An education researcher would like to test whether 2nd graders retain or lose knowledge during the summer when they are presumably not in school. She asks nine 2nd graders to take a comprehension exam at the end of the school year (May), and then asks those same students to come back after the summer (late August) to retake a different but equivalent exam, to see if their level of comprehension has changed. Using the data below, test this hypothesis using an (alpha level of 0.05.)

May August
95 100
72 80
78 95
50 65
89 85
92 98
75 70
90 96
65 87

Required:
What is the appropirate test?

Answer: [tex]10,000\ lb.ft[/tex]

Step-by-step explanation:

Given

Initial weight of the bucket is [tex]145\ lb[/tex]

It is lifted at constant rate and rate of sand escaping is [tex]0.5\ lb/ft[/tex]

At any height weight of the sand is [tex]w(h)=145-0.5h[/tex]

Work done is given by the product of applied force and displacement or the area under weight-displacement graph

from the figure area is given by

[tex]\Rightarrow W=\int_{0}^{80}\left ( 145-0.5h \right )dh\\\\\Rightarrow W=\left | 145h-\dfrac{0.5h^2}{2} \right |_0^{80}\\\\\Rightarrow W=\left [ 145\times 80-\dfrac{0.5(80))^2}{2} \right ]-0\\\\\Rightarrow W=11,600-1600\\\\\Rightarrow W=10,000\ lb.ft[/tex]

Answer:

7x is monomial according to question.

Answer:

17690

Step-by-step explanation:

The amount Michael owes is given by :

y = - 290x +18,560

x = number of months ; y = amount owed

Amount owed after 3 months :

Put x = 3 in the equation :

y= - 290(3) +18,560

y = - 870 + 18560

y = 17690

According to the general form of a linear equation :

y = mx + c

Where, m = slope ; c = intercept

The slope = - 290 ; Intercept = 18560

Slope is the change in y per unit change in x ; The slope means that amount owed, y decreases by 290 per month

The intercept shows that the original amount borrowed is 18560

Amount Michael owes his father decreases by 290 per month.

9514 1404 393

Answer:

Step-by-step explanation:

A) If we let x represent "a number", then "seven times a number" is 7x. When 5 is subtracted from that, we have ...

  7x -5 . . . . . 5 subtracted from 7 times a number

This is said to be 9, so ...

  7x -5 = 9 . . . . . . . 5 from 7 times a number is 9. Your equation.

__

B) To solve this, we can add 5 to both sides. This eliminates the constant term we don't want on the left.

  7x = 14

Now, we can divide by 7, the coefficient of x that we don't want.

  x = 14/7

  x = 2

Solution :

We have given two baskets :

[tex]$H_1$[/tex] : 8 apples + 2 bananas + 6 cantaloupes = 16 fruits

[tex]$H_2$[/tex] : 6 apples + 4 bananas = 10 fruits

A fair coin is made to flipped. If the [tex]\text{flip}[/tex] results is head, then the fruit is selected from a basket [tex]$H_1$[/tex].

If the flip results in tail, then the fruit is selected from the basket [tex]$H_2$[/tex].

Probability of head P(H) = [tex]1/2[/tex]

Probability of tail P(T) = [tex]1/2[/tex]

if given event is :

B = selected fruit is BANANA

We have to calculate : P(T|B)

Probability of banana if the flip results is head P(B|H) = [tex]$\frac{2}{16}$[/tex]

Probability of banana if the flip results is tail P(B|T) = [tex]$\frac{4}{10}$[/tex]

From the Bayes' theorem :

Probability of flip results is tail when selected fruit is BANANA.

[tex]$P(T|B) = \frac{P(B|T)\ P(T)}{P(B|T) \ P(T) + P(B|H)\ P(H)}$[/tex]

            [tex]$=\frac{\frac{4}{10} \times \frac{1}{2}} {\frac{4}{10} \times \frac{1}{2} + \frac{2}{16} \times \frac{1}{2}}$[/tex]

            [tex]$=\frac{\frac{1}{5}}{\frac{1}{5}+\frac{1}{16}}$[/tex]

            [tex]$=\frac{\frac{1}{5}}{\frac{21}{80}}$[/tex]

            [tex]$=\frac{16}{21}$[/tex]

∴  [tex]$P(\ T|B\ )=\frac{16}{21}$[/tex]

Answer:

a) 0.55 = 55% probability that the person is traveling on business

b) 0.14 = 14% probability that the person is traveling for business on a privately owned plane.

c) 0.2545 = 25.45% probability that the person arrived on a privately owned plane, given that the person is traveling for business reasons.

d) 0.2 = 20% probability that the person is traveling on business, given that the person is flying on a commercially owned plane.

Step-by-step explanation:

Conditional Probability

We use the conditional probability formula to solve this question. It is

[tex]P(B|A) = \frac{P(A \cap B)}{P(A)}[/tex]

In which

P(B|A) is the probability of event B happening, given that A happened.

[tex]P(A \cap B)[/tex] is the probability of both A and B happening.

P(A) is the probability of A happening.

Question a:

50% of 60%(major airlines)

70% of 20%(privately owned airplanes)

80% of 100 - (60+20) = 20%(comercially owned airplanes). So

[tex]p = 0.5*0.5 + 0.7*0.2 + 0.8*0.2 = 0.55[/tex]

0.55 = 55% probability that the person is traveling on business.

Question b:

70% of 20%, so:

[tex]p = 0.7*0.2 = 0.14[/tex]

0.14 = 14% probability that the person is traveling for business on a privately owned plane.

Question c:

Event A: Traveling for business reasons.

Event B: Privately owned plane.

0.55 = 55% probability that the person is traveling on business.

This means that [tex]P(A) = 0.55[/tex]

0.14 = 14% probability that the person is traveling for business on a privately owned plane.

This means that [tex]P(A \cap B) = 0.14[/tex]

Desired probability:

[tex]P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.14}{0.55} = 0.2545[/tex]

0.2545 = 25.45% probability that the person arrived on a privately owned plane, given that the person is traveling for business reasons.

Question d:

Event A: Commercially owned plane.

Event B: Business

80% of those arriving on other commercially owned planes are traveling for business reasons.

This means that:

[tex]P(B|A) = 0.2[/tex]

0.2 = 20% probability that the person is traveling on business, given that the person is flying on a commercially owned plane.

Answer:

-49

8

Step-by-step explanation:

-196/4 =

-49

(3+u)^2

-------------

8

Let u=5

(3+5)^2

-------------

8

(8)^2

-------------

8

64

-----

8

8

Answer:

The answer is A

Step-by-step explanation:

(-2,-3) (3,-4) (0,-1) (-7,3)

inverse means the opposite signs

(2,3) (-3,4) (0,1) (7,-3)

Answer:

a) 0.18 = 18% probability a randomly chosen business customer flying with Global is on a jumbo jet.

b) 0.3 = 30% probability a randomly chosen non-business customer flying with Global is on an ordinary jet.

Step-by-step explanation:

Conditional Probability

We use the conditional probability formula to solve this question. It is

[tex]P(B|A) = \frac{P(A \cap B)}{P(A)}[/tex]

In which

P(B|A) is the probability of event B happening, given that A happened.

[tex]P(A \cap B)[/tex] is the probability of both A and B happening.

P(A) is the probability of A happening.

Question a:

Event A: Jumbo

Event B: Business

60% of its capacity is provided on jumbo jets.

This means that [tex]P(A) = 0.6[/tex]

On jumbo jets, 30% of the passengers are on business

This means that [tex]P(B|A) = 0.3[/tex]

Desired probability:

We want to find [tex]P(A \cap B)[/tex], so:

[tex]P(B|A) = \frac{P(A \cap B)}{P(A)}[/tex]

[tex]P(A \cap B) = P(B|A)P(A) = 0.6*0.3 = 0.18[/tex]

0.18 = 18% probability a randomly chosen business customer flying with Global is on a jumbo jet.

b) What is the probability a randomly chosen non-business customer flying with Global is on an ordinary jet?

Event A: Ordinary

Event B: Non-business

60% of its capacity is provided on jumbo jets.

So 100 - 60 = 40% are ordinary, which means that [tex]P(A) = 0.4[/tex]

On ordinary jets 25% of the passengers are on business.

So 100 - 25 = 75% are non-business, that is [tex]P(B|A) = 0.75[/tex]

Desired probability:

We want to find [tex]P(A \cap B)[/tex], so:

[tex]P(B|A) = \frac{P(A \cap B)}{P(A)}[/tex]

[tex]P(A \cap B) = P(B|A)P(A) = 0.75*0.4 = 0.3[/tex]

0.3 = 30% probability a randomly chosen non-business customer flying with Global is on an ordinary jet.

Answer:

ACB = JCB

Step-by-step explanation:

ASA means angle - (included) side - angle.

we have one angle confirmed.

we have the connected side BC confirmed (the diagram shows that the side is shared, so it is not only congruent, it is actually identical).

now we need confirmation for the angle at the other end point of that side.

Answer:

The x-intercept is -8, the y-intercept is 4, and the slope is 1/2.

Step-by-step explanation:

The x-intercept is -8, the y-intercept is 4, and the slope is 1/2.

Answer: See below

Concept:

There are different forms of linear equations:

Solve:

Given: y - 3 = 1/2 (x + 2)

Here, we can see the linear equation is in the form of point-slope form.

x₁ = -2

y₁ = 3

m = 1/2

Point included in the graph = (-2, 3)

Slope of the graph = 1/2

Hope this helps!! :)

Please let me know if you have any questions

Step-by-step explanation:

We determined above that the greatest perfect square from the list of all factors of 96 is 16.

Step-by-step explanation:

1 since a tank holds 121/2 gallons of gas 121/2 multiplied by two it gives 25 -30 it gives 5 450 divided by two it gives 225-25 i guess.

Hello there!

Previously, we learnt that to solve the equation, we have to isolate the sin, cos, tan, etc first.

First Question

The first question has sin both sides. Notice that if we move sin(theta) to left. We get:-

[tex] \displaystyle \large{2 {sin}^{2} \theta - sin \theta = 0}[/tex]

We can common factor out the expression.

[tex] \displaystyle \large{sin \theta(2sin \theta - 1) = 0}[/tex]

It is a trigonometric equation in quadraric pattern.

We consider both equations:-

First Equation

[tex] \displaystyle \large{sin \theta = 0}[/tex]

Remind that sin = y. When sin theta = 0. It means that it lies on the positive x-axis.

We know that 0 satisfies the equation, because sin(0) is 0.

Same goes for π as well, but 2π does not count because the interval is from 0 ≤ theta < 2π.

Hence:-

[tex] \displaystyle \large { \theta = 0,\pi}[/tex]

Second Equation

[tex] \displaystyle \large{2sin \theta - 1 = 0}[/tex]

First, as we learnt. We isolate sin.

[tex] \displaystyle \large{sin \theta = \frac{1}{2} }[/tex]

We know that, sin is positive in Quadrant 1 and 2.

As we learnt from previous question, we use π - (ref. angle) to find Q2 angle.

We know that sin(π/6) is 1/2. Hence π/6 is our reference angle. Since π/6 is in Q1, we only have to find Q2.

Find Quadrant 2

[tex] \displaystyle \large{\pi - \frac{\pi}{6} = \frac{6\pi}{6} - \frac{\pi}{6} } \\ \displaystyle \large{ \frac{5\pi}{6} }[/tex]

Hence:-

[tex] \displaystyle \large{ \theta = \frac{\pi}{6} , \frac{5\pi}{6} }[/tex]

Since both first and second equations are apart of same equation. Therefore, mix both theta from first and second.

Therefore, the solutions to the first question:-

[tex] \displaystyle \large \boxed{ \theta = 0,\pi, \frac{\pi}{6} , \frac{5\pi}{6} }[/tex]

Second Question

This one is a reciprocal of tan, also known as cot.

[tex] \displaystyle \large{cot3 \theta = 1}[/tex]

For this, I will turn cot to 1/tan.

[tex] \displaystyle \large{ \frac{1}{tan3 \theta} = 1}[/tex]

Multiply whole equation by tan3 theta, to get rid of the denominator.

[tex] \displaystyle \large{ \frac{1}{tan3 \theta} \times tan3 \theta = 1 \times tan3 \theta } \\ \displaystyle \large{ 1= tan3 \theta }[/tex]

We also learnt about how to deal with number beside theta.

We increase the interval, by multiplying with the number.

Since our interval is:-

[tex] \displaystyle \large{0 \leqslant \theta < 2\pi}[/tex]

Multiply the whole interval by 3.

[tex] \displaystyle \large{0 \times 3 \leqslant \theta \times 3 < 2\pi \times 3} \\ \displaystyle \large{0 \leqslant 3 \theta < 6\pi }[/tex]

We also know that tan is positive in Quadrant 1 and Quadrant 3.

and tan(π/4) is 1. Therefore, π/4 is our reference angle and our first theta value.

When we want to find Quadrant 3, we use π + (ref. angle).

Find Q3

[tex] \displaystyle \large{\pi + \frac{\pi}{4} } = \frac{5\pi}{4} [/tex]

Hence, our theta values are π/4 and 5π/4. But that is for [0,2π) interval. We want to find theta values over [0,6π) interval.

As we learnt previously, that we use theta + 2πk to find values that are in interval greater than 2π.

As for tangent, we use:-

[tex] \displaystyle \large{ \theta + \pi k = \theta}[/tex]

Because tan is basically a slope or line proportional graph. So it gives the same value every π period.

Now imagine a unit circle, and make sure to have some basic geometry knowledge. Know that when values addition by 180° or π would give a straight angle.

We aren't using k = 1 for this because we've already found Q3 angle.

Since we know Q1 and Q3 angle in [0,2π).

We can also use theta + 2πk if you want.

First Value or π/4

[tex] \displaystyle \large{ \frac{\pi}{4} + 2\pi = \frac{9\pi}{4} } \\ \displaystyle \large{ \frac{\pi}{4} + 4\pi = \frac{17\pi}{4} }[/tex]

Second Value or 5π/4

[tex] \displaystyle \large{ \frac{5\pi}{4} + 2\pi = \frac{13\pi}{4} } \\ \displaystyle \large{ \frac{5\pi}{4} + 4\pi = \frac{21\pi}{4} }[/tex]

Yes, I use theta + 2πk for finding other values.

Therefore:-

[tex] \displaystyle \large{3 \theta = \frac{\pi}{4} , \frac{5\pi}{4} , \frac{9\pi}{4}, \frac{17\pi}{4} , \frac{13\pi}{4} , \frac{21\pi}{4} }[/tex]

Then we divide every values by 3.

[tex] \displaystyle \large \boxed{\theta = \frac{\pi}{12} , \frac{5\pi}{12} , \frac{9\pi}{12}, \frac{17\pi}{12} , \frac{13\pi}{12} , \frac{21\pi}{12} }[/tex]

Let me know if you have any questions!

Answer:

6+7=13

13+8=21

32+21=52

11+34=46

31+03=34

Step-by-step explanation:

im not sure in the 31+03

It looks like you are trying to compute the improper integral,

[tex]I = \displaystyle\int_1^\infty \dfrac{\mathrm dx}{x(9+\ln^2(x))}[/tex]

or some flavor of this. If this interpretation is correct, substitute u = ln(x) and du = dx/x. Then

[tex]I = \displaystyle\int_0^\infty \dfrac{\mathrm du}{9+u^2} \\\\ = \frac13\arctan\left(\frac u3\right)\bigg|_{u=0}^{u\to\infty} \\\\ = \frac13\lim_{u\to\infty}\arctan\left(\frac u3\right) \\\\ = \frac13\times\frac\pi2 = \boxed{\frac\pi6}[/tex]

Answer:

15

Step-by-step explanation:

1-0 =1

3-1 =2

6-3=3

10-6=4

We are adding 1 more each time

10+5 = 15

Step-by-step explanation:

Fraction =

[tex] \frac{10}{100} = \frac{1}{10} [/tex]

Ratio is 1 : 10

Answer:

1/52

Step-by-step explanation:

Answer: 6.9

Step-by-step explanation:

Answer:

4 scoops of ice cream

Step-by-step explanation:

Plug in the total cost, number of scoops of nuts, and number of liquid toppings into the formula. Then, solve for s:

C = 0.50s + 0.35n + 0.25t

3.55 = 0.50s + 0.35(3) + 0.25(2)

3.55 = 0.50s + 1.05 + 0.5

3.55 = 0.50s + 1.55

2 = 0.50s

4 = s

So, the sundae had 4 scoops of ice cream.

Answer:

can you please send the picture of the diagram.

Step-by-step explanation:

Answer:

ABC:

AB-5 units

BC-12.65 units

AC-15 units

FGH:

FG-5 units

GH-12.65 units

FH-15 units

Step-by-step explanation:

PLATO SAMPLE ANSWER

Answer:

-2.9 > w

Step-by-step explanation:

1.8>4.7+w

Subtract 4.7 from each side

1.8-4.7>4.7-4.7+w

-2.9 > w

Answer:

w = -2.9

Step-by-step explanation:

Step-by-step explanation:

the answer is regular octagon

i think

Answer:

V = 34,13*π   cubic units

Step-by-step explanation: See Annex

We find the common points of the two curves, solving the system of equations:

y²  = 2*x                           x = 2*y  ⇒  y = x/2

(x/2)² = 2*x

x²/4 = 2*x

x  =  2*4         x  = 8      and   y = 8/2       y = 4

Then point  P ( 8 ;  4 )

The other point Q is  Q ( 0; 0)

From these  two points, we get the integration limits for dy ( 0 , 4 )are the integration limits.

Now with the help of geogebra we have: In the annex segment ABCD is dy then

V = π *∫₀⁴ (R² - r² ) *dy   =  π *∫₀⁴ (2*y)² - (y²/2)² dy =  π * ∫₀⁴ [(4y²) - y⁴/4 ] dy

V = π * [(4/3)y³ - (1/20)y⁵] |₀⁴

V =  π * [ (4/3)*4³ - 0 - 1/20)*1024 + 0 )

V = π * [256/3  - 51,20]

V = 34,13*π   cubic units

Answer:

The missing side length is of 5.

Step-by-step explanation:

The sides are proportional, which means that the missing side can be found using a rule of three.

x - 10

3 - 6

Applying cross multiplication:

[tex]6x = 30[/tex]

[tex]x = \frac{30}{6} = 5[/tex]

The missing side length is of 5.

Answer:

The solution is:

(1) 0.0104

(2) 0.0008

Step-by-step explanation:

Given:

Mean,

[tex]\mu = 43.7[/tex]

Standard deviation,

[tex]\sigma = 1.6[/tex]

(1)

⇒ [tex]P(X<40) = P(\frac{x-\mu}{\sigma}<\frac{40-43.7}{1.6} )[/tex]

                      [tex]=P(z< - 2.3125)[/tex]

                      [tex]=P(z<-2.31)[/tex]

                      [tex]=0.0104[/tex]

(2)

As we know,

n = 15

⇒ [tex]P(\bar X > 45)= P(\frac{\bar x - \mu}{\frac{\sigma}{\sqrt{n} } } >\frac{45-43.7}{\frac{1.6}{\sqrt{15} } } )[/tex]

                      [tex]=P(z> 3.15)[/tex]

                      [tex]=1-P(z<3.15)[/tex]

                      [tex]=1-0.9992[/tex]

                      [tex]=0.0008[/tex]

Answer:

probability of the product of the chosen integers being a multiple of 3 is P(E)= 1 - (91/285)

=194/285 or 0.6807.

Step-by-step explanation:

The sample space of all possible choices of three integers from the set {1,2,…..,20} has C(20,3) = (20×19×18)/3! elements.

The complement of the event space E consists of all possible choices of three integers from the complement of the set of all the multiples of 3 in the above set because the product of the chosen integers is a multiple of 3 if and only if at least one of them is a multiple of 3. Hence we have to choose three elements from the set {1,2,4,5,7,8,10,11,13,14,16,17,19,20} which has 14 elements.

Hence

|E'| = C(14,3)

= 14×13×12/3!.

Therefore probability P(E')

= |E'|/|S|

= (14×13×12)/(20×19×18)

= (14×13×2)/(20×19×3)

=(7×13)/(5×19×3)

= 91/285.

Therefore the required probability of the product of the chosen integers being a multiple of 3 is P(E)= 1 - (91/285)=194/285 or 0.6807.

Answer:

40141

Step-by-step explanation:

Answer:

A.

Step-by-step explanation:

A. is the answer because they list all of the sandwiches, soups, and beverages with every possible combination.