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2 years ago

Jos

Mathematics

1 answer:

Dahasolnce [82]2 years ago

5 0

Answer:

$7,200

Step-by-step explanation:

If 5400 is 75% then 5400/3 is 1800 or 25%. Add 5400 and 1800 and you get 7200. (Double-check to make sure I'm correct)

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In ΔRST, s = 58 inches, ∠R=115° and ∠S=19°. Find the length of r, to the nearest 10th of an inch.

Lorico [155]

Answer:

365.9 Inches

Step-by-step explanation:

To find the value of Sin A can be found using the sine rule. the rule states that the ratio of the length facing an angle in a triangle to the sine of the angle is constant. As such,

a/Sin A = b/Sin B = c/Sin C

Where is a is the side facing angle A, b is the side facing angle B and c is the side facing angle C.

Hence,

58/Sin 19 = r/Sin 115

r = 58 Sin 115/ Sin 19

= 365.9 Inches

8 0

3 years ago

Use the Trapezoidal Rule, the Midpoint Rule, and Simpson's Rule to approximate the given integral with the specified value of n.

Otrada [13]

I guess the "5" is supposed to represent the integral sign?

$I=\displaystyle\int_1^4\ln t\,\mathrm dt$

With $n=10$ subintervals, we split up the domain of integration as

[1, 13/10], [13/10, 8/5], [8/5, 19/10], ... , [37/10, 4]

For each rule, it will help to have a sequence that determines the end points of each subinterval. This is easily, since they form arithmetic sequences. Left endpoints are generated according to

$\ell_i=1+\dfrac{3(i-1)}{10}$

and right endpoints are given by

$r_i=1+\dfrac{3i}{10}$

where $1\le i\le10$ .

a. For the trapezoidal rule, we approximate the area under the curve over each subinterval with the area of a trapezoid with "height" equal to the length of each subinterval, $\dfrac{4-1}{10}=\dfrac3{10}$ , and "bases" equal to the values of $\ln t$ at both endpoints of each subinterval. The area of the trapezoid over the $i$ -th subinterval is

$\dfrac{\ln\ell_i+\ln r_i}2\dfrac3{10}=\dfrac3{20}\ln(ell_ir_i)$

Then the integral is approximately

$I\approx\displaystyle\sum_{i=1}^{10}\frac3{20}\ln(\ell_ir_i)\approx\boxed{2.540}$

b. For the midpoint rule, we take the rectangle over each subinterval with base length equal to the length of each subinterval and height equal to the value of $\ln t$ at the average of the subinterval's endpoints, $\dfrac{\ell_i+r_i}2$ . The area of the rectangle over the $i$ -th subinterval is then

$\ln\left(\dfrac{\ell_i+r_i}2\right)\dfrac3{10}$

so the integral is approximately

$I\approx\displaystyle\sum_{i=1}^{10}\frac3{10}\ln\left(\dfrac{\ell_i+r_i}2\right)\approx\boxed{2.548}$

c. For Simpson's rule, we find a quadratic interpolation of $\ln t$ over each subinterval given by

$P(t_i)=\ln\ell_i\dfrac{(t-m_i)(t-r_i)}{(\ell_i-m_i)(\ell_i-r_i)}+\ln m_i\dfrac{(t-\ell_i)(t-r_i)}{(m_i-\ell_i)(m_i-r_i)}+\ln r_i\dfrac{(t-\ell_i)(t-m_i)}{(r_i-\ell_i)(r_i-m_i)}$

where $m_i$ is the midpoint of the $i$ -th subinterval,

$m_i=\dfrac{\ell_i+r_i}2$

Then the integral $I$ is equal to the sum of the integrals of each interpolation over the corresponding $i$ -th subinterval.

$I\approx\displaystyle\sum_{i=1}^{10}\int_{\ell_i}^{r_i}P(t_i)\,\mathrm dt$

It's easy to show that

$\displaystyle\int_{\ell_i}^{r_i}P(t_i)\,\mathrm dt=\frac{r_i-\ell_i}6(\ln\ell_i+4\ln m_i+\ln r_i)$

so that the value of the overall integral is approximately

$I\approx\displaystyle\sum_{i=1}^{10}\frac{r_i-\ell_i}6(\ln\ell_i+4\ln m_i+\ln r_i)\approx\boxed{2.545}$

4 0

3 years ago

Manufacture of a certain component requires three different machining operations. Machining time for each operation has a normal

tigry1 [53]

Answer:

0.0314 = 3.14% probability that it takes at most 1 hour of machining time to produce a randomly selected component

Step-by-step explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

Sum of normal variables:

When normal variables are added, the mean is the sum of the means while the standard deviation is the square root of the sum of the variances.

The mean values are 20, 15, and 30 min, respectively:

This means that $\mu = 20 + 15 + 30 = 65$

The standard deviations are 1, 2, and 1.5 min, respectively.

This means that $\sigma = \sqrt{1^2 + 2^2 + 1.5^2} = 2.6926$

What is the probability that it takes at most 1 hour of machining time to produce a randomly selected component?

Mean and standard deviation are in minutes, so this is the pvalue of Z when X = 60.

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

$Z = \frac{60 - 65}{2.6926}$

$Z = -1.86$

$Z = -1.86$ has a pvalue of 0.0314

0.0314 = 3.14% probability that it takes at most 1 hour of machining time to produce a randomly selected component

7 0

2 years ago

Help a bab out please?

Volgvan

Answer:

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

2. On Monday, I bought 3 dogs and 5 cats for a total cost of $8.25. On Tuesday, I bought 3 dogs and 6 cats for a total cost of $

Ratling [72]

I don’t know if you need to round, but you can do so on the answer for mondays if necessary

On Monday, the animals cost $1.03125.
On Tuesday, the animals cost exactly $1.

7 0

2 years ago