1) IQ scores have a mean of 100 and a standard deviation of 16. Albert Einstein reportedly had an IQ of 160. What is the difference between Einsteins IQ and the mean?
2) IQ scores have a mean of 100 and a standard deviation of 16. Albert Einstein reportedly had an IQ of 160. How many standard deviations is that from the mean?
3) IQ scores have a mean of 100 and a standard deviation of 16. Albert Einstein reportedly had an IQ of 160. Convert Einsteins IQ score to a z score.
4) IQ scores have a mean of 100 and a standard deviation of 16. Albert Einstein reportedly had an IQ of 160. If we consider usual IQ scores to be those that convert z scores between -2 and 2, is Einstein's IQ usual or unusual?
Z-scores
Given a population that is normally distributed with mean μ and standard deviation σ, the z-score of a value x from the population is the distance, measured in standard deviation units, between x and μ. It is computed by applying the following formula:

\((\stackrel{x_1}{-2}~,~\stackrel{y_1}{10})\qquad (\stackrel{x_2}{-9}~,~\stackrel{y_2}{-5}) \\\\\\ \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{-5}-\stackrel{y1}{10}}}{\underset{run} {\underset{x_2}{-9}-\underset{x_1}{(-2)}}} \implies \cfrac{-15}{-9 +2} \implies \cfrac{ -15 }{ -7 } \implies \cfrac{ 15 }{ 7 }\)

\(\begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{10}=\stackrel{m}{ \cfrac{ 15 }{ 7 }}(x-\stackrel{x_1}{(-2)}) \implies y -10 = \cfrac{ 15 }{ 7 } ( x +2) \\\\\\ y-10=\cfrac{ 15 }{ 7 }x+\cfrac{ 30 }{ 7 }\implies y=\cfrac{ 15 }{ 7 }x+\cfrac{ 30 }{ 7 }+10\implies {\Large \begin{array}{llll} y=\cfrac{ 15 }{ 7 }x+\cfrac{100}{7} \end{array}}\)

The probability for all five answers is greater than 0.

The normal distribution is a type of continuous probability distribution, and probability can never equal 0 for a continuous distribution. The probability of a particular value (in this case, 8) is the area under the normal curve at that point. Since the area under a continuous distribution is always greater than 0, the probability of a specific value (A) is not 0.

The probability of a range of values (D) is calculated by subtracting the area under the normal curve to the left of the lower value, from the area under the normal curve to the right of the higher value. This calculation would produce a value greater than 0.

The probability of values greater than a given number (B) is calculated by subtracting the area under the normal curve to the left of the given number, from the area under the normal curve to the right of the given number. This calculation would produce a value greater than 0.

The probability of values less than a given number (C) is calculated by subtracting the area under the normal curve to the right of the given number, from the area under the normal curve to the left of the given number. This calculation would produce a value greater than 0.

Finally, the probability of values less than one number or greater than another number (E) is calculated by subtracting the area under the normal curve to the left of the lower number, from the area under the normal curve to the right of the higher number. Again, this calculation would produce a value greater than 0.

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The required,  measure of the unknowns is given as x = 3  and y = 3/√2

A right triangle is given, to determine the unknowns x and y by applying the trig ratio:

cos45 = [3√2/2]/x
x =  [3√2/2]/cos45
x = 3

Similarly,
sin45 = y / x
1/√2 = y / 3
y = 3/√2 or [3√2/2]

Thus, the required,  measure of the unknowns is given as x = 3  and y = 3/√2

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Statement 1: Some cats like turkey, the form is I, the subject class is Cats, and the predicate class is Turkey, statement 2: There are burglars coming in the window, the form is E, the subject class is Burglars, and the predicate class is Not coming in the window and statement 3: Everyone will be robbed, the form is A, the subject class is Everyone, and the predicate class is Being robbed.

The given categorical propositions and their forms are as follows:

(1) Some cats like turkey - Form: I:

Subject class: Cats,

Predicate class: Turkey

(2) There are burglars coming in the window - Form: E:

Subject class: Burglars,

Predicate class: Not coming in the window

(3) Everyone will be robbed - Form: A:

Subject class: Everyone,

Predicate class: Being robbed

In the first statement:

Some cats like turkey, the form is I, the subject class is Cats, and the predicate class is Turkey.

In the second statement:

There are burglars coming in the window, the form is E, the subject class is Burglars, and the predicate class is Not coming in the window.

In the third statement:

Everyone will be robbed, the form is A, the subject class is Everyone, and the predicate class is Being robbed.

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When a range of values is used to estimate a population parameter, it is called an interval estimate. The correct answer is option D.

An interval estimation is the evaluation of a parameter of a population by computing an interval, or range of values, within which the parameter is most likely to be located. An interval estimate gives you a range of values where the parameter is expected to lie. In other words, an interval estimate is a range in which the population parameter probably falls for a given level of confidence. We have to establish a level of confidence because we can never be 100 percent sure that the population mean falls within our confidence interval. A confidence interval is the most common type of interval estimate.

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Answer:

\(Given~inequality~is\\2(x-3) -5x<x-2\\or, 2x-6-5x<x-2\\or, -3x-6<x-2\\Subtracting~x~on~both~sides,\\-3x-x-6<x-x-2\\or, -4x-6<-2\\Adding~'6'~on~both~sides,\\-4x-6+6<-2+6\\or, -4x<4\\Dividing~'-4'~on~both~sides,\\\frac{-4x}{-4}>\frac{4}{-4} \\[Sign ~of~ inequality ~is~ reversed ~on ~dividing~ on~multiplying ~both ~sides~ by~ a~ negative~ number]\\or, x>-1 \\\)

So, the required answer is C. [Look at the screenshot]

Answer:

49

Step-by-step explanation:

90-51=49

The cost of a new home, built of standard materials, with standard techniques and design replacement cost and reproduction cost may be the same.

When analyzing the cost of a new home built with standard materials, techniques, and design, the relationship between replacement cost and reproduction cost may vary. Option 4) Replacement cost and reproduction cost may be the same, is the most accurate statement.

Replacement cost refers to the amount required to replace or rebuild a property, while reproduction cost refers to the cost of constructing an exact replica of the original property. These costs can be the same in some cases, but factors such as market conditions, construction costs, and availability of materials can affect the costs differently. Cost and value are not always the same, as value considers factors such as location, demand, and economic conditions.

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Answer:

  San Francisco

Step by step explanation:

Given that 1 US gallon of petrol costs $3.34 in San Francisco, and 1 litre of petrol costs 107.6p in Manchester, with conversion rates being 3.785 litres per gallon and $1.46 per 1£, you want to know the city where the value is better.

We can compare the costs by converting the rate in Manchester to a dollar cost per gallon.

The cost of 1 litre in Manchester is £(107.6/100) = £1.076. The cost of 1 US gallon of petrol in Manchester is 3.785 × £1.076 = £4.07226. Converted to dollars, that is 4.07266 × $1.46 ≈ $5.95.

In San Francisco, the cost of 1 US gallon is $3.34.Therefore, the cost of 1 US gallon of petrol in San Francisco is a better value. The price of petrol is higher in Manchester than in San Francisco.

Answer:

\(8.8 ~ \dfrac{kg}{m^3}\)

Step-by-step explanation:

\(\text{Note that,}\\1~ kg = 1000~ g\\\\1 ~m = 100 ~cm \\\\\text{Now,}\\\\0.0088 \dfrac{g}{cm^3}\\\\\\=\dfrac{0.0088~ g}{ (1~ cm)^3}\\\\\\=\dfrac{0.0088\times 10^{-3}~ kg}{(1 \times 10^{-2}~ m)^3 }\\\\\\=\dfrac{0.0088 \times 10^{-3}~ kg}{1 \times 10^{-6} ~m^3}\\\\\\\)

\(=0.0088 \times 10^{-3+6} ~ \dfrac{kg}{m^3}\\\\\\=0.0088 \times 10^{3} ~\dfrac{kg}{m^3}\\\\\\= 8.8 ~ \dfrac{kg}{m^3}\)

Answer:

Negative

Step-by-step explanation:

Answer:

Negative

Step-by-step explanation:

A positive divided by a negative is a negative.

The statements that relate each function are given as follows:

1. At 3 seconds, the toy rocket is higher than the drone: f(3) > g(3).

2. At the start, the toy rocket is 25ft above the drone: f(0) - g(0) = 25.

The four inequality symbols, along with their meaning, are presented as follows:

The functions for this problem are given as follows:

At 3 seconds, we have that the toy rocket is higher than the drone, that is, f(3) is greater than g(3), hence the inequality is given as follows:

f(3) > g(3).

At the start, the heights are given as follows:

The difference is of 25 ft, hence the expression is given as follows:

f(0) - g(0) = 25.

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Converting 1,000 mg to 1 gram would require moving the decimal place zero places to the left.

Unit conversion is a process with multiple steps that involves multiplication or division by a numerical factor or, particularly a conversion factor. The process may also require selection of the correct number of significant digits, and rounding. Different units of conversion are used to measure different parameters. By definition conversion of units means the conversion between different units and measurements of the same quantity done by the process of multiplication or division. In maths, conversion is the process of changing the value of one form to another for example inches to millimeters, or liters to gallons. Units are used for measuring length, measuring weight, measuring capacity, measuring temperature, and measuring speed.

If we want to calculate how many Grams are 1000 Milligrams we have to multiply 1000 by 1 and divide the product by 1000. So for 1000, we have (1000 × 1) ÷ 1000 = 1000 ÷ 1000 = 1 Gram.

So finally 1000 mg = 1 g

Thus, converting 1,000 mg to 1 gram would require moving the decimal place zero places to the left.

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Answer:

5+4÷8-7x1

Just do out the equations :)

The total amount assessed on the cargo weighing 6,520 tons at the Marin Port Of Entry (POE) was $391.20.

According to the given information,

The cargo weighs 6,520 tons and is assessed a fee of 6 cents per ton.

We can set up the formula to calculate the total amount assessed,

⇒ Total amount assessed = Weight of cargo x Fee per ton

We can substitute the given values into the formula,

⇒ Total amount assessed = 6,520 tons x $0.06/ton

To simplify this calculation,

we can first multiply the weight of the cargo by the fee per ton,

⇒ 6,520 tons x $0.06 = $391.20

Therefore,

The required total amount assessed on the cargo weighing was $391.20.

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Answer:

1/6

Explain:

The reciprocal of the slope is its perpendicular.

6/1 --> 1/6

m<1 is 50

m<2 is 130

m<3 is 50

m<4 is 130

m<5 is 40

m<6 is 140

m<7 is 90

m<8 is 140

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Answer:

32

Step-by-step explanation:

3 x 8 = 24, 24 + 8 = 32

Answer:

Step-by-step explanation:

Given

The equation for the interest:

So $17500 invested at 6.5% and $12500 at 4.2%

Answer:

yes

Step-by-step explanation:

Yes, we can conclude that there is a positive correlation between the success of college students (measured by cumulative grade point average) and the income of their respective families.

For testing whether there is a significant correlation between two variables, we need to calculate the correlation coefficient r.

Given that the sample size (n) is 20, and the correlation coefficient (r) is 0.40. The test statistic value, t can be calculated using the formula:

(\(t = (r * \sqrt{n - 2} /\sqrt{1 - r^2} )\))

Therefore, substituting the values,

(\(t = (0.40 *\sqrt{20 - 2} / \sqrt{1 - 0.4^2} )\))

= 2.53

Using the t-table with 18 degrees of freedom (df = n - 2 = 20 - 2 = 18) at a significance level of 0.01, we find that the critical value of t is 2.878.

Since the calculated value of t is less than the critical value of t, we fail to reject the null hypothesis.

Therefore, we can conclude that there is a positive correlation between the success of college students (measured by cumulative grade point average) and the income of their respective families.

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Answer:

5/64

Step-by-step explanation:

5/8 : 8 = 5/8 * 1/8 = 5/64

Answer:

12 4/5

Step-by-step explanation:

simplified

The given expressions, when A=1, B=0, and C=1, X evaluates to 1.001; when A=B=C=1, X evaluates to 1.111; and when A=B=C=0, X evaluates to 0.000. These evaluations are based on the given values of A, B, and C, and the notation ¯¯¯¯¯¯BC represents the complement of BC.

To evaluate the expression X = A.¯¯¯¯¯¯BC, we substitute the given values of A, B, and C into the expression.

(a) For A = 1, B = 0, and C = 1:

X = 1.¯¯¯¯¯¯01

To find the complement of BC, we replace B = 0 and C = 1 with their complements:

X = 1.¯¯¯¯¯¯01 = 1.¯¯¯¯¯¯00 = 1.001

(b) For A = B = C = 1:

X = 1.¯¯¯¯¯¯11

Similarly, we find the complement of BC by replacing B = 1 and C = 1 with their complements:

X = 1.¯¯¯¯¯¯11 = 1.¯¯¯¯¯¯00 = 1.111

(c) For A = B = C = 0:

X = 0.¯¯¯¯¯¯00

Again, we find the complement of BC by replacing B = 0 and C = 0 with their complements:

X = 0.¯¯¯¯¯¯00 = 0.¯¯¯¯¯¯11 = 0.000

In conclusion, when A = 1, B = 0, and C = 1, X evaluates to 1.001. When A = B = C = 1, X evaluates to 1.111. And when A = B = C = 0, X evaluates to 0.000. The evaluation of X is based on substituting the given values into the expression A.¯¯¯¯¯¯BC and finding the complement of BC in each case.

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A drug administered 50 mg every three hours and the drug decays 12% per hour, the limiting value is approximately 416.67 mg.

To find the limiting value, we need to determine the amount of the drug remaining after each administration and observe how it approaches a stable value over time.

First, let's calculate the decay factor per hour. The drug decays by 12% per hour, which means it retains 88% of its previous value after each hour.

Decay factor = 1 - 0.12 = 0.88

Now, let's calculate the amount of drug remaining after each administration:

After 1st administration: 50 mg

After 2nd administration: 50 mg * 0.88 = 44 mg

After 3rd administration: 44 mg * 0.88 = 38.72 mg

After 4th administration: 38.72 mg * 0.88 = 34.04 mg

After 5th administration: 34.04 mg * 0.88 = 29.92 mg

As we can see, the amount of drug remaining decreases with each administration, approaching a limiting value. To find this limiting value, we can continue the pattern or use a formula.

The formula for the limiting value of a drug administered every three hours is:

Limiting value = dosage / (1 - decay factor)

In this case, the dosage is 50 mg, and the decay factor is 0.88.

Limiting value = 50 mg / (1 - 0.88) = 50 mg / 0.12 ≈ 416.67 mg (rounded to two decimal places)

Therefore, the limiting value is approximately 416.67 mg.

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The probability that the dog is still lost at the end of the second day is 0.41

To solve this problem, we can use Bayes' theorem, which allows us to update the probability of an event based on new information.

Let A be the event that the dog is in forest A, and B be the event that the dog is in forest B. Let F be the event that the dog is found within two days.

We want to calculate P(F'), the probability that the dog is still lost at the end of the second day. We can use the law of total probability to express this as

P(F') = P(F'|A) × P(A) + P(F'|B) × P(B)

where P(F'|A) is the probability of not finding the dog within two days given that the dog is in forest A, P(F'|B) is the probability of not finding the dog within two days given that the dog is in forest B, and P(A) and P(B) are the prior probabilities of the dog being in forest A and forest B, respectively.

Using the information given in the problem, we can calculate

P(F'|A) = 1 - 0.5 = 0.5 (the probability of not finding the dog in forest A within two days is 1 - 0.5 = 0.5)

P(F'|B) = 1 - 0.8 = 0.2 (the probability of not finding the dog in forest B within two days is 1 - 0.8 = 0.2)

P(A) = 0.7 (the prior probability of the dog being in forest A)

P(B) = 0.3 (the prior probability of the dog being in forest B)

Plugging these values into the formula above, we get

P(F') = 0.5 × 0.7 + 0.2 × 0.3 = 0.41

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The given question is incomplete, the complete question is:

Oscar has lost his dog; there is a 70% probability it is in forest A and a 30% chance it is forest B. If the dog is in forest A and Oscar looks there for a day, he has a 50% change of finding the dog. If the dog is in forest B and Oscar looks there for a day, he has an 80% chance of finding the dog. what is the probability that the dog is still lost at the end of the second day?

Answer:

Step-by-step explanation:

radius r=16/2=8 in

l=17 in

curved surface=πrl=π×8×17=136 π in²

area of circular base=πr²=π×8²=64 πin²

total surface area=136 π+64 π=200 π in²

Answer:

m = 2

Step-by-step explanation:

2m + 2 = 6

2m = 6 - 2

2m/2 = 4/2

m = 2

m=2

Subtract 2 from each side and then divide 2

The iron pipe weighs 0.9375 kg.

In arithmetic, a product is the result of multiplication or an expression that identifies elements to be accelerated.

In math, to multiply method to feature the same agencies. when we multiply, the number of factors inside the organization increases. The two factors and the product are parts of a multiplication hassle. inside the multiplication problem, 6 × nine = 54, the numbers 6 and 9 are the elements, even as the variety 54 is the product.

2/5 m = 0.4 m

3/8 kg = 0.375 kg

2/5 m = 3/8 kg

0.4 m = 0.375 kg

Multiplying both sides by 2.5,

1 m = 0.9375 kg

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