The mean weekly earnings of a sample of 30 construction workers was $759, with a standard deviation of $73, and the mean weekly earnings of a sample of 28 manufacturing workers was $658, with a standard deviation of $65. Construct a 85% confidence interval for the difference between the mean weekly earnings for construction workers and the mean weekly earnings for manufacturing workers. Round to the second decimal place.

Answer:

Probability that it wins at least 3 of its final 5 games = .02387

Step-by-step explanation:

Given -

The probability of win the weekend game = 0.5

The probability of loose  the weekend game = 0.5

If he wins the game this weekend then it will play its final 5 games in the upper bracket of its league

In this case,  probability of success is (p) = 0.3

probability of failure is (q) = 1 - p = 0.7

Let X be number of game won out of last five games

probability that it wins at least 3 of its final 5 games

( 1 )

[tex]P(X\geq3)[/tex] = [tex]P(X\geq3/first\; game\; won)[/tex] ( probability of first game won )

               =   [tex]0.5\times[/tex]P( X =3 ) + [tex]0.5\times[/tex]P( X =4) + [tex]0.5\times P(X = 5)[/tex]

                =  [tex]0.5\times\binom{5}{3}(0.3)^{3}(0.7)^{2} + 0.5\times\binom{5}{4}(0.3)^{4}(0.7)^{1}[/tex] + [tex]0.5\times\binom{5}{5}(0.3)^{5}(0.7)^{0}[/tex]

                 = [tex]0.5\times\frac{5!}{(3!)(2!)}\times(0.3)^{3}\times(0.7)^{2} + 0.5\times\frac{5!}{(4!)(1!)}\times(0.3)^{4}\times(0.7)^{1}[/tex] + [tex]0.5\times\frac{5!}{(5!)(0!)}\times(0.3)^{5}\times(0.7)^{0}[/tex]= = .065 + .014 + .001215  = .080

If he loose the game this weekend then it will play its final 5 games in the lower bracket of its league

In this case,  probability of success is (s) = 0.4

probability of failure is (t) = 1 - s = 0.6

( 2 )

[tex]P(X\geq3/first\; game\; lost)[/tex] ( probability of first game lost )

= [tex]0.5\times P(X = 3) + 0.5\times P(X = 4)[/tex] + [tex]0.5\times P(X=5)[/tex]

= [tex]\binom{5}{3}(0.4)^{3}(0.6)^{2} + 0.5\times\binom{5}{4}(0.4)^{4}(0.6)^{1}[/tex]+ [tex]0.5\times\binom{5}{5}(0.4)^{5}(0.6)^{0}[/tex]

= [tex]0.5\times\frac{5!}{(3!)(2!)}\times(0.4)^{3}\times(0.6)^{2} + 0.5\times\frac{5!}{(4!)(1!)}\times(0.4)^{4}\times(0.6)^{1}[/tex] + [tex]0.5\times\frac{5!}{(5!)(0!)}\times(0.4)^{5}\times(0.6)^{0}[/tex] = = .1152 + .0384 + .00512 = .1587

Required probability = ( 1 ) + ( 2 ) = .02387

Answer:

The probability that at least two-third of vehicles in the sample turn is 0.4207.

Step-by-step explanation:

Let X = number of vehicles that turn left or right.

The proportion of the vehicles that turn is, p = 2/3.

The nest n = 50 vehicles entering this intersection from the east, is observed.

Any vehicle taking a turn is independent of others.

The random variable X follows a Binomial distribution with parameters n = 50 and p = 2/3.

But the sample selected is too large and the probability of success is close to 0.50.

So a Normal approximation to binomial can be applied to approximate the distribution of X if the following conditions are satisfied:

Check the conditions as follows:

[tex]np=50\times \frac{2}{3}=33.333>10\\\\n(1-p)=50\times \frac{1}{3}= = 16.667>10[/tex]

Thus, a Normal approximation to binomial can be applied.

So, [tex]X\sim N(np, np(1-p))[/tex]

Compute the probability that at least two-third of vehicles in the sample turn as follows:

[tex]P(X\geq \frac{2}{3}\times 50)=P(X\geq 33.333)=P(X\geq 34)[/tex]

                        [tex]=P(\frac{X-\mu}{\sigma}>\frac{34-33.333}{\sqrt{50\times \frac{2}{3}\times\frac {1}{3}}})[/tex]

                        [tex]=P(Z>0.20)\\=1-P(Z<0.20)\\=1-0.5793\\=0.4207[/tex]

Thus, the probability that at least two-third of vehicles in the sample turn is 0.4207.

The solution is, Volume of prism = 1/2x³ + x², is the expression which represents the volume of the prism, in cubic units.

Volume can be stated as the space taken by an object. Volume is a measure of three-dimensional space. Volume is defined as the space occupied within the boundaries of an object in three-dimensional space. It is also known as the capacity of the object.

here, we have,

Given that:

The oblique prism below has an isosceles right triangle base and the length of the base is x

=> the area of the base: 1/2 × x × x = 1/2

The vertical height of the prism is (x + 2)

=> The volume of the oblique prism is:

V = the base area * the vertical height

<=> V = 1/2* x² * (x + 2)

<=> V =  1/2x³ + x²

Hence, The solution is, Volume of prism = 1/2x³ + x², is the expression which represents the volume of the prism, in cubic units.

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Answer:

z-test

Step-by-step explanation:

-A z-test is a test used to determine the difference in two population means of known variances.

-The sample size has to be large enough, [tex]n\geq 30[/tex](n=71).

-And, the population must follow a normal distribution.

-Since our sample meets all the above conditions, the hypothesis test used is a z-test

Answer:

25 Inches

Step-by-step explanation:

Given the table:

[tex]\left|\begin{array}{c|c|c|c|c}\text{Number of Inches}&2&10&&40\\\text{Number of Centimeters}&5.08&25.04&63.5&101.6\end{array}\right|[/tex]

We want to determine the missing value on the table.

Let the missing value be x.

1 inch = 2.54 cm

x inch = 63.5

Expressing the above as a ratio

[tex]\dfrac{1}{x}=\dfrac{2.54}{63.5} \\$Cross Multiply$\\2.54x=63.5\\$Divide both sides by 2.54$\\x=25 \:Inches[/tex]

Therefore, the missing value is 25.

Answer:

25 Inches

Step-by-step explanation:

hope this helps :))

Answer:

i believe the answer is 1/27

Step-by-step explanation:

you take the fractions and multiply them all together.

1x1x1 equals 1

and 3x3x3 equals 27

meaning the answer is 1/27 :)

To calculate 1/3 x 1/3 x 1/3, you're effectively cubing 1/3, which results in (1/3)^3 or 1^3/3^3, simplifying to 1/27.

The student is asking about the multiplication of fractions and exponentiation rules in algebra. To solve 1/3 x 1/3 x 1/3, you multiply the fractions normally. When multiplying identical fractions, we simply raise the fraction to the power of the number of times it is being multiplied by itself. So 1/3 x 1/3 x 1/3 is equivalent to (1/3)^3. When you raise a fraction to an exponent, you raise both the numerator and the denominator to that power. Therefore, (1/3)^3 equals 1^3/3^3, which simplifies to 1/27.

The example given with 3².35 relates to the rules of exponents, which state that when multiplying exponential terms with the same base, you can add the exponents (x^p x x^q = x^(p+q)). For the concept of cubing of exponentials, you would cube the base and multiply the existing exponent by 3 to execute the operation effectively.

Answer:

Therefore the rate at which water level is dropping is [tex]\frac{11}{21}[/tex] in per minute.

Step-by-step explanation:

Given that,

The diameter of cylindrical bucket = 12 in.

Depth is increasing at the rate of = 4.0 in per minutes.

i.e [tex]\frac{dh_1}{dt}=4[/tex]

[tex]h_1[/tex] is depth of the bucket.

The volume of the bucket is V = [tex]\pi r^2 h[/tex]

                                                 [tex]=\pi \times 6^2\itimes h_1[/tex]

[tex]\therefore V=36\pi h_1[/tex]

Differentiating with respect yo t,

[tex]\frac{dV}{dt}=36\pi \frac{dh_1}{dt}[/tex]

Putting  [tex]\frac{dh_1}{dt}=4[/tex]

[tex]\therefore\frac{dV}{dt}=36\pi\times 4[/tex]

The rate of volume change of the bucket = The rate of volume change of the aquarium .

Given that,The aquarium measures 24 in × 36 in × 18 in.

When the water pumped out from the aquarium, the depth of the aquarium only changed.

Consider h be height of the aquarium.

The volume of the aquarium is V= ( 24× 36 ×h)

V= 24× 36 ×h

Differentiating with respect to t

[tex]\frac{dV}{dt}=24\times 36 \times \frac{dh}{dt}[/tex]

Putting [tex]\frac{dV}{dt}=36\pi\times 4[/tex]

[tex]36\pi\times 4= 24\times 36\times \frac{dh}{dt}[/tex]

[tex]\Rightarrow \frac{dh}{dt}=\frac{36\pi \times 4}{24\times 36}[/tex]

[tex]\Rightarrow \frac{dh}{dt}=\frac{11}{21}[/tex]

Therefore the rate at which water level is dropping is [tex]\frac{11}{21}[/tex] in per minute.

To find the rate at which the aquarium water level is dropping, calculate the volume of water being pumped into the bucket per minute and the volume of the aquarium. Then, divide the volume of water by the volume of the aquarium to find the rate.

To find the rate at which the aquarium water level is dropping, we first need to determine the flow rate of water being pumped into the cylindrical bucket.

Now, let's find the rate at which the aquarium water level is dropping:

By calculating the above expression, you can find the rate at which the aquarium water level is dropping.

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The water level in the aquarium is dropping at an average rate of π/6 inches per minute. This is obtained by relating the volumes and their rates of change in both the bucket and the aquarium.

The subject of this question is calculus, specifically, related rates. The situation describes two rates: one at which the depth of water in the bucket is increasing, and one at which the water level in the aquarium is dropping - a classic related rates problem.

To comprehend this, let us calculate the volume rates of each one: the bucket and the aquarium. The bucket is referred as cylindrical with a radius of 6 inches (half of 12 inches). The volume of a cylinder is V = πr²h. The depth or height, h, is increasing at the rate of 4 inches per minute, so dh/dt = 4 in/min. The rate at which the volume has been changing, dV/dt = πr²dh/dt = π * (6 in)² * 4 in/min = 144π in³/min.

The aquarium is a rectangular prism, thus its volume can be calculated by V = l*w*h. Therefore, the rate at which the water level is dropping is dV/dt / (l*w) which equals 144π in³/min divided by the area of the aquarium (24in*36in = 864 in²), which gives dh/dt = 144π / 864, which reduces to dh/dt = π/6 in/min downwards. Hence, the water level in the aquarium is dropping at π/6 in/min on an average.

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Answer:

a. (4, 650), (8, 400)

b. -62.5

Step-by-step explanation:

a......................

Data points are (4, 650) and (8, 400) since price is independent variable and weight of coffee is dependent

b......................

Slope = (y2-y1)/(x2-x1)

Slope = (400 - 650)/(8-4) = -250/4 = - 62.5

The data points representing the price and quantity of coffee sold are ($4, 650) and ($8, 400). The slope of the line connecting these points, which shows the relationship between price and quantity, is -62.5.

This question is about mathematical calculations used in economics, specifically related to pricing and demand. Let's turn these scenarios into data points, with price ($4 and $8) as the independent variable and quantity (650 lbs and 400 lbs) as the dependent variable. So, your data points are ($4, 650) and ($8, 400).

To find the slope of the line connecting these points, use the slope formula. The slope(m) is equal to the difference in the y-values divided by the difference in the x-values. So, m = (400-650) / (8-4) = -250/4 = -62.5.

Thus, the slope of the line connecting these two points, representing the relationship between the price and quantity of coffee sold, is -62.5.

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Answer:

[tex]\Delta s \approx 754.579\,m[/tex] (See explanation below).

Step-by-step explanation:

Each floor has a height of 3 meters. Then, the number of floors of the cylinder is:

[tex]n = \frac{30\,m}{3\,m}[/tex]

[tex]n = 10\,floors[/tex]

Let consider that spiral makes a revolution per floor. Then, the parametric equations of the spiral are:

[tex]x = r\cdot \cos \theta[/tex]

[tex]y = r\cdot \sin \theta[/tex]

[tex]z = \Delta h \cdot \frac{\theta}{2\pi}[/tex]

Length of the staircase can be modelled by using the formula for arc length:

[tex]\Delta s = \int\limits^{20\pi}_{0} {\sqrt{\left(\frac{dx}{d\theta} \right) ^{2}+\left(\frac{dy}{d\theta} \right)^{2}+\left(\frac{dz}{d\theta}\right)^{2}}} \, d\theta[/tex]

[tex]\Delta s = \int\limits^{20\pi}_{0} {\sqrt{\left(-r\cdot \sin \theta\right)^{2}+\left(r\cdot \cos \theta\right)^{2}+\left(\frac{\Delta h}{2\pi} \right)^{2}} } \, d\theta[/tex]

[tex]\Delta s = \int\limits^{20\pi}_{0} {\sqrt{r^{2}+\frac{(\Delta h)^{2}}{4\pi^{2}} }} \, d\theta[/tex]

[tex]\Delta s = \sqrt{(12\,m)^{2}+\frac{(3\,m)^{2}}{4\pi^{2}} } \cdot (20\pi-0)[/tex]

[tex]\Delta s \approx 754.579\,m[/tex]

Step-by-step explanation:

a) $1,000 or more?

As the data is usually distributed, using the 3-step approach to solve the problem: draw the image (I'm not going to do it here), locate the z-point, and identify the region on the graph.

(1000-951)/96 = .5104 = 0.51 = Z, so area = 1 -0.6950 = 0.3050

b) Between $900 and $1,100?

(900-951)/96 = -0.53 = Z, and (1100-951)/96 = 1.55 = Z So area = 0.9394 - 0.2981 = 0.6413

c) Between $825 and $925?

(825-951)/96 = -1.31 = Z, and (925-951)/96 = -0.27 = Z So area = 0.3936 - 0.0951 = 0.2985

d) Less than $700?

(700-951)/96 = -2.61 = Z, so area = 0.0045

The percentage increase in apartment supply is 30%, and the price sensitivity or price elasticity of supply is 3.90, indicating a relatively elastic supply to price changes.

The question deals with the concept of price elasticity of supply, which measures the responsiveness of the quantity supplied to a change in price. To determine the percentage increase in apartment supply, we calculate the change in quantity supplied, divide it by the original quantity supplied, and then multiply by 100 to convert to a percentage:

The original quantity supplied was 10,000 units, and the new quantity supplied is 13,000 units. So the change in quantity supplied is 13,000 - 10,000 = 3,000 units.

The percentage increase in supply is (3,000 units / 10,000 units) × 100% = 30%

Next, we calculate the price sensitivity, also known as price elasticity of supply, using the formula:

Price Elasticity of Supply (Es) = (% Change in Quantity Supplied) / (% Change in Price)

The price increased from $650 to $700, which is an increase of $700 - $650 = $50. The percentage change in price is ($50 / $650) × 100% = 7.69%

Then the price elasticity of supply is 30% / 7.69% = 3.90, indicating that the supply is relatively elastic.

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Answer:

7

Step-by-step explanation:

The surface area is the lateral area plus the base area. The base area is 9*5=45. Since there are two bases, its 90. Subtract that from 286 to get the lateral surface area, which is 196. The lateral surface area is the base perimeter*height. The base perimeter is 9+9+5+5= 28. 196/28=height. The height is 7

Answer:

Height= 7in

Step-by-step explanation:

Use the surface area formula and pulg in.

2(wl)(hl)(hw)=286

divide by 2 to get it to the other side.

143=45+9h+5h

subtract 45 and divide by 14 (9+5)

the height is 7

The restaurant pay $24,192 for 9 pounds of truffles.

The unitary technique involves first determining the value of a single unit, followed by the value of the necessary number of units.

For example, Let's say Ram spends 36 Rs. for a dozen (12) bananas.

12 bananas will set you back 36 Rs. 1 banana costs 36 x 12 = 3 Rupees.

As a result, one banana costs three rupees. Let's say we need to calculate the price of 15 bananas.

This may be done as follows: 15 bananas cost 3 rupees each; 15 units cost 45 rupees.

Given:

A restaurant buys 9 pounds of truffles.

as, 1 pound = 16 ounce

So, the restaurant buys

= 16 x 9

= 144 ounce of truffles.

If truffles cost $168 per ounce.

So, the cost for 9 pounds ruffles is

=  144 x 168

= $ 24,192

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Answer:

  (-2, 3)

Step-by-step explanation:

In the form ...

  y -k = a(x -h)^2

the vertex is (h, k).

Your equation has k = 3, a = -1, h = -2, so the vertex is ...

  (h, k) = (-2, 3)

This mathematics question from middle school algebra asks to solve a system of equations to find the price of pool passes and gym memberships at a recreation center. The problem is partially incomplete due to a cutoff in the additional information needed to set up the second equation. Consequently, we cannot provide a final answer without this information.

The question involves a linear system of equations, which is a topic in algebra within the field of mathematics. The problem presents a scenario where a family purchases a combination of pool passes and gym memberships for a total cost, and another individual purchases a different combination of the same items for a different total cost. To solve, we set up two equations based on the given information.

Step-by-step solution:

Without the full information, this question is incomplete, and thus we cannot provide a final answer.

Answer: gang in la

Step-by-step explanation:

Answer:

The model car is 10 inches long

Step-by-step explanation:

To solve this question, we use conversion of units

Feet to inches.

Each feet has 12 inches.

The car is 15ft long.

So the car has 15*12 = 180 inches.

.5 inches = 9 in.

Rule of three

.5 inches - 9 inches

x inches - 180 inches

[tex]9x = 180*0.5[/tex]

[tex]9x = 90[/tex]

[tex]x = \frac{90}{9}[/tex]

[tex]x = 10[/tex]

The model car is 10 inches long

To find the length of Uni's model car, we convert the actual car's length to inches, set up a proportion with the given scale, and cross-multiply to solve for the model car's length, resulting in a model that is 10 inches long.

The subject matter of the question is related to scale models, which falls under the field of Mathematics. To solve this problem, we need to find the length of the model car based on the given scale and the actual length of the car.

The scale given is 0.5 inches = 9 inches. Firstly, we need to convert the actual length of the car from feet to inches, so we can work in the same units. There are 12 inches in a foot, so a 15 feet long car is 15 x 12 inches long, which is 180 inches. Now, we need to set up a proportion to find the length of the model car:

Actual car length (inch) : Model car length (inch) = Actual scale (inch) : Model scale (inch) 180 inches (actual car length) : x inches (model car length) = 9 inches (actual scale) : 0.5 inches (model scale)

By cross-multiplying, we get:

(180 inches x 0.5 inches) = (x inches x 9 inches)

Dividing both sides by 9 inches, we get:

x inches = (180 inches x 0.5 inches) / 9 inches

So, the length of the model car is:

x inches = 10 inches.

Therefore, Uni's model car is 10 inches long.

Answer:

z statistic = 2.82

Step-by-step explanation:

Sample size = 1200

Null hypothesis, [tex]H_{0}[/tex]: p = 0.58

Alternative hypothesis, [tex]H_{a}[/tex]: p > 0.58

From the null and alternative hypothesis, we can derive that Hypothesized proportion, [tex]p_{0}[/tex] = 0.58 = 58%

Significance level = 2.5% = 0.025

Sample proportion, [tex]p_{1}[/tex] = 0.62 = 62%

Test statistic, z:

[tex]z_{statistic} = \frac{p_{1} -p_{0} }{\sqrt{\frac{p_{0}( 1 -p_{0} )}{n} } }[/tex]

[tex]z_{statistic} = \frac{0.62-0.58}{\sqrt{\frac{0.58(1-0.58)}{1200} } }[/tex]

[tex]z statistic =\frac{0.04}{\sqrt{0.000203} }[/tex] [tex]= \frac{0.04}{0.0142}[/tex]

[tex]z statistic = 2.82[/tex]

The value of the test statistic 'z' for the given hypothesis test is approximately 5.65, calculated using the provided sample proportion, hypothesized population proportion, and sample size.

The question asks for the calculation of the test statistic 'z' for a hypothesis test where the null hypothesis states that the population proportion is equal to .58 and the alternative hypothesis says that the population proportion is greater than .58. This is tested at a 2.5% significance level with a sample size of 1200 and sample proportion of .62.

To calculate the test statistic z, use the formula:

z = (p' - p) / √(p(1 - p)/n)

where:

Now, plug in the values:

z = (0.62 - 0.58) / √(0.58(1 - 0.58)/1200)

z ≈ 5.65

This is the value of the test statistic z for conducting the hypothesis test.

Answer:

90% confidence interval for the true mean speed of all cars on this particular stretch of highway is [68.9517 miles per hour , 72.4483 miles per hour].

Step-by-step explanation:

We are given that a sample of 37 cars traveling on a particular stretch of highway revealed an average speed of 70.7 miles per hour with a standard deviation of 6.3 miles per hour.

Firstly, the pivotal quantity for 90% confidence interval for the true mean is given by;

                            P.Q. = [tex]\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }[/tex]  ~ [tex]t_n_-_1[/tex]

where, [tex]\bar X[/tex] = sample average speed of cars = 70.7 miles per hour

             s = sample standard deviation = 6.3 miles per hour

             n = sample of cars = 37

             [tex]\mu[/tex] = true mean speed

Here for constructing 90% confidence interval we have used One-sample t test statistics as we know don't about population standard deviation.

So, 90% confidence interval for the true mean, [tex]\mu[/tex] is ;

P(-1.688 < [tex]t_3_6[/tex] < 1.688) = 0.90  {As the critical value of t at 36 degree of

                                 freedom are -1.688 & 1.688 with P = 5%}  

P(-1.688 < [tex]\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }[/tex] < 1.688) = 0.90

P( [tex]-1.688 \times {\frac{s}{\sqrt{n} } }[/tex] < [tex]{\bar X-\mu}[/tex] < [tex]1.688 \times {\frac{s}{\sqrt{n} } }[/tex] ) = 0.90

P( [tex]\bar X-1.688 \times {\frac{s}{\sqrt{n} } }[/tex] < [tex]\mu[/tex] < [tex]\bar X+1.688 \times {\frac{s}{\sqrt{n} } }[/tex] ) = 0.90

90% confidence interval for [tex]\mu[/tex] = [ [tex]\bar X-1.688 \times {\frac{s}{\sqrt{n} } }[/tex] , [tex]\bar X+1.688 \times {\frac{s}{\sqrt{n} } }[/tex] ]

                    = [ [tex]70.7-1.688 \times {\frac{6.3}{\sqrt{37} } }[/tex] , [tex]70.7+1.688 \times {\frac{6.3}{\sqrt{37} } }[/tex] ]

                    = [68.9517 miles per hour , 72.4483 miles per hour]

Therefore, 90% confidence interval for the true mean speed of all cars on this particular stretch of highway is [68.9517 miles per hour , 72.4483 miles per hour].

The interpretation of the above interval is that we are 90% confident that the true mean speed of all cars will lie between 68.9517 miles per hour and 72.4483 miles per hour.

Answer:

$220262

Step-by-step explanation:

We are given that an exponential function

[tex]g(x)=190000\cdot(1.03)^x[/tex]

Where x=Number of years

We have to find the value of Evie's house after 5 years .

Substitute the values in the given function

[tex]g(5)=190000\cdot(1.03)^5[/tex]

[tex]g(5)=220262.07[/tex]

[tex]g(5)\approx 220262[/tex]

Hence, the value of Evie's house after 5 years=$220262

Answer:

base : 9

three points : (1,0), (9,1), (81,2)

domain : x>0

range : all real number

asymptote : x=0

Answer:

x=0

Step-by-step explanation:

I need to write a 5-paragraph eassy, so please help me it is base on an article name "Schools in Maryland Allow Elementary Students to Carry Cellphones, by Amanda Lenhart, The Washington Post" here are some pic. I just need help writing two paragraph. I already have my Introduction, Body Paragraph #1 and my Body paragraph #2 just need my Body paragraph #3 and my Conclusion I will give brainlis and 30 pnt.

Prompt:

Write an argumentative essay answering the questions: Should students be allowed to carry cellphones on campus? You must support your claim with evidence from the text. You may also use relevant examples from your own experience, observations, and other readings.

Directions:

Before you begin, read the text below, which presents information about the advantages and disadvantages of carrying a cell phone at school. Use the Student Writing Checklist on the back of this page to plan and write a multi-paragraph essay that addresses the prompt. Use your own words, except when quoting directly from the text.

PLEASE DONT WAST THEM

Answer:

A rectangle and a triangle

Step-by-step explanation:

When you have a negative exponent, you move the base with the negative exponent to the other side of the fraction to make the exponent positive.

For example:

[tex]\frac{1}{2y^{-3}} =\frac{y^3}{2}[/tex]    ("y" is the base with the negative exponent)

[tex]x^{-5}[/tex] or [tex]\frac{x^{-5}}{1} =\frac{1}{x^5}[/tex]

When you multiply an exponent directly to a base with an exponent, you multiply the exponents together.

For example:

[tex](y^3)^2=y^{(3*2)}=y^6[/tex]

[tex](x^2)^4=x^{(2*4)}=x^8[/tex]

[tex](2n)^3[/tex] or [tex](2^1n^1)^3=2^{(1*3)}n^{(1*3)}=2^3n^3=8n^3[/tex]

When you have an exponent of 0, the result will always equal 1

For example:

[tex]x^0=1[/tex]

[tex]5^0=1[/tex]

[tex]y^0=1[/tex]

[tex](6m^{-1}*n^0)^{-3}[/tex]      I think you should first make the exponents positive

[tex]\frac{1}{(\frac{6}{m^1} *n^0)^3}[/tex]    

Since you know:

m = 3

n = -5    Substitute/plug it into the equation

[tex]\frac{1}{(\frac{6}{(3)^1}*(-5)^0)^3 }[/tex]

[tex]\frac{1}{(2*1)^3}[/tex]

[tex]\frac{1}{2^3}[/tex]

[tex]\frac{1}{8}[/tex]      

Answer:

(y-12)/4

Step-by-step explanation:

If g(x) is the inverse of f(x)

and f(x) = 4x + 12

f⁻¹(x) = g(x)

let f(x) be represented as y

f(x) = y

y = 4x + 12

subtract 12 from both sides

y-12= 4x

divide both sides by 4

(y-12)/4 = x

so f  ⁻¹ (y)=  (y-12)/4 so g(x) =  (x-12)/4

Answer:

0.1667

Step-by-step explanation:

There are 6! ways to arrange the hats. The number of ways for which no guest will receive the proper hat is 5! (since there are 5 wrong hats for the first guest, 4 for the second guest, and so on). The probability that no guest will receive the proper hat is:

[tex]P=\frac{5!}{6!}=0.1667[/tex]

The probability is 0.1667.

Answer: 8/10 or 4/5

Step-by-step explanation:

10/10 - 2/10 = 8/10

Answer:

Since 10 - 2 = 8

The fraction of the remaining oceans would be 8/10

And if you simplify both 8 and 10 by 2

Meaning you divide them by two

8 ÷ 2 = 4

10 ÷ 2 = 5

Our new fraction is 4/5

~DjMia~

Answer:

[tex]t=\frac{18.7-20}{\frac{5}{\sqrt{80}}}=-2.326[/tex]    

[tex]df=n-1=80-1=79[/tex]  

[tex]p_v =P(t_{(79)}<-2.326)=0.0113[/tex]  

Step-by-step explanation:

Information given

[tex]\bar X=18.7[/tex] represent the sample mean for the content of active ingredient

[tex]s=5[/tex] represent the sample standard deviation for the sample  

[tex]n=80[/tex] sample size  

[tex]\mu_o =20[/tex] represent the value that we want to test

t would represent the statistic

[tex]p_v[/tex] represent the p value for the test

System of hypothesis

We need to conduct a hypothesis in order to check if the true mean for the active agent is at least 20 mg, the system of hypothesis would be:  

Null hypothesis:[tex]\mu \geq 20[/tex]  

Alternative hypothesis:[tex]\mu < 20[/tex]  

The statistic would be:

[tex]t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}[/tex]  (1)  

Now we can calculate the statistic:

[tex]t=\frac{18.7-20}{\frac{5}{\sqrt{80}}}=-2.326[/tex]    

P value

The degrees of freedom are calculated like this:

[tex]df=n-1=80-1=79[/tex]  

Since is a one left tailed test the p value would be:  

[tex]p_v =P(t_{(79)}<-2.326)=0.0113[/tex]  

Option (e) 0.0113 is correct. The approximate p-value for the lab's one-sample t-test on fluoxetine is 0.0113.

To determine the approximate p-value for the lab's test on fluoxetine, we can use a one-sample t-test. Here's a step-by-step explanation:

(bar{x} - μ) / (s/√n)

where bar{x} is the sample mean, μ is the population mean, s is the sample standard deviation, and n is the sample size.

bar{x} = 18.7, μ = 20, s = 5, n = 80

t = (18.7 - 20) / (5/√80)

t = -1.3 / (5/8.944)

t = -1.3 / 0.559

t = -2.325

Therefore, the correct answer is e. 0.0113.

Complete question:

Fluoxetine, a generic anti-depressant, claims to have, on average, at least 20 milligrams of active ingredient. An independent lab tests a random sample of 80 tablets and finds the mean content of active ingredient in this sample is 18.7 milligrams with a standard deviation of 5 milligrams. If the lab doesn't believe the manufacturer's claim, what is the approximate p-value for the suitable test?

a. 0.0226

b. 0.4885

c. 0.5115

d. 0.15

e. 0.0113

Answer:

So the Null hypothesis is rejected in this case

Step-by-step explanation:

  The number of celebrities is  n = 348

So to solve this we would assume that p is the percentage of people that died on the month preceding their birth month  

  Generally if there is no death postponement then p will be mathematically evaluated as

            [tex]p = \frac{1}{12}[/tex]  

This implies the probability of date in one month out of the 12 months

Now from the question we can deduce that the hypothesis we are going to be testing is  

  [tex]Null Hypothesis \ \ H_0 : p = 0.083[/tex]

This is a hypothesis is stating that a celebrity  dies in the month preceding their birth

  [tex]Alternative \ Hypothesis H_1 : p < 0.083[/tex]

   This is a hypothesis is stating that a celebrity does not die in the month preceding their birth

       is c is the represent probability for each celebrity which either c = 0 or c = 1

Where c = 0 is that the probability  that the celebrity does not die on the month preceding his/ her birth month

     and  c =  1  is that the probability  that the  celebrity dies on the month preceding his/ her birth month

  Then it implies that

   for  

       n= 1 + 2 + 3 + .... + 348  celebrities

Then the sum of c for each celebrity would be  [tex]c_s = 16[/tex]

i.e The number of celebrities that died in the month preceding their birth month

We are told that the significance level is  [tex]\alpha = 0.05[/tex], the the z value of [tex]\alpha[/tex] is

              [tex]z_{\alpha } = 1.65[/tex]

This is obtained from the z-table

Since this test is carried out on the left side of the area under the normal curve then the critical value will be

                 [tex]z_{\alpha } = - 1.65[/tex]

So what this implies is that  [tex]H_o[/tex] will be rejected if

                [tex]z \le -1.65[/tex]

Here z is the test statistics

Now z is mathematically evaluated as follows

                  [tex]z = \frac{c - np}{\sqrt{np_o(1- p_o)} }[/tex]

                [tex]z = \frac{16 - (348 *0.083)}{\sqrt{348*0.083 (1- 0.083)} }[/tex]

                [tex]z =-2.50[/tex]

From our calculation we see that the value of z is less than [tex]-1.65[/tex] so the Null hypothesis will be rejected

   Hence this tell us that the  evidence provided is not enough to conclude  that 16 celebrities died a month to their birth month

The question involves statistical hypothesis testing where the null hypothesis (H0) suggests no significant increase in celebrity deaths before their birth month, and the alternative hypothesis (H1) suggests a significant increase. Using significance level 0.05 and the provided data, the p-value is compared to decide on H0.

The question provided relates to setting up and testing a null hypothesis (H0) against a one-sided alternative hypothesis (H1) in the context of statistical hypothesis testing. Specifically, it involves determining whether the occurrence of celebrity deaths in the month preceding their birth month is statistically significant using a significance level of 0.05. To address this, the null hypothesis would state that there is no significant increase in the frequency of deaths in the month before the celebrities' birth month compared to any other month. The alternative hypothesis would state that there is a significant increase in deaths in the month preceding the birth month of celebrities. We would use the data provided (16 out of 348 celebrities dying in the month before their birth month) to calculate the p-value and compare it with the alpha level of 0.05 to decide whether to reject the null hypothesis or not.

Answer:

60

Step-by-step explanation:

1/2 of 80 is 40.1/4 is 20.40 plus 20 equals 60.

Answer:

80x

Step-by-step explanation:

The probability is needed for this, so I put it as x.

If it was .25, he could expect to score 20 times.

Answer:

We conclude that the college students spend less time on extra-curricular activities than they did ten years ago.

Step-by-step explanation:

We are given that Ten years ago, college students spent an average of 120 hours per semester on extra-curricular activities.

A simple random sample of 100 college students found that in the past year the average number of hours spent per semester in extracurricular activities was 107 hours with a standard deviation of 45 hours.

Let [tex]\mu[/tex] = average time spent on extra-curricular activities.

So, Null Hypothesis, [tex]H_0[/tex] : [tex]\mu[/tex] [tex]\geq[/tex] 120 hours     {means that the college students spend more or equal time on extra-curricular activities than they did ten years ago}

Alternate Hypothesis, [tex]H_A[/tex] : [tex]\mu[/tex] < 120 hours     {means that the college students spend less time on extra-curricular activities than they did ten years ago}

The test statistics that would be used here One-sample t test statistics as we don't know about the population standard deviation;

                       T.S. =  [tex]\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }[/tex]  ~ [tex]t_n_-_1[/tex]

where, [tex]\bar X[/tex] = sample average number of hours spent per semester = 107 hrs

             s = sample standard deviation = 45 hours

             n = sample of college students = 100

So, test statistics  =  [tex]\frac{107-120}{\frac{45}{\sqrt{100} } }[/tex]  ~  [tex]t_9_9[/tex]

                               =  -2.889

The value of t test statistics is -2.889.

Now, the P-value of the test statistics is given by following formula;

               P-value = P( [tex]t_9_9[/tex] < -2.889) = 0.00314

Since, the P-value is less than the level of significance as 0.05 > 0.00314, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which we reject our null hypothesis.

Therefore, we conclude that the college students spend less time on extra-curricular activities than they did ten years ago.

Answer:

For the maximize the total area, the all wire i.e. 26 m should be used for the square.

Step-by-step explanation:

Length of the wire = 26 m

Let Amount of wire cut for square = x

Amount of wire cut for triangle  = 26 - x

Side of the square = [tex]\frac{x}{4}[/tex]

Area of the square = [tex]\frac{x^{2} }{16}[/tex]  ------ (1)

Side of the triangle is given by

[tex]a = \frac{26 - x}{3}[/tex]

Side of the triangle is [tex]a = \frac{26 - x}{3}[/tex]

Area of the triangle is given by

[tex]A = \frac{\sqrt{3} }{4} a^{2}[/tex]

Area of the triangle is

[tex]A =\frac{\sqrt{3} }{36} (26 - x)^{2}[/tex]  ------- (2)

Now the total area = Area of square + Area of triangle

The total area = [tex]\frac{x^{2} }{16} + \frac{\sqrt{3} }{36} (26 - x)^{2}[/tex]  ------- (3)

Differentiate above equation with respect to x we get

[tex]A' = \frac{x}{8} - \frac{\sqrt{3} }{18} (26 - x)[/tex]

Take [tex]A' = 0[/tex]

[tex]\frac{x}{8} - \frac{\sqrt{3} }{18} (26 - x) = 0[/tex] ------- (4)

By solving the above equation we get

x = 11.31 m

Again take [tex]A''[/tex] by differentiating equation  (4)

[tex]\frac{x}{8} + \frac{\sqrt{3} }{18} (26)[/tex]

Which is greater than zero. so the value x = 11.31 m gives the area minimum.

Thus for the maximize the total area, the all wire i.e. 26 m should be used for the square.