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xz_007 [3.2K]
2 years ago
7

there are 104 students in the fourth grade at Allisons school. one day 15 4th graders were absent how many 4th graders were at s

chool that day
Mathematics
1 answer:
Rama09 [41]2 years ago
8 0
<span>This is a simple subtraction question. To find the answer, you simple need to subtract 15 from 104. This equals 89, so there were 89 4th graders at school that day in total. In the instance you're unsure that a subtraction equation is right, you can also add your answers back together to double check, so 89 + 15 = 104.</span>
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Evaluate the line integral by the two following methods. xy dx + x2y3 dy C is counterclockwise around the triangle with vertices
nadezda [96]

Answer:

a)

\frac{2}{3}

b)

\frac{2}{3}

Step-by-step explanation:

a) The first part requires that we use line integral to evaluate directly.

The line integral is

\int_C xydx +  {x}^{2}  {y}^{3} dy

where C is counterclockwise around the triangle with vertices (0, 0), (1, 0), and (1, 2)

The boundary of integration is shown in the attachment.

Our first line integral is

L_1 = \int_ {(0,0)}^{(1,0)} xydx +  {x}^{2}  {y}^{3} dy

The equation of this line is y=0, x varies from 0 to 1.

When we substitute y=0 every becomes zero.

\therefore \: L_1 =0

Our second line integral is

L_2 = \int_ {(1,0)}^{(1,2)} xydx +  {x}^{2}  {y}^{3} dy

The equation of this line is:

x = 0 \implies \: dx = 0

y varies from 1 to 2.

We substitute the boundary and the values to get:

L_2 = \int_ {1}^{2}1 \cdot y(0) +  {1}^{2}   \cdot \: {y}^{3} dy

L_2 = \int_ {1}^2 {y}^{3} dy =  \frac{8}{3}

The 3rd line integral is:

L_3 = \int_ {(1,2)}^{(0,0)} xydx +  {x}^{2}  {y}^{3} dy

The equation of this line is

y = 2x \implies \: dy = 2dx

x varies from 0 to 1.

We substitute to get:

L_3 = \int_ {1}^{0} x \cdot \: 2xdx +  {x}^{2}  {(2x)}^{3}(2 dx)

L_3 = \int_ {1}^{0} 8 {x}^{5}  + 2 {x}^{2} dx  =  - 2

The value of the line integral is

L = L_1 + L_2 + L_3

L = 0 +  \frac{8}{3}  +  - 2 =  \frac{2}{3}

b) The second part requires the use of Green's Theorem to evaluate:

\int_C xydx +  {x}^{2}  {y}^{3} dy

Since C is a closed curve with counterclockwise orientation, we can apply the Green's Theorem.

This is given by:

\int_C \: Pdx +Q  \: dy =  \int \int_ R \: Q_y -  P_x \: dA

\int_C \: xydx + {x}^{2} {y}^{3}   \: dy =  \int \int_ R \: 3 {x}^{2}  {y}^{2}  -  y \: dA

We choose our region of integration parallel to the y-axis.

\int_C \: xydx + {x}^{2} {y}^{3}   \: dy =  \int_ 0^{1} \int_ 0^{2x}  \: 3 {x}^{2}  {y}^{2}  -  y \: dydx

\int_C \: xydx + {x}^{2} {y}^{3}   \: dy =  \int_ 0^{1} \:  {x}^{2}  {y}^{3}  -   \frac{1}{2}  {y}^{2} |_ 0^{2x}  dx

\int_C \: xydx + {x}^{2} {y}^{3}   \: dy =  \int_ 0^{1} \:  8{x}^{5} -  2 {x}^{2}   dx =  \frac{2}{3}

8 0
2 years ago
Drag each situation to show whether it can be modeled using a linear or an exponential function. If the situation cannot be mode
lawyer [7]

Answer:

Step-by-step explanation:

A). An investment account earns 2.8% simple interest.

    Since investment amount increases every year linearly, therefore, the modeled situation represents a LINEAR FUNCTION.

B). The price of a stock varies by 2.8% each week.

    Since price of the stock may increase or decrease every week, therefore, this situation can't be modeled by any function.

Therefore, the answer is NEITHER.

C). An investment account earns 2.8% compound interest, compounded monthly.

Formula to get the value of the final amount in the account is,

Final value = Initial value × (1+\frac{.028}{100})^t

Here 't' = Duration of investment

It's an EXPONENTIAL FUNCTION.

3 0
2 years ago
A recycling truck begins its weekly route at the recycling plant at point A, as pictured on the coordinate plane below. It trave
mr_godi [17]

Answer:

41

Step-by-step explanation:

5 0
2 years ago
Read 2 more answers
Suppose that your business is operating at the 4.5-Sigma quality level. If projects have an average improvement rate of 50% annu
Luda [366]

Answer:

  11.75 years

Step-by-step explanation:

If we ignore the fact that "6-sigma" quality means the error rate corresponds to about -4.5σ (3.4 ppm) and simply go with ...

  P(z ≤ -6) ≈ 9.86588×10^-10

and

  P(z ≤ -4.5) ≈ 3.39767×10^-6

the ratio of these error rates is about 0.000290372. We're multiplying the error rate by 0.5 each year, so we want to find the power of 0.50 that gives this value:

  0.50^t = 0.000290372

  t·log(0.50) = log(0.00290372) . . . . take logarithms

  t = log(0.000290372)/log(0.50) ≈ -3.537045/-0.301030

  t ≈ 11.75

It will take about 11.75 years to achieve Six Sigma quality (0.99 ppb error rate).

_____

<em>Comment on Six Sigma</em>

A 3.4 ppm error rate is customarily associated with "Six Sigma" quality. It assumes that the process may have an offset from the mean of up to 1.5 sigma, so the "six sigma" error rate is P(z ≤ (1.5 -6)) = P(z ≤ -4.5) ≈ 3.4·10^-6.

Using that same criteria for the "4.5-Sigma" quality level, we find that error rate to be P(z ≤ (1.5 -4.5)) = P(z ≤ -3) ≈ 1.35·10^-3.

Then the improvement ratio needs to be only 0.00251699, and it will take only about ...

  t ≈ log(0.00251699)/log(0.5) ≈ 8.6 . . . . years

5 0
2 years ago
Please Help Me! It would be very much appreciated!
skad [1K]
The answer should be B) an=3+5(n-1)

hope i helped :)
7 0
2 years ago
Read 2 more answers
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