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3 years ago

Name the coordinates for the reflection. Reflect point (4, -1) over the x-axis.

Mathematics

1 answer:

inysia [295]3 years ago

6 0

Answer:

i just took the test its (4,1)

Step-by-step explanation:

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nataly862011 [7]

Answer:

120 cubic cm

Step-by-step explanation:

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2 years ago

Read 2 more answers

Which one do I have to chose

DIA [1.3K]

Answer:

The top one I think

Step-by-step explanation:

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3 years ago

Graph the solution set of the inequality 3(1-x)<9.

zloy xaker [14]

Step by step solution:

3(1-x)<9
3-3x<9
-3x<6
x>-2

8 0

4 years ago

Compute the following binomial probabilities directly from the formula for b(x; n, p). (Round your answers to three decimal plac

yuradex [85]

<h2>Answer with explanation:</h2>

We know that the binomial theorem for finding the probability of x success out of a total of n experiments is given by:

$b(x;n;p)=n_C_x\cdot p^x\cdot (1-p)^{n-x}$

(a)

b(5; 8, 0.25)

is given by:

$8_C_5\cdot (0.25)^5\cdot (1-0.25)^{8-5}\\\\=8_C_5\cdot (0.25)^5\cdot (0.75)^3\\\\=56\cdot (0.25)^5\cdot (0.75)^3\\\\=0.023$

Hence, the answer is: 0.023

(b)

b(6; 8, 0.65)

i.e. it is calculated by:

$=8_C_6\cdot (0.65)^6\cdot (1-0.65)^{8-6}\\\\=8_C_6\cdot (0.65)^6\cdot (0.35)^2\\\\=0.259$

Hence, the answer is: 0.259

(c)

P(3 ≤ X ≤ 5) when n = 7 and p = 0.55

$P(3\leq x\leq 5)=P(X=3)+P(X=4)+P(X=5)$

Now,

$P(X=3)=7_C_3\cdot (0.55)^3\cdot (1-0.55)^{7-3}\\\\P(X=3)=7_C_3\cdot (0.55)^3\cdot (0.45)^{4}\\\\P(X=3)=0.239$

$P(X=4)=7_C_4\cdot (0.55)^4\cdot (1-0.55)^{7-4}\\\\P(X=4)=7_C_4\cdot (0.55)^4\cdot (0.45)^{3}\\\\P(X=4)=0.292$

$P(X=5)=7_C_5\cdot (0.55)^5\cdot (1-0.55)^{7-5}\\\\P(X=3)=7_C_5\cdot (0.55)^5\cdot (0.45)^{2}\\\\P(X=3)=0.214$

Hence,

$P(3\leq x\leq 5)=0.745$

(d)

P(1 ≤ X) when n = 9 and p = 0.1 .613

$P(1\leq X)=1-P(X=0)$

Also,

$P(X=0)=9_C_0\cdot (0.1)^{0}\cdot (1-0.1)^{9-0}\\\\i.e.\\\\P(X=0)=1\cdot 1\cdot (0.9)^9\\\\P(X=0)=0.387$

i.e.

$P(1\leq X)=1-0.387$

Hence, we get:

$P(1\leq X)=0.613$

6 0

3 years ago

As a result of radioactive decay, heat is generated uniformly throughout the interior of the earth at a rate of around 30 watts

qaws [65]

Answer:

a) $div F = 27 \frac{W}{km^3}$

b) $\alpha=\frac{27 W/km^3}{3}= 9\frac{W}{Km^3}$

c) $T(0,0,0)=-\frac{10}{2(27000)} (0) +7605.185=7605.185$

Step-by-step explanation:

(a) Suppose that the actual heat generation is 27W/km3 What is the value of div F? div F- Include units)

For this case the value for div F correspond to the generation of heat.

$div F = 27 \frac{W}{km^3}$

(b) Assume the heat flows outward symmetrically. Verify that $F= \alpha r$ where $r=xi +yj+zk$ . Find a α, (Include units.)

For this case we can satisfy this condition:

$div[\alpha (xi +yj +z k)]]=\alpha(1+1+1)=3\alpha$

And since we have the value for the $div F$ we can find the value of $\alpha$ like this:

$\alpha=\frac{27 W/km^3}{3}= 9\frac{W}{Km^3}$

(c) Let T (x,y,z) denote the temperature inside the earth. Heat flows according to the equation F= -k grad T where k is a constant. If T is in °C then k=27000 C/km. Assuming the earth is a sphere with radius 6400 km and surface temperature 20°C, what is the temperature at the center? 27,0 C/km.

For this case we have this:

$F =-k grad T$