</em><em>.</em><em> </em>

<em>a</em><em>.</em><em> </em><em>To</em><em> </em><em>find</em><em> </em><em>perimeter</em><em> </em><em>we</em><em> </em><em>just</em><em> </em><em>add</em><em> </em><em>every</em><em> </em><em>side</em><em> </em><em>together</em><em> </em><em>(</em><em>or</em><em> </em><em>since</em><em> </em><em>every</em><em> </em><em>side</em><em> </em><em>is</em><em> </em><em>the</em><em> </em><em>same</em><em> </em><em>size</em><em>,</em><em> </em><em>we</em><em> </em><em>multiply</em><em> </em><em>by</em><em> </em><em>8</em><em>)</em>

10 × 8 = 80

<em>b</em><em>.</em> <em>The</em><em> </em><em>formula</em><em> </em><em>for</em><em> </em><em>finding</em><em> </em><em>the</em><em> </em><em>area</em><em> </em><em>of</em><em> </em><em>an</em><em> </em><em>octagon</em><em> </em><em>is</em><em> </em><em>2</em><em>(</em><em>1</em><em> </em><em>+</em><em> </em><em>radical</em><em> </em><em>2</em><em>)</em><em>a</em><em>²</em><em> </em><em>where</em><em> </em><em>a</em><em> </em><em>is</em><em> </em><em>the</em><em> </em><em>length</em><em> </em><em>of</em><em> </em><em>a</em><em> </em><em>side</em><em>.</em><em> </em><em>By</em><em> </em><em>plugging</em><em> </em><em>in</em><em> </em><em>to</em><em> </em><em>a</em><em> </em><em>calculator</em><em> </em><em>we</em><em> </em><em>get</em><em>:</em>

2(1 + radical 2)(10)² = 482.843

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How do you recognize a linear equation
weqwewe [10]
<span>We look at the exponents. If the highest exponent on a single variable is one, then the </span>equation<span> is a </span>linear equation<span>.</span>
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