COHEA 8th Grade Test 1

A superintendent of schools wonders if scores on a test of science achievement differ for female and male 8th grade students. To

lina2011 [118]

Answer:

b. 2-independent sample t-test

Correct for this case is the best test since the result from each group are independent and we can compare the sample means between th two groups

Step-by-step explanation:

For this case the intention is try to test if scores on a test of science achievement differ for female and male 8th grade students. let's analyze the possible options for this case:

a. 2-dependent sample t-test

False is not possible since the nature of the gender is not possible to conclude tha the results from boys and girls are dependent

b. 2-independent sample t-test

Correct for this case is the best test since the result from each group are independent and we can compare the sample means between the two groups

c. Correlation

False we can't compare the means of interest with a correlation coefficient since that's not the purpose of the study

d. 2-way ANOVA

False we have just one variable between two groups so is not possible to apply a 2 way ANOVA

5 0

2 years ago

The average sunflower has 34 petals. What is the best estimate of the total number of petals on 57 sunflowers?

Marianna [84]

The best estimate would be to assume that they all have an average number of petals.

An estimate of 1,938 petals

4 0

2 years ago

The square of a number decreased by 3 times the number 28 find all possible values for the number

stealth61 [152]

Question:

The square of a number decreased by 3 times the number is 28 find all possible values for the number

Answer:

The possible values of number are 7 and -4

Solution:

Given that the square of a number decreased by 3 times the number is 28

To find: all possible values of number

Let "a" be the unknown number

From given information,

square of a number decreased by 3 times the number = 28

$a^2 - 3a = 28$

$a^2 - 3a - 28 = 0$

Let us solve the above quadratic equation

$\text {For a quadratic equation } a x^{2}+b x+c=0, \text { where } a \neq 0$

$x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$

Using the above formula,

$\text { For } a^{2}-3 a-28=0 \text { we have } a=1, b=-3, c=-28$

$\begin{aligned}&a=\frac{-(-3) \pm \sqrt{(-3)^{2}-4(1)(-28)}}{2 \times 1}\\\\&a=\frac{3 \pm \sqrt{9+112}}{2}\\\\&a=\frac{3 \pm \sqrt{121}}{2}=\frac{3 \pm 11}{2}\\\\&a=\frac{3+11}{2} \text { or } a=\frac{3-11}{2}\\\\&a=7 \text { or } a=-4\end{aligned}$

Thus the possible values of number are 7 and -4

5 0

2 years ago

The library charged Monica $50 for a book that was 10 days overdue

barxatty [35]

Erm what’s your question?

8 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