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iragen [17]
2 years ago
15

Please help me 33/2 + 3y/5 = 7y/10 + 15

Mathematics
1 answer:
8090 [49]2 years ago
8 0

We are given with an equation in <em>variable y</em> and we need to solve for <em>y</em> . So , now let's start !!!

We are given with ;

{:\implies \quad \sf \dfrac{33}{2}+\dfrac{3y}{5}=\dfrac{7y}{10}+15}

Take LCM on both sides :

{:\implies \quad \sf \dfrac{165+6y}{10}=\dfrac{7y+150}{10}}

<em>Multiplying</em> both sides by <em>10</em> ;

{:\implies \quad \sf \cancel{10}\times \dfrac{165+6y}{\cancel{10}}=\cancel{10}\times \dfrac{7y+150}{\cancel{10}}}

{:\implies \quad \sf 165+6y=7y+150}

Can be <em>further written</em> as ;

{:\implies \quad \sf 7y+150=165+6y}

Transposing <em>6y </em>to<em> LHS</em> and <em>150</em> to<em> RHS </em>

{:\implies \quad \sf 7y-6y=165-150}

{:\implies \quad \bf \therefore \quad \underline{\underline{y=15}}}

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The estimated daily living costs for an executive traveling to various major cities follow. The estimates include a single room
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Answer:

\bar x = 260.1615

\sigma = 70.69

The confidence interval of standard deviation is: 53.76 to 103.25

Step-by-step explanation:

Given

n =20

See attachment for the formatted data

Solving (a): The mean

This is calculated as:

\bar x = \frac{\sum x}{n}

So, we have:

\bar x = \frac{242.87 +212.00 +260.93 +284.08 +194.19 +139.16 +260.76 +436.72 +355.36 +.....+250.61}{20}

\bar x = \frac{5203.23}{20}

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\bar x = 260.16

Solving (b): The standard deviation

This is calculated as:

\sigma = \sqrt{\frac{\sum(x - \bar x)^2}{n-1}}

\sigma = \sqrt{\frac{(242.87 - 260.1615)^2 +(212.00- 260.1615)^2+(260.93- 260.1615)^2+(284.08- 260.1615)^2+.....+(250.61- 260.1615)^2}{20 - 1}}\sigma = \sqrt{\frac{94938.80}{19}}

\sigma = \sqrt{4996.78}

\sigma = 70.69 --- approximated

Solving (c): 95% confidence interval of standard deviation

We have:

c =0.95

So:

\alpha = 1 -c

\alpha = 1 -0.95

\alpha = 0.05

Calculate the degree of freedom (df)

df = n -1

df = 20 -1

df = 19

Determine the critical value at row df = 19 and columns \frac{\alpha}{2} and 1 -\frac{\alpha}{2}

So, we have:

X^2_{0.025} = 32.852 ---- at \frac{\alpha}{2}

X^2_{0.975} = 8.907 --- at 1 -\frac{\alpha}{2}

So, the confidence interval of the standard deviation is:

\sigma * \sqrt{\frac{n - 1}{X^2_{\alpha/2} } to \sigma * \sqrt{\frac{n - 1}{X^2_{1 -\alpha/2} }

70.69 * \sqrt{\frac{20 - 1}{32.852} to 70.69 * \sqrt{\frac{20 - 1}{8.907}

70.69 * \sqrt{\frac{19}{32.852} to 70.69 * \sqrt{\frac{19}{8.907}

53.76 to 103.25

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