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Elodia [21]
3 years ago
13

Perform the following computations.You may use13≈0.333333,34= 0.75 and100301= 0.332226.(i). Compute13+34by using five significan

t digits rounding arithmetic and compute the absoluteand relative errors.(ii) Compute13−100301by using five significant digits chopping arithmetic and compute the absoluteand relative errors
Mathematics
1 answer:
White raven [17]3 years ago
4 0

Answer:

a. 1.0833

Absolute Error = 0.416667

Relative Error = 1.250002

b. 0.0011070

Absolute Error = 0.0011070

Relative Error = 0.003321

Step-by-step explanation:

Given

1/3 = 0.333333

3/4 = 0.75

100/301 = 0.332226

a.

1/3 + 3/4

= 0.333333 + 0.75

= 1.083333

= 1.0833 ------ Approximated to 5 significant digits

Absolute Error = |Real Value - Estimated Value|

Relative Error = Absolute Error/Real Value

Assume 1/3 to be the real value and 3/4 to be the estimated value

Absolute Error = |0.333333 - 0.75|

Absolute Error = |-0.416667|

Absolute Error = 0.416667

Relative Error = 0.416667/0.333333

Relative Error = 1.250002

b.

1/3 - 100/301

= 0.333333 - 0.332226

= 0.001107

= 0.0011070 ----- Approximated to 5 significant digits

Assume 1/3 to be real value and 100/301 to be estimated value

Absolute Error = 0.333333 - 0.332226

Absolute Error = 0.0011070

Relative Error = 0.0011070/0.333333

Relative Error = 0.003321

Absolute and relative errors are approximation errors and they are due to the discrepancy between an exact value and some approximation to them.

The absolute error is the magnitude of the difference between the exact value and the approximation. The relative error is the absolute error divided by the magnitude of the exact value

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Translate words into numbers.

13/(1 3/7)

First convert to improper fraction.

13/(10/7)

Then, note that division is the same as multiplying by the divisor's reciprocal.

13x7/10=9.1

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3 years ago
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Construct a 90% confidence interval for μ1-μ2 with the sample statistics for mean calorie content of two​ bakeries' specialty pi
DIA [1.3K]

Answer:

The 90% confidence interval for the difference in mean (μ₁ - μ₂) for the two bakeries is; (<u>49</u>) < μ₁ - μ₂ < (<u>289)</u>

Step-by-step explanation:

The given data are;

Bakery A

\overline x_1<em> </em>= 1,880 cal

s₁ = 148 cal

n₁ = 10

Bakery B

\overline x_2<em> </em>= 1,711 cal

s₂ = 192 cal

n₂ = 10

\left (\bar{x}_1-\bar{x}_{2}  \right ) - t_{c}\cdot \hat \sigma \sqrt{\dfrac{1}{n_{1}}+\dfrac{1}{n_{2}}}< \mu _{1}-\mu _{2}< \left (\bar{x}_1-\bar{x}_{2}  \right ) + t_{c}\cdot \hat \sigma \sqrt{\dfrac{1}{n_{1}}+\dfrac{1}{n_{2}}}

df = n₁ + n₂ - 2

∴ df = 10 + 18 - 2 = 26

From the t-table, we have, for two tails, t_c = 1.706

\hat{\sigma} =\sqrt{\dfrac{\left ( n_{1}-1 \right )\cdot s_{1}^{2} +\left ( n_{2}-1 \right )\cdot s_{2}^{2}}{n_{1}+n_{2}-2}}

\hat{\sigma} =\sqrt{\dfrac{\left ( 10-1 \right )\cdot 148^{2} +\left ( 18-1 \right )\cdot 192^{2}}{10+18-2}}= 178.004321469

\hat \sigma ≈ 178

Therefore, we get;

\left (1,880-1,711  \right ) - 1.706\times178 \sqrt{\dfrac{1}{10}+\dfrac{1}{18}}< \mu _{1}-\mu _{2}< \left (1,880-1,711  \right ) + 1.706\times178 \sqrt{\dfrac{1}{10}+\dfrac{1}{18}}

Which gives;

169 - \dfrac{75917\cdot \sqrt{35} }{3,750} < \mu _{1}-\mu _{2}< 169 + \dfrac{75917\cdot \sqrt{35} }{3,750}

Therefore, by rounding to the nearest integer, we have;

The 90% C.I. ≈ 49 < μ₁ - μ₂ < 289

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