Steven bagged 52 pounds of potatoes. About what is that measure in kilograms?

Antawn Jamison is shooting free throws. Making or missing free throws doesn't change the probability that

scoray [572]

Answer: 0.73^9

Step-by-step explanation: Khan

5 0

2 years ago

A survey of the students in Lance’s school found that 58% of the respondents want the school year lengthened, while 42% think it

alina1380 [7]

The margin of error of a given statistic is an amount that is allowed for in case of miscalculation or change of circumstances.

It is usually the radius or half of the width of the confidence interval of that statistic.

Given that a survey of the students in Lance’s school found that 58% of the respondents want the school year lengthened, while 42% think it should remain the same. The margin of error of the survey is ±10%.

This means that 58% ± 10% of the respondents want the school year lengthened, while 42% ± 10% think it should remain the same.

Thus, from 48% to 68% of the respondents want the school year lengthened, while from 32% to 52% think it should remain the same. 

Therefore, according to the survey data, at least 32% of students want the duration of the school year to remain unchanged, and at least 48% want the school year to be lengthened.

7 0

2 years ago

4. The data shows the number of siblings each student in a 9th-grade class has.

antiseptic1488 [7]

Answer:

The correct option is Mean and standard deviation.

Step-by-step explanation:

The data set provided, arranged in ascending order is:

S = {0 , 0 , 0 , 1 , 1 , 1 , 1 , 1 , 1 , 2 , 2 , 2 , 2 , 2 , 2 , 2 , 3 , 3 , 3 , 4 , 4}

Now, it is provided that Macy was absent when the data was collected and the class guessed she had 4 siblings. When she returned, they found out she actually has 9 siblings.

So, the new data set is:

X = {0 , 0 , 0 , 1 , 1 , 1 , 1 , 1 , 1 , 2 , 2 , 2 , 2 , 2 , 2 , 2 , 3 , 3 , 3 , 4 , 9}

The mean of a data set is the average value representing the entire data.

$\bar X=\frac{1}{n}\sum X_{i}$

The median is the middle value of the data set.

The inter-quartile range is the difference between the first and the third quartile.

The standard deviation is the value that represents how dispersed the data values are.

$SD=\sqrt{\frac{1}{n}\sum (X_{i}-\bar X)^{2}}$

From all the above statistic, the mean and standard deviation are the only ones that uses all the observations in the data set to compute their value.

So, on changing one of the 4s by a 9, the set of measures that will be affected are the mean and standard deviation.

Thus, the correct option is Mean and standard deviation.

3 0

2 years ago

(1 point) Let P(t) be the performance level of someone learning a skill as a function of the training time t. The derivative dPd

monitta

Answer:

$P(t)=M+Ce^{-kt}$

Step-by-step explanation:

Given the differential model

$\dfrac{dP}{dt}=k[M-P(t)]$

We are required to solve the equation for P(t).

$\dfrac{dP}{dt}=kM-kP(t)\\$Add kP(t) to both sides\\\dfrac{dP}{dt}+kP(t)=kM\\$Taking the integrating factor\\e^{\int k dt} =e^{kt}\\$Multiply all through by the integrating factor\\\dfrac{dP}{dt}e^{kt}+kP(t)e^{kt}=kMe^{kt}\\\dfrac{dP}{dt}e^{kt}=kMe^{kt}\\(Pe^{kt})'=kMe^{kt} dt\\$Take the integral of both sides with respect to t\\\int (Pe^{kt})'=\int kMe^{kt} dt\\Pe^{kt}=kM \int e^{kt} dt\\Pe^{kt}=\dfrac{kM}{k} e^{kt} + C_0, C_0$ a constant of integration$

$Pe^{kt}=Me^{kt} + C\\$Divide both side by e^{kt}\\P(t)=M+Ce^{-kt}\\P(t)=M+Ce^{-kt}\\$Therefore:\\P(t)=M+Ce^{-kt}$

4 0

1 year ago

The geometric average of -12%, 20%, and 25% is _________.

Grace [21]

20.28% is the answer

8 0

1 year ago