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umka21 [38]
3 years ago
8

Heyo random person scrolling.Please help me?

Mathematics
1 answer:
Ierofanga [76]3 years ago
6 0

Let's do

\\ \sf\longmapsto 2sin^260cos60tan^245

\\ \sf\longmapsto 2\left(\dfrac{\sqrt{3}}{2}\right)^2\times \dfrac{1}{2}\times (1)^2

\\ \sf\longmapsto 2\times \dfrac{(\sqrt{3})^2}{2^2}\times \dfrac{1}{2}\times 1

\\ \sf\longmapsto \dfrac{3}{2}\times \dfrac{1}{4}

\\ \sf\longmapsto \dfrac{3}{4}

<u>Trigono</u><u>metric</u><u> </u><u>values</u>

\Large{ \begin{tabular}{|c|c|c|c|c|c|} \cline{1-6} \theta & \sf 0^{\circ} & \sf 30^{\circ} & \sf 45^{\circ} & \sf 60^{\circ} & \sf 90^{\circ} \\ \cline{1-6} $ \sin $ & 0 & $\dfrac{1}{2 }$ & $\dfrac{1}{ \sqrt{2} }$ & $\dfrac{ \sqrt{3}}{2}$ & 1 \\ \cline{1-6} $ \cos $ & 1 & $ \dfrac{ \sqrt{ 3 }}{2} } $ & $ \dfrac{1}{ \sqrt{2} } $ & $ \dfrac{ 1 }{ 2 } $ & 0 \\ \cline{1-6} $ \tan $ & 0 & $ \dfrac{1}{ \sqrt{3} } $ & 1 & $ \sqrt{3} $ & $ \infty $ \\ \cline{1-6} \cot & $ \infty $ &$ \sqrt{3} $ & 1 & $ \dfrac{1}{ \sqrt{3} } $ &0 \\ \cline{1 - 6} \sec & 1 & $ \dfrac{2}{ \sqrt{3}} $ & $ \sqrt{2} $ & 2 & $ \infty $ \\ \cline{1-6} \csc & $ \infty $ & 2 & $ \sqrt{2 } $ & $ \dfrac{ 2 }{ \sqrt{ 3 } } $ & 1 \\ \cline{1 - 6}\end{tabular}}

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Assume that adults have IQ scores that are normally distributed with a mean of mu equals 100μ=100 and a standard deviation sigma
Ksivusya [100]

Answer:

51.60% probability that a randomly selected adult has an IQ between 86 and 114.

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

\mu = 114, \sigma = 86

Find the probability that a randomly selected adult has an IQ between 86 and 114.

Pvalue of Z when X = 114 subtracted by the pvalue of Z when X = 86. So

X = 114

Z = \frac{X - \mu}{\sigma}

Z = \frac{114 - 100}{20}

Z = 0.7

Z = 0.7 has a pvalue of 0.7580

X = 86

Z = \frac{X - \mu}{\sigma}

Z = \frac{86 - 100}{20}

Z = -0.7

Z = -0.7 has a pvalue of 0.2420

0.7580 - 0.2420 = 0.5160

51.60% probability that a randomly selected adult has an IQ between 86 and 114.

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4 years ago
Factorise FULLY and show working out (2)
Klio2033 [76]
7x^2-63=7(x^2-9)=7(x-3)(x+3)
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3 years ago
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the larger of two number is 15 more than three times the smaller number. If the sum of the two numbers is 63, find tge numbers.
irakobra [83]
X is the smaller number. 3x + 15 is the larger number. So x + 3x + 15 = 63. 4x + 15 = 63.
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4 years ago
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Rhombus has all sides equal. The total amount of paper that will be needed to cover the wall is 396 in².

Given to us

Each rhombus will be 6 inches tall and 4 inches wide.

A.)

The area of each rhombus can be found using the formula,

Area = \dfrac{1}{2} \times d_1 \times d_2

We know that diagonals of the rhombus are of length 6 in and 4 in. therefore,

Area = \dfrac{1}{2} \times 6 \times 4\\\\Area = 12\ in^2

Hence, the area of the single rhombus is 12 in².

We know that the wall is 11 feet long, also, 1 foot = 12 inches, therefore,

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Now, since the wall is 132 in long and the width of the rhombus is 4 in, therefore, the number of rhombi that will be needed,

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We know the area of each paper rhombus, also, we know the number of rhombi that will be needed. Therefore,

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Hence, the total amount of paper that will be needed to cover the wall is 396 in².

Learn more about Rhombus:

brainly.com/question/14462098

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Answer:

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Step-by-step explanation:

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