Which of the following could be appropriate values for an equation to be in Standard Form (Ax + By = C)?

In expanded notation, the hexadecimal 74AF16 is (7*4096) + (4*256) + (A*16) + (F*1). When converting from hexadecimal to decimal

Feliz [49]

Answer:

F is assigned the value of 15

$74AF_{16} = 29871_{10}$

Step-by-step explanation:

Hexadecimal number system is base 16 and it contain the following numbers:

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E, F

A has a value of 10

B has a value of 11

C has a value of 12

D has a value of 13

E has a value of 14

F has a value of 15

By completing the expanded notation:

$(7*4096) + (4*256) + (A * 16) + (F *1)\\= (7*4096) + (4*256) + (10 * 16) + (15 *1)\\= 28672 + 1024 + 160 + 15\\= 29871$

6 0

2 years ago

1. WHEELS Zack is designing wheels for a concept car.

mars1129 [50]

Answer:

9 inches

Step-by-step explanation:

Diameter=D

Radius=R

D=(2)R

(18)=(2)R

18/2=2/2 R

9=R

4 0

2 years ago

Suppose the probability of an athlete taking a certain illegal steroid is 10%. A test has been developed to detect this type of

Travka [436]

Answer:

93.25% probability that they have taken this steroid

Step-by-step explanation:

Bayes Theorem:

Two events, A and B.

$P(B|A) = \frac{P(B)*P(A|B)}{P(A)}$

In which P(B|A) is the probability of B happening when A has happened and P(A|B) is the probability of A happening when B has happened.

In this question:

Event A: Positive test

Event B: Taking the steroid.

Suppose the probability of an athlete taking a certain illegal steroid is 10%.

This means that $P(B) = 0.1$

Given that the athlete has taken this steroid, the probability of a positive test result is 0.995.

This means that $P(A|B) = 0.995$

Positive test:

99.5% of 10%(If the athlete has taken).

100-99.2 = 0.8% of 100-10 = 90%(Athlete has not taken)

Then

$P(B) = 0.995*0.1 + 0.008*0.9 = 0.1067$

Given that a positive test result has been observed for an athlete, what is the probability that they have taken this steroid

$P(B|A) = \frac{P(B)*P(A|B)}{P(A)} = \frac{0.1*0.995}{0.1067} = 0.9325$

93.25% probability that they have taken this steroid

4 0

2 years ago

A researcher is estimating the mean income of residents in a large city. The income variable is usually skewed to the right. She

Gnom [1K]

Answer: We are 95% confident that the mean income for all residents of this city is between $26700 and $35400.

Step-by-step explanation:

We know that a 95% confidence interval given an interval of values that we can be 95% sure , that it contains the true mean of the population, not 95% of data lies in it.

Given : A researcher is estimating the mean income of residents in a large city. The income variable is usually skewed to the right. She collects a random sample of 25 people.

The resulting 95% confidence interval is ($26700, $35400).

Then, valid conclusion will be : We are 95% confident that the mean income for all residents of this city is between $26700 and $35400.

5 0

2 years ago

Suppose a research paper states that the distribution of the daily sea-ice advance/retreat from each sensor is similar and is ap

strojnjashka [21]

Answer:

The value of the parameter is λ is 0.03692

Step-by-step explanation:

Consider the provided function.

$f(x) = 0.5\lambda e^{-\lambda |x|}$ for −∞ < x < ∞.

It is given that standard deviation is given as 38.3 km.

Now we need to calculate the value of parameter λ.

The general formula for the probability density function of the double exponential distribution is: $f(x)=\frac{e^{-|\frac{x-\mu}{\beta}|}}{2\beta}$

Where μ is the location parameter and β is the scale parameter.

Compare the provided equation with the above formula we get.

$\lambda=\frac{1}{\beta}$ and μ = 0.

Standard deviation = √2β

$S.D=\sqrt{2} \beta\\\beta=\frac{38.3}{\sqrt{2}}\\\beta=27.08219$

Now substitute the value of β in $\lambda=\frac{1}{\beta}$ .

$\lambda=\frac{1}{27.08219}=0.03692$

Hence, the value of the parameter is λ is 0.03692

7 0

2 years ago