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2 years ago

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In this problem, the ratios are inversely proportional. Find the missing value. If R1 = 6, R2 = 8, and I1 = 12, what is the valu

e of I2?

1 answer:

Drupady [299]2 years ago

5 0

Answer:

$I_2=9$

Step-by-step explanation:

We have been told that the ratios are inversely proportional in our given problem. We are asked to find the missing value.

We know that two inversely proportional quantities are in form $y=\frac{k}{x}$ , where, y is inversely proportional to x and k is the constant of proportionality.

Let us find constant of proportionality using $R_1 = 6$ and $I_1 = 12$ in above equation.

$6=\frac{k}{12}$

$6*12=\frac{k}{12}*12$

$72=k$

Now, we will use $72=k$ and $R_2 = 8$ in our equation to find $I_2$ as:

$8=\frac{72}{I_2}$

$I_2=\frac{72}{8}$

$I_2=9$

Therefore, the value of $I_2$ is 9.

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A thin sheet of ice is in the form of a circle. If the ice is melting in such a way that the area of the sheet is decreasing at

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Answer:

dx/dt = 0,04 m/sec

Step-by-step explanation:

Area of the circle is:

A(c) =π*x² where x is a radius of the circle

Applying differentiation in relation to time we get:

dA(c)/dt = π*2*x* dx/dt

In this equation we know:

dA(c)/dt = 0,5 m²/sec

And are looking for dx/dt then

0,5 = 2*π*x*dx/dt when the area of the sheet is 12 m² (1)

When A(c) = 12 m² x = ??

A(c) = 12 = π*x² ⇒ 12 = 3.14* x² ⇒ 12/3.14 = x²

x² = 3,82 ⇒ x = √3,82 ⇒ x = 1,954 m

Finally plugging ths value in equation (1)

0,5 = 6,28*1,954*dx/dt

dx/dt = 0,5 /12.28

dx/dt = 0,04 m/sec

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1 year ago

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The heights of the trees for sale at two nurseries are shown below. Heights of trees at Yard Works in feet : 7, 9, 7, 12, 5 Heig

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Answer:

The mean of tree heights at Yard Works is 8 feet and at Grow Station is 9 feet.The mean absolute deviation of the tree heights at both yards is 2.

Step-by-step explanation:

1). Height of the trees at Yard Works are = 7,9,7,12,5 feet

So mean height of the trees = (7+9+7+12+5)÷5

= 40÷5 =8 feet

Standard deviation of the trees at Yard works = ∑(║(height of the tree-mean height of the tree))║/(number of trees)

(height of the tree-mean height of the tree)= ║(7-8)║+║(9-8)║+║(7-8)║+║(12-8)║+║(5-8)║ = (1)+1+(1)+4+(3)= 10

Therefore standard deviation = (10)/(5) =2

2). In the same way mean height of the trees at Grow Station=(9+11+6+12+7)/5= 45/5 = 9

Now we will calculate the mean deviation of the tress at Grow Station

= ∑║(height of the tree-mean height of the tree)║/(number of trees)

= ║(9-9)║+║(11-9)║+║(6-9)║+║(12-9)║+║(7-9)║/(5)

= (0+2+3+3+2)/5

= 10/5 =2

Therefore The mean of tree heights at Yard Works is 8 feet and at Grow Station is 9 feet.The mean absolute deviation of the tree heights at both yards is 2.

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2 years ago

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Suppose that lower production costs increases the supply of wheat, such that more wheat is supplied at each price level. After t

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When the demand and supply curve intersect, that is, where the quantity demanded and quantity supplied are equal, the market is said to be in equilibrium. Thus, the given quantity is equilibrium quantity.

From the graph, we see that when the production cost of wheat is $4, the equilibrium quantity is 600 units.

When the production cost lowers from $4 to $3, the supply of wheat increases, such that the equilibrium quantity increases from 600 units to 800 units.

Thus, after an increase in supply, the equilibrium quantity increases.

So, Option A is the correct answer.

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2 years ago

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4. Hydrocarbons in the cab of an automobile were measured during trips on the New Jersey Turnpike and trips through the Lincoln

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Answer:

No these these result do not differ at 95% confidence level

Step-by-step explanation:

From the question we are told that

The first concentrations is $c _1= 30.0 \ g/m^3$

The second concentrations is $c _2 = 52.9 \ g/m^3$

The first sample size is $n_1 = 32$

The second sample size is $n_2 = 32$

The first standard deviation is $\sigma_1 = 30.0$

The first standard deviation is $\sigma_1 = 29.0$

The mean for Turnpike is $\= x _1 = \frac{c_1}{n} = \frac{31.4}{32} = 0.98125$

The mean for Tunnel is $\= x _2 = \frac{c_2}{n} = \frac{52.9}{32} = 1.6531$

The null hypothesis is $H_o : \mu _1 - \mu_2 = 0$

The alternative hypothesis is $H_a : \mu _1 - \mu_2 \ne 0$

Generally the test statistics is mathematically represented as

$t = \frac{\= x_1 - \= x_2}{ \sqrt{\frac{\sigma_1^2}{n_1} +\frac{\sigma_2^2}{n_2} }}$

$t = \frac{0.98125 - 1.6531}{ \sqrt{\frac{30^2}{32} +\frac{29^2}{32} }}$

$t = - 0.0899$

Generally the degree of freedom is mathematically represented as

$df = 32+ 32 - 2$

$df = 62$

The significance $\alpha$ is evaluated as

$\alpha = (C - 100 )\%$

=> $\alpha = (95 - 100 )\%$

=> $\alpha =0.05$

The critical value is evaluated as

$t_c = 2 * t_{0.05 , 62}$

From the student t- distribution table

$t_{0.05, 62} = 1.67$

So

$t_c = 2 * 1.67$

=> $t_c = 3.34$

given that

$t_c > t$ we fail to reject the null hypothesis so this mean that the result do not differ

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Zacharias is using the quadratic formula to solve the equation 0 = –2x2 5x – 3. he begins by substituting as shown.quadratic for

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Answer:

D is the answer

Step-by-step explanation:

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