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3 years ago

Ten times as much as 3000

Mathematics

2 answers:

aev [14]3 years ago

8 0

That would be 30,000. Because since it is 10 times and 10 has one zero you would add one zero to 3,000 to get 30,000.

chubhunter [2.5K]3 years ago

3 0

Multiply 3000 by 10

10*3000=30000

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It took 7.2 minutes to upload 8 digital photographs from your computer to a website. At this rate, how long will it take to uplo

kenny6666 [7]

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Spoilers for answer:
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3 years ago

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Step-by-step explanation:

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3 years ago

Suppose a normal distribution has a mean of 222 and a standard deviation of 16. What is the probability that a data value is bet

valentina_108 [34]

Mean of the distribution = u = 222
Standard Deviation = s = 16

We have to find the probability that a value lies between 190 and 230.

First we need to convert these data values to z score.

$z= \frac{x-u}{s}$

For x = 190,
$z= \frac{190-222}{16}=-2$

For x = 230
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3 years ago

Can anyone help me? If f (x) = -2x +5, what is f (-3x) equal to?

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Answer:

Step-by-step explanation:

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2 years ago

Approximate the integral integral integral integral f(x, y) dA by dividing the rectangle R with vertices (0, 0), (4, 0), (4, 2),

amm1812

Answer:

Step-by-step explanation:

Approximate the integral $\int\int\limits_R {f(x,y)} \, dA$ by dividing the region $R$ with vertices (0,0),(4,0),(4,2) and (0,2) into eight equal squares.

Find the sum $\sum\limits^8_{i=1}f(x_i,y_i)\delta A_i$

Since all are equal squares, so $\delta A_i=1$ for every $i$

$\sum\limits^8_{i=1}f(x_i,y_i)\delta A_i=f(x_1,y_1)\delta A_1+f(x_2,y_2)\delta A_2+f(x_3,y_3)\delta A_3+f(x_4,y_4)\delta A_4+f(x_5,y_5)\delta A_5+f(x_6,y_6)\delta A_6+f(x_7,y_7)\delta A_7+f(x_8,y_8)\delta A_8\\\\=f(0.5,0.5)(1)+f(1.5,0.5)(1)+f(2.5,0.5)(1)+f(3.5,0.5)(1)+f(0.5,1.5)(1)+f(1.5,1.5)(1)+f(2.5,1.5)(1)+f(3.5,1.5)(1)\\\\=0.5+0.5+1.5+0.5+2.5+0.5+3.5+0.5+0.5+1.5+1.5+1.5+2.5+1.5+3.5+1.5\\\\=24$

Thus, $\sum\limits^8_{i=1}f(x_i,y_i)\delta A_i=24$

Evaluating the iterate integral $\int\limits^4_0 \int\limits^2_0 {(x+y)} \, dydx=\int\limits^4_0 {[xy+\frac{y^2}{2} ]}\limits^2_0 \, dx =\int\limits^4_0 {[2x+2]}dx\\\\=[x^2+2x]\limits^4_0=24.$

Thus, $\int\limits^4_0 \int\limits^2_0 {(x+y)} \, dydx=24$

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3 years ago