The range of sound intensities that the human ear can detect is so large that a special decibel scale (named after Alexander Graham Bell) is used to measure and compare sound intensities. The decibel level (dB) is given by I dB(I) = 10 log Io log() where Io is the intensity of sound that is barely audible to the human ear. Use the decibel level formula to find the decibel level for the following sounds. Round to the nearest tenth of a decibel. I = 1.58 x 108 · 10 (a) Automobile traffic, dB I = 10,800 · Io (b) Quiet conversation, dB (c) Fender guitar, dB I = 3.16 * 1011.10 (d) Jet engine, dB I = 1.58 x 1015. Io

Answer:

x = 56

Step-by-step explanation:

[tex]\frac{3}{7}[/tex] x + [tex]\frac{1}{4}[/tex] x = 38  The common denominator would be 28

[tex]\frac{12}{28}[/tex] x + [tex]\frac{7}{28}[/tex] x =38                                                                                                                                                

[tex]\frac{19}{28}[/tex] x = 38  Multiply both sides by [tex]\frac{28}{19}[/tex]

[tex](\frac{28}{19})[/tex][tex]\frac{19}{28}[/tex] x = [tex](\frac{28}{19})[/tex][tex]\frac{38}{1}[/tex]  Cross cancel the 19 and 38.  

x = [tex](\frac{28}{1})[/tex][tex]\frac{2}{1}[/tex]

x = 56

In the two period consumer model , the net effect from a decrease in the interest rate for a net borrower is increase in present consumption (C1 increases) , future consumption decreases (C2 decreases) and savings decreases.

What is two period consumer model?

Hence, the net effect from a decrease in the interest rate for a net borrower is increase in present consumption (C1 increases) , future consumption decreases (C2 decreases) and savings decreases.

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Answer:

El ecuación es y = 3x + 2

Step-by-step explanation:

Lo primero que tienes que hacer es poner 3x - y = 8 en la forma y = mx + b para para saber que 'm'. Porque la ecuación que intentamos escribir es paralela a 3x - y = 8, entonces tienen la misma pendiente (m).

3x - y = 8

-3x        -3x

-y = -3x + 8

/-1    /-1     /-1                   dividir todo por -1 para hacer 'y' positivo.

y = 3x - 8

Y como sabemos que la ecuación pasa por (7, 2), sabemos que pasa por 2 en el y-axis (vertical). Entonces 2 es el intercepto en y (b).

Answer:

sqrt2

Step-by-step explanation:

csc (x)= 1/sin(x)

sin(x)= (sqrt22)/(sqrt22*sqrt2)= 1/sqrt2

1/(1/sqrt2)=sqrt2/1=sqrt2

The graph of the inequality y > 4x - 3  , 3y - x ≤ 9 . is shown below.

In the question ,
the inequalities are given ,

we have to graph given inequalities ,

the inequalities are y > 4x - 3

3y - x ≤ 9 .

in the first inequality , y > 4x - 3

we put x = 0 , y is = -3 .

we put y = 0 , x is = 3/4 .

So ,the points for the line will be  (0,-3) and (3/4,0) .

since the inequality does not have equal to  sign , the line will be a dotted line .

in the second inequality ,3y - x ≤ 9 .

we put x = 0 ,  y is = 3 .

we put y = 0 ,  x is = -9 .

the points for the line will be (0,3) and (-9,0).

Since the inequality has equal to sign . Thus, line will be solid line .

Therefore , the for the given inequality is shown below.

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Step-by-step explanation:

-1/2h(2h+3) = 0

Seperate the equation:

-1/2 = 0

h = 0

(2h+3) = 0

Subtract 3 from both sides:

2h+3 = 0

2h+3-3 = 0

Simply:

2h = -3

Final Answer:

h = 0

h = -3/2

Answer:

a) (7-2x) +3 = 10 - 2X

b) x = y + 3 <=> y = x-3

c) (x-3).-2 +7 = -2x +6 +7 => -10 + 13 = 3

the orthogonal projection of v onto the subspace W spanned by the vectors UI. v = 1 2 3 , u1 = 1 −1 1 , u2 = −1 1 2 then The orthogonal projection of v onto W is (1/6) (4, 4, 9).

To find the orthogonal projection of v onto W, we need to first find the orthogonal basis vectors for W. Since the vectors u1 and u2 are given to be orthogonal, we can use them as the basis vectors for W. We can then calculate the projection of v onto each of these vectors using the dot product. This gives us the components of the projection vector. Finally, we can multiply each component by the appropriate weighting factor to obtain the orthogonal projection vector.

Let v = (2, 3, 4)

Let u1 = (1, 0, 0) and u2 = (0, 1, 0)

The projection of v onto u1 is given by:

(2, 3, 4) • (1, 0, 0) = 2

The projection of v onto u2 is given by:

(2, 3, 4) • (0, 1, 0) = 3

The orthogonal projection vector of v onto W is then given by:

2u1 + 3u2 = (2, 3, 0)

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The probability that the mean number of hours spent studying is less than 70 hours is; 0.023

We are given that:

Population mean; μ = 75

Population Standard deviation; σ = 25

Sample size; n = 100

The formula for the z-score is;

z = (x' - μ)/(σ/√n)

We want to find P(x < 70). Thus;

z = (70 - 75)/(25/√100)

z = -5/2.5

z = -2

From online p-value from z-score calculator, we have;

P(Z < -2) = 0.023

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For the best cases there will be 6+operations, The number of operations are best cases 6 and the worst cases are 10.

Given that,

The following pseudocode snippet performs the maximum number of + operations. Assume that the probability of each potential piece of data is equal. Preconditions: X = {x₁, x₂, x₃, x₄, x₅} ⊆ {10, 20, 30, 40, 50, 60, 70, 80}, where x1 < x2 < x3 < x4 < x5. t ← 0 i ← 1 while t < 101 do t ← t + xi i ← i + 1

We know that,

Here,

X = {x₁, x₂, x₃, x₄, x₅} ⊆ {10, 20, 30, 40, 50, 60, 70, 80}

By doing the iteration method

Iteration process till 4th iteration we get 6

Therefore, For the best cases there will be 6+operations, The number of operations are best cases 6 and the worst cases are 10.

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The grade needed by Kim on the fourth test to have the average of 80% is 89%  , the correct option is (d)  .

In the question ,

it is given that ,

the scores that Kim got in the first three tests are 82% , 75% , 74% .

let the grade required by Kim to score average of 80% be = x% .

Since Kim needs to score average of 80% in all the four tests ,

that means ,

(82 + 75 + 74 + x)/4 = 80

After Simplifying further further ,

we get ,

(231 + x)/4 = 80

231 + x = 80 × 4

231 + x = 320

x = 89% .

Therefore , Kim needs to score 89% in the fourth test .

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If it is important to detect a mean difference in score of one point with the probability of at least 0.90 , then the number of pairs that should be used is  10 .

It is given that ,

the probability is at least 0.90 ,

So , in the paired t test ,

testing mean paired difference is = 0

alpha = 0.05 , the assumed standard deviation of paired difference = 0.441

So the output is

Difference = 1 , Size = 5 power(probability) = 0.90 ,

So , the actual probability is = 0.95190

From the output above , the required sample size is n = 5 .

We observe that , under the given conditions the sample size is = 5 .

but the researcher  considered 10 pairs ,

So , the sample size 10 is used for this study .

Therefore , 10 pairs should be used .

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The given question is incomplete ,the complete question is

It is important to detect a mean difference in score of one point with a probability of at least 0.90. how many pairs should have been used ?

The slope of a line that is perpendicular to it Undefined.

What is slope?

A line's steepness can be determined by looking at its slope. Slope is calculated mathematically as "rise over run" (change in y divided by change in x).

That is the only slope which cannot be defined by a number.

A horizontal line with a slope of 0 has a change in y that is always 0 for any change in x.
As long as x is not 0, m=03, 08, 0x.

The line that runs perpendicular to this is vertical and has a "undefined" slope. We cannot divide by zero since the change in x for every change in y is always 0.
m=60,−50,y0 etc.
The slope remains undefined.

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a) The maximum mean weight of the passengers is 187.5 lb.,b) The probability that the mean weight exceeds 187.5 lb is 0.0062.

The mean weight of each passenger is 198 lb and the standard deviation is 42 lb. The water taxi has a stated capacity of 25 passengers and a load limit of 3750 lb. The maximum mean weight of the passengers is 187.5 lb, which is determined by the load limit of 3750 lb divided by the stated capacity of 25 passengers. The probability that the mean weight exceeds 187.5 lb is 0.0062, which is calculated using the normal distribution table. If the capacity is revised to 20 passengers, the maximum mean weight is still 187.5 lb. The probability of the mean weight exceeding 187.5 lb is still 0.0062. The probability that an individual passenger exceeds the maximum load limit of 3750 lb is 0.0228, which is calculated using the normal distribution table.

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The domain of the function is: [0,∞), the range of the function is: [3500,∞)

Domain and Range: The set which contains all the first elements of all the ordered pairs of relation R is known as the domain of the relation. The set which contains all the second elements, on the other hand, is known as the range of the relation.

The potential input and output values of a function are its domain and range, respectively.

The formula for the function is C=130S+3500.

the region

This is a list of the function's potential S values.

Due to the fact that S refers to a physical quantity, it cannot be less than 0. (i.e. tons of sugar)

The domain of the function is [0,∞)

Since S can have any value larger than 0.

The variety

This represents the function's potential C values.

The formula for the function is C=130S+3500.

Assuming S = 0, the following is true: C=130*0+3500=3500.

C must be more than 3500.

Any number higher than 3500 can be used as the value of C.

Consequently, the function's range is [3500,∞)

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By finding the combinations C(15, 3), we wills see that there are 455 different sets of prizes.

Basically, we want to see how many different sets of 3 eggs can Trevor pick out of the set of 15 eggs.

So we want to find the combinations, remember that for a set of N elements, the number of different subsets of K elements is given by:

C(N. K) = N!/(K!*(N - K)!)

Here we have:

N = 15

K = 3

Then:

C(15, 3) = 15!/(3!*12!) = 15*14*13/3*2 = 455

There are 455 different sets of prizes that Trevor could pick.

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There are 455 different sets of prizes.

The confidence interval for the random sample of 12 residents of the state of Montana is found as: 1.764 ≤ μ ≤ 2.636.

The width of the gap and the likelihood that the population parameter will fall beyond the projected range of values increase with increasing confidence level.

The stated data;

Degree of freedom

Df = n - 1

Df = 12 - 1

Df = 11

The critical value of t.

t critical = t(α/2,Df)

             = t(0.05, 11)

Using the t distribution.

t critical = ± 1.796

The confidence interval:

μ = x ± t.s / √n

 = 2.2 ± (1.796).(0.84)/√12

= 2.2 + 0.4355

1.764 ≤ μ ≤ 2.636

Thus, the confidence interval for the random sample of 12 residents of the state of Montana is found as: 1.764 ≤ μ ≤ 2.636.

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The volume of the composite solid is equal to 72π cubic centimeters.

The volume of the composite solid shown in the figure is the result of the sum of the volumes of two solids: a cylinder and a cone. The volume formula of each element is shown below:

Volume of a cylinder

V = π · r² · h

Volume of a cone

V = (π / 3) · r² · h'

Where:

Volume of the composite solid

V = π · r² · h + (π / 3) · r² · h'

If we know that r = 3 cm, h = 7 cm and h' = 3 cm, then the volume of the composite solid is:

V = π · (3 cm)² · (7 cm) + (π / 3) · (3 cm)² · (3 cm)

V = 72π cm³

The composite solid has a volume of 72π cubic centimeters.

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By using the method of eigenvalues and eigenvectors, the long-term probability that a given marked rat is in room 4 is 1/3

Here we have given that Forty rats are placed at random in a house having 4 rooms. There is one door between rooms 1 and 2, one door between rooms 1 and 4, one door between rooms 2 and 4, one door between rooms 2 and 3, and two doors between rooms 3 and 4 see the book for a picture). There are no doors between rooms 1 and 3. After each minute, a rat may change rooms, or stay still.

And we need to find the the long-term probability that a given marked rat is in room 4.

By using the eigenvalues method, we know that

Room 4 has the 3 exits that is from Room 1, Room 2 and Room 3.

So, here we have 3 possibilities for the exit so, the rat can choose any one these three ways,

So, the probability can be written as,

=> 1/3

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Answer:

20(3-2y)

Step-by-step explanation:

60−40y =

10(6-4y) =   ==> both 60 and 40 are multiples of 10

2(10(3-2y)) =  ==> both 6 and 4 are factors of 2

2*10(3-2y) = ==> simplify

20(3-2y)

The response variable was the weight of the package's standard deviation

a)the generator for this design is E=-ABCD

b)the resolution of this design is I=-ABCDE Therefor the resolution of the design is V.

c)estimate the factor effects there is 3 larger effcts namely E=-0.4700,BE=-0.4050,DE=-0.3150

d)Construct a linear regrassion model.The constant is estimated by the grand average and the regression coefficent are estimated by one-half the corresponding effect estimates.

If the underlying assumptions have any issues, the residual analysis will show them.

Y=1.22625+0.04375x₂-0.01875x₄+0.2350x₅-0.08125x₂x₄-0.1575x₄x₅

e)The standard deveation of package weight is affected the most by the dealy between mixing and packing,factor E.Also, its intreaction with the temperature factor B,BE and with the batch weight factor D,DE are important.

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Answer:

12 row plans

Step-by-step explanation:

the first seat has two possible options , the two directors. The second seat has another two options. With each director comes 2 options for the second seat so 2×2=4 possible options for the first and second seat . The 3rd seat has three options, that means for each combination of the 1st and 2nd seat three possible options 2×3=12 combinations

The required equation of the given function in the respective form is given as g(x) = 2(x - 4)² - 7, and the vertex is (4, -7).

The graph is a demonstration of curves that gives the relationship between the x and y-axis.

Here,
g(x) = 2x² - 16x + 25
g(x) = 2[x² - 8x] + 25
g(x) = 2[x² - 8x + 16 -16] + 25
g(x) = 2(x - 4)²  - 32 + 25
g(x) = 2(x - 4)² - 7

Now, the vertex is given as (h, k) = (4, -7).
And the graph of the given function is shown.

Thus, the required equation of the given function in the respective form is given as g(x) = 2(x - 4)² - 7, and the vertex is (4, -7).

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The score does Dana need on the fourth examination to raise his average to 87 is 98.

What is the average of numbers?

Average by adding a group of numbers, dividing by their count, and then summing the results, the arithmetic mean is determined. For instance, the sum of 2, 3, 3, 5, 7, and 10 is equal to 30 divided by 6, which equals 5. Median the central number in a set of numbers.

Given: Dana received scores of  85, 92, and 73.

We have to find the score does Dana need on the fourth examination to raise his average to 87.

Suppose the score on the fourth examination is x.

The average of scores is 87.

⇒

[tex]87 = \frac{85 + 92 + 73 + x}{4} \\348 = 250 + x\\x = 348 - 250\\x = 98[/tex]

Hence, the score does Dana need on the fourth examination to raise his average to 87 is 98.

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Answer:

26

Step-by-step explanation:

To find the value of the expression 2|x-4|-3y³ when x=3 and y=-2, we need to substitute the values of x and y into the expression and then simplify. The absolute value part of the expression, |x-4|, is the distance between the number x and 4 on the number line. When x=3, this distance is |3-4|=1. So the expression becomes 2*1-3(-2)³ = 2-3(-8) = 2+24 = 26. Therefore, when x=3 and y=-2, the value of the expression 2|x-4|-3y³ is 26.

(a) At 95% confidence interval z- value is 1.96 and t-value is 2.447.

(b) At 80% confidence interval z- value is 1.282 and t-value is 1.440.

(c) At 90% confidence interval z- value is 1.645 and t-value is 1.943.

Given that,

Let's say you randomly choose n=7 measurements from a normal distribution. If you were constructing the following confidence intervals, compare the standard normal z values with the appropriate t values.

We have to find

(a) At 95% confidence interval what is z- value and t-value.

(b) At 80% confidence interval what is z- value and t-value.

(c) At 90% confidence interval what is z- value and t-value.

We know that,

Sample size = n = 7

Degrees of freedom = df = n - 1 = 7 - 1 = 6

(a)  At 95% confidence level

α = 1 - 95%  

α = 1 - 0.95 =0.05

α/2 = 0.025

Zα/2 = Z0.025  = 1.96

z = 1.96

At 95% confidence level

α= 1 - 95%

α =1 - 0.95 =0.05

α/2 = 0.025

tα/2,df = t0.025,6 = 2.447

t = 2.447

(b)  At 80% confidence level

α = 1 - 80%  

α = 1 - 0.80 =0.20

α /2 = 0.10

Zα /2 = Z0.10  = 1.282

z = 1.282

At 80% confidence level

α = 1 - 80%

α =1 - 0.80 =0.20

α /2 = 0.10

tα /2,df = t0.10,6 = 1.440

t = 1.440

(c) At 90% confidence level

α = 1 - 90%  

α = 1 - 0.90 =0.10

α /2 = 0.05

Zα /2 = Z0.05  = 1.645

z = 1.645

At 90% confidence level

α = 1 - 90%

α =1 - 0.90 =0.10

α /2 = 0.05

tα /2,df = t0.05,6 = 1.943

t = 1.943

Therefore,

(a) At 95% confidence interval z- value is 1.96 and t-value is 2.447.

(b) At 80% confidence interval z- value is 1.282 and t-value is 1.440.

(c) At 90% confidence interval z- value is 1.645 and t-value is 1.943.

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The transformed graph is attached

The domain is (-∝, ∝) and the range is (-3, ∝)

From the question, we have the following parameters that can be used in our computation:

y = eˣ

To transform the graph to y = e⁻ˣ - 3, we make use of the following rules:

See attachment for the graph of y = e⁻ˣ - 3

From the attached graph, we have:

The graph extends across the x-axis i.e. domain = (-∝, ∝)

The graph opens upward from y > -3 i.e. range = (-3, ∝)

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The distribution is not normal. The points show a systematic pattern that is not a straight-line pattern curved over mean.

A normal distribution is a type of data distribution in which the data points form a symmetric, bell-shaped curve around the mean.

The data points in the example provided show a systematic pattern that is not a straight-line pattern, indicating that the data does not come from a normally distributed population.

This is because the points are not reasonably close to a straight line, which is a characteristic of a normal distribution.

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