A football team starts on the 10 yard line moving toward the 50 yard line so they can score on the other side of the field. In three plays they gain 14 yards, lose 12 yards, and gain 4 more yards. What yard line do they start their fourth play?

Answers

Answer 1

Answer:

16 yard line

Step-by-step explanation:

The football team is starting on the 10 yard line. In the first play, they move up to the 24 yard line. Then in the second play, they go back to the 12 yard line since they lost 12 yards. Then in the third play, they gain 4 yards so you add 4 to 12. They end up at the 16 yard line after the third play. This means that they're going to start their fourth play at the 16 yard line.

Answer 2

Answer:

16 yards.

Step-by-step explanation:

They start at 10 yards. They are moving towards the 50 yard line, so gaining yards will add to the 10 yards instead of subtract from the 10 yards.

In the first play, they gain 14 yards. 10 + 14 = 24 yards.

In the second play, they lose 12 yards. 24 - 12 = 12 yards.

In the third play, they gain 4 yards. 12 + 4 = 16 yards, which is where they start their fourth play.

Hope this helps!


Related Questions

a sample of 25 workers with employer provided health insurance paid an average premium of $6600 eith a sample standard deviation of $800. Construct a 95% confidence interval for the mean premium amount paid by all workers who have employer provided health insurance g

Answers

Answer:

$6284.4≤μ≤$6313.6

Step-by-step explanation:

Using the formula for calculating confidence interval as shown:

CI = xbar ± Z×S/√n

xbar is the average premium

Z is the z-score at 95% confidence

S is the standard deviation

n is the sample size

Given parameters

xbar = $6600

Z score at 95% CI = 1.96

S = $800

n = 25

Substituting this parameters in the formula we have;

CI = 6600±1.96×800/√25

CI = 6600±(1.96×800/5)

CI = 6600±(1.96×160)

CI = 6600±313.6

CI = (6600-313.6, 6600+313.6)

CI = (6284.4, 6913.6)

Hence the 95% confidence interval for the mean premium amount paid by all workers who have employer provided health insurance is $6284.4≤μ≤$6313.6

A manager wants to determine an appropriate learning percentage for processing insurance claims for storm damage. Toward that end, times have been recorded for completion of each of the first six repetitions:
Repetition 1 2 3 4 5 6
Time (minutes) 46 39 35 33 32 30
a. Determine the approximate learning percentage. (Round your answer to the nearest whole percent. Omit the "%" sign in your response.)
P %
b. Using your answer from part a, estimate the average completion time per repetition assuming a total of 30 repetitions are planned. (Round your answer to 3 decimal places.)

Answers

Answer:

Step-by-step explanation:

The approximate learning percentage can be estimated by using a doubling method.

If we breakdown the repetitions into three consecutive parts, we have:

1 - 2

2 - 4

3 - 6

then

1 - 2        →         46P = 39

P =39/46

P = 0.8478

P = 84.8%

2 - 4        →       39P = 33

P = 33/39

P = 0.84615

P = 84.6%

3  -  6    →       35P = 30

P = 30/35

P = 0.8571

P = 85.7%

The average value of P = (84.8 + 84.6 + 85.7)/3 = 85.03%

[tex]\simeq[/tex] 85%

From the tables of Learning Curves coefficient

The values are likened against times derived from 85% table factors at T[tex]_1[/tex] = 46

Unit                 1                2           3                    4                5            6

Date                46            39         35                33                32          30

Computed       -              39.1       35.56          33.26        31.56       30.22

b. Using your answer from part a, estimate the average completion time per repetition assuming a total of 30 repetitions are planned. (Round your answer to 3 decimal places.)

The average completion time = [tex]\mathtt{\dfrac{T_1 \times \ Total \ time\ factor}{n}}[/tex]

At the total time factor 30, from the learning curves table , n(30) = 17.091

Thus:

The average completion time = [tex]\mathtt{\dfrac{46 \times \ 17.091}{30}}[/tex]

The average completion time = [tex]\mathtt{\dfrac{786.186}{30}}[/tex]

The average completion time = [tex]\mathtt{26.2062}[/tex]

Varia is studying abroad in Europe. She is required pay $3,500 (in US dollars) per year to the university; however, she must pay in euros. How many euros can Varia expect to pay per month to the university?

Answers

Answer: 247.92 euros

Step-by-step explanation:

Given:  Varia is required pay $3,500 (in US dollars) per year to the university.

So, [tex]$3500\div 12 \approx\$291.67[/tex]

i.e. She will pay $ 291.67 per month.

Recent currency value: 1 US dollar = 0.85 euro

∴ $291.67 = ( 0.85 ×291.67) euros

= 247.92 euros  [Round to the nearest cent]

She can expect 247.92 euros to pay per month to the university.

Question 1 (5 points)
The line segment AB with endpoints A(-3, 6) and B(9, 12) is dilated with a scale
factor 2/3 about the origin. Find the endpoints of the dilated line segment.
OA) (-2, 4), (6,8)
B) (2, 4). (6,8)
OC) (4, -2), (6,8)
OD) (-2,4), (8,6)​

Answers

Answer: A) (-2, 4), (6,8)

Step-by-step explanation:

When a point (x,y) is dilated by a scale factor of k , then the new points is given by (kx,ky).

Given: The line segment AB with endpoints A(-3, 6) and B(9, 12) is dilated with a scale factor [tex]\dfrac23[/tex] about the origin.

Let A' and B' b the endpoints of the dilated line segment.

Then, [tex]A'(\dfrac{2}{3}(-3), \dfrac23(6))=A'(-2,4)[/tex]

[tex]B'(\dfrac{2}{3}(9), \dfrac23(12))=B'(6,8)[/tex]

Hence, the correct option is A) (-2, 4), (6,8)

Question 3 Rewrite in simplest rational exponent form √x • 4√x. Show each step of your process.

Answers

Answer:

4x

Step-by-step explanation:

Given:

√x • 4√x

Required:

Simplify in rational exponent form

SOLUTION:

Recall => [tex] \sqrt{a} = a^{\frac{1}{2}} [/tex]

Thus,

[tex] \sqrt{x}*4\sqrt{x} [/tex] can be expressed in exponent form as [tex] x^{\frac{1}{2}}*4x^{\frac{1}{2} [/tex]

When multiplying 2 bases having exponents together, their exponents should be added together, while you multiply the bases. I.e. [tex]x^m*x^n = x^{m+n }[/tex]

[tex] = x*4^{\frac{1}{2}+\frac{1}{2} [/tex]

[tex] = 4x^{1} = 4x [/tex]

The answer is 4x

Vhat is the volume of the right rectangular prism?
Will mark brainliest

Answers

Answer:

432 mm³

Step-by-step explanation:

Volume of a Rectangular Prism: V = lwh

Step 1: Define variables

l = 8

w = 6

h = 9

Step 2: Plug into formula

V = 8(6)(9)

Step 3: Evaluate

V = 48(9)

V = 432

And we have our answer!

HELP PLEASE!! (math)

Answers

Answer:

Hey there!

We can write: -2+9=7.

Let me know if this helps :)

Answer:

[tex]\large \boxed{{-2+9=7}}[/tex]

Step-by-step explanation:

-2 is also 0-2

The arrow goes from 0 to -2.

-2 + 9 = 7

The arrow goes from -2 to 7.

Translate this sentence into an equation. 43 is the sum of 11 and Carlos age. Use the variable c to represent Carlos age.

Answers

Answer:

c + 11 = 43

Step-by-step explanation:

C = Carlos age

11 = The number added

43 = The number added plus carlos' age

c +11 = 43

c = 43 - 11

c = 32

Carlos' age is 32 years.

Answer:

C+11=43

Step-by-step explanation:

C= Carlos age

11= added number

43= Carlos age +added number

C+11=43

C=43-11

C=32

Age of Carlos 32.  :)

Write 8:18 as a fraction in simplest form.

Ratio as a Fraction:

Fraction in Simplest Form:

Answers

Answer:

[tex]\text{Ratio as a fraction - \: \boxed{\frac{8}{18}}}[/tex]

[tex]\text{Fraction in simplest form - \boxed{\frac{4}{9}}}[/tex]

Step-by-step explanation:

Part 1: Writing a ratio as a fraction

A fraction and a ratio are the same thing - just a different name. Therefore, the colon in a ratio is the same as a divisor line in a fraction. Therefore, to write a ratio as a fraction,

Replace the colon with a divisor line or the divisor line with a colon (use the first portion to transform a ratio into a fraction and the second form to transform a fraction into a ratio).

Therefore, 8:18 as a fraction is 8/18.

Part 2: Fraction in simplest form

To put a fraction in simplest form, first divide the numerator by the denominator. If it contains a remainder, you cannot use this step to verify it.

8 only goes into 18 twice and leaves a 2 as a remainder, so this method does not work.

Instead, if both numbers are even, divide by 2.

8/2 = 4

18/2 = 9

Check to see if the new numerator and denominator can reduce any further.

4/9 = 4/9

The fraction in simplest form is 4/9.

Musah stands at the centre of a rectangular field. He first takes 50 steps north, then 25 steps

west and finally 50 steps on a bearing of 3150

.

i. Sketch Musah’s movement

ii. How far west is Musah’s final point from the centre?
iii. How far north is Musah’s final point from the centre?

iv. Describe how you would guide a JHS student to find the bearing and distance of

Musah’s final point from the centre. ​

Answers

Answer:

ii. 75 steps

iii. 75 steps

iv. 106 steps, and [tex]315^{0}[/tex]

Step-by-step explanation:

Let Musah's starting point be A, his waiting point after taking 50 steps northward and 25 steps westward be B, and his stopping point be C.

ii. From the second attachment, Musah's distance due west from A to C (AD) can be determined as;

bearing at B = [tex]315^{0}[/tex], therefore <BCD = [tex]45^{0}[/tex]

To determine distance AB,

[tex]/AB/^{2}[/tex] = [tex]/50/^{2}[/tex]   +  [tex]/25/^{2}[/tex]

          = 25000 + 625

          = 3125

AB = [tex]\sqrt{3125}[/tex]

     = 55.90

AB ≅ 56 steps

Thus, AC = 50 steps + 56 steps

               = 106 steps

From ΔACD,

Sin [tex]45^{0}[/tex] = [tex]\frac{x}{106}[/tex]

⇒ x = 106 × Sin [tex]45^{0}[/tex]

      = 74.9533

     ≅ 75 steps

Musah's distance west from centre to final point is 75 steps

iii. From the secon attachment, Musah's distance north, y, can be determined by;

Cos [tex]45^{0}[/tex] = [tex]\frac{y}{106}[/tex]

⇒ y = 106 × Cos [tex]45^{0}[/tex]

      = 74.9533

      ≅ 75 steps

Musah's distance north from centre to final point is 75 steps.

iv. Musah's distance from centre to final point is AC = AB + BC

                                     = 50 steps + 56 steps

                                     = 106 steps

From ΔACD,

Tan θ = [tex]\frac{75}{75}[/tex]

          = 1.0

θ = [tex]Tan^{-1}[/tex]  1.0

 = [tex]45^{0}[/tex]

Musah's bearing from centre to final point = [tex]45^{0}[/tex] + [tex]270^{0}[/tex]

                                                           =  [tex]315^{0}[/tex]

WILL GIVE BRAINLYEST AND 30 POINTS Which of the followeing can be qritten as a fraction of integers? CHECK ALL THAT APPLY 25, square root of 14, -1.25, square root 16, pi, 0.6

Answers

Answer:

25 CAN be written as a fraction.

=> 250/10 = 25

Square root of 14 is 3.74165738677

It is NOT POSSIBLE TO WRITE THIS FULL NUMBER AS A FRACTION,  but if we simplify the decimal like: 3.74, THEN WE CAN WRITE THIS AS A FRACTION

=> 374/100

-1.25 CAN be written as a fraction.

=> -5/4 = -1.25

Square root of 16 CAN also be written as a fraction.

=> sqr root of 16 = 4.

4 can be written as a fraction.

=> 4 = 8/2

Pi = 3.14.........

It is NOT POSSIBLE TO WRITE THE FULL 'PI' AS A FRACTION, but if we simplify 'pi' to just 3.14, THEN WE CAN WRITE IT AS A FRACTION

=> 314/100

.6 CAN be written as a fraction.

=> 6/10 = .6

How many g of amino acids are in a 2,000mL total parenteral nutrition of 4.25% travesol (amino acids and 20%dextrose?

Answers

Answer:

The  mass of  amino acid present is  85 g

Step-by-step explanation:

From the question we are told that

   The  volume of the total  parenteral nutrition is  [tex]V = 2000 mL[/tex]

   

Generally the volume of the amino acid present is  

                [tex]V_a = 0.0425 * 2000[/tex]

                [tex]V_a = 85 mL[/tex]

Generally  

               [tex]mass \ \ \alpha \ \ volume[/tex]

So the mass of amino acid present is 85g


How many cubic inches of a milkshake can you fit up to the brim of this cup without letting it overflow? The
cup is 10 inches tall, and the rim of the cup is 4 inches across. (Hint: the radius is half of the diameter.)

Answers

Assuming the cup is a right circular cylinder, it's volume is [tex]V=\pi r^2 h[/tex]

$h=10$, $r=\frac 42$

So the volume is $\pi\cdot(2)^2\cdot10=125.66$

hence you can fill up to 125.66 cubic Inches of milkshake

.

Suppose a 99% confidence interval for the mean salary of college graduates in a town in Mississippi is given by [$34,393, $47,207]. The population standard deviation used for the analysis is known to be $14,900.

Required:
a. What is the point estimate of the mean salary for all college graduates in this town?
b. Determine the sample size used for the analysis.

Answers

Answer: a. $40,800 b. 36

Step-by-step explanation:

Given : a 99% confidence interval for the mean salary of college graduates in a town in Mississippi is given by [$34,393, $47,207].

[tex]\sigma= \$14,900[/tex]

a. Since Point estimate of of the mean = Average of upper limit and lower limit of the interval.

Therefore , the point estimate of the mean salary for all college graduates in this town = [tex]\dfrac{34393+47207}{2}=\dfrac{81600}{2}[/tex]

= 40,800

hence, the point estimate of the mean salary for all college graduates in this town = $40,800

b.  Since  lower limit = Point estimate - margin of error, where Margin of error is the half of the difference between upper limit and lower limit.

Margin of error[tex]=\dfrac{47207-34393}{2}=6407[/tex]

Also, margin of error = [tex]z\times\dfrac{\sigma}{\sqrt{n}}[/tex], where z= critical z-value for confidence level and n is the sample size.

z-value for 99% confidence level  = 2.576

So,

[tex]6407=2.576\times\dfrac{14900}{\sqrt{n}}\\\\\Rightarrow\ \sqrt{n}=2.576\times\dfrac{14900}{6407}=5.99\\\\\Rightarrow\ n=(5.99)^2=35.8801\approx 36[/tex]

The sample size used for the analysis =36

The scores for all the Algebra 1 students at Miller High on a test are normally distributed with a mean of 82 and a standard deviation of 7. What percent of students made scores above 89?

Answers

Answer:

15.7% of students made above an 89.

Step-by-step explanation:

If the data is normally distributed, the standard deviation is 7, and the mean is 82, then about 68.2% of students made between 75 and 89. 13.6% made between 90 and 96, and 2.1% made over 96. 13.6+2.1=15.7%

This expression represents the average cost per game, in dollars, at a bowling alley, where n represents the number of games:

3n+7/n


What is the average cost per game if James bowls 4 games?

Answers

Answer:

13.75 dollars

Step-by-step explanation:

n=4

3n+7/n =3(4) + 7/4

=12 + 1.75

=$13.75

Answer: A) 4.75

Step-by-step explanation:

You have 9kg of oats and cup scales that gears of 50g and 200g. How − in three weighings− can you measure 2kg of the oats?

Answers

Answer: You will need 8 cup scales

Step-by-step explanation:

kg=1000 grams

2000/250=8

BRAINLIEST IF CORRECT!!! and 15 points solve for z -cz + 6z = tz + 83

Answers

Answer:

z = 83/( -c+6-t)

Step-by-step explanation:

-cz + 6z = tz + 83

Subtract tz from each side

-cz + 6z -tz= tz-tz + 83

-cz + 6z - tz = 83

Factor out z

z( -c+6-t) = 83

Divide each side by ( -c+6-t)

z( -c+6-t)/( -c+6-t)  = 83/( -c+6-t)

z = 83/( -c+6-t)

Find all solutions to the equation. 2sin theta - squareroot 3 = 0
Write your answer in radians in terms of pi, and use the "or" button as necessary.
Example: theta = pi/5 + 2 k pi, k element Z or theta = pi/7 + k pi, k element Z

Answers

Answer:

[tex]\theta[/tex] =2mπ + π/3 for m ∈ Z.

Step-by-step explanation:

Given the equation [tex]2sin\theta - \sqrt{3} = 0[/tex], we are to find all the values of [tex]\theta[/tex] that satisfies the equation.

[tex]2sin\theta - \sqrt{3} = 0\\\\2sin\theta = \sqrt{3} \\\\sin\theta = \sqrt{3}/2 \\\\\theta = sin{-1} \sqrt{3}/2 \\\\\theta = 60^0[/tex]

General solution for sin[tex]\theta[/tex] is [tex]\theta[/tex] = nπ + (-1)ⁿ ∝, where n ∈ Z.

If n is an even number say 2m, then [tex]\theta[/tex] = (2m)π + ∝ where ∝ = 60° = π/3

Hence, the general solution to the equation will be [tex]\theta[/tex] = 2mπ + π/3 for m ∈ Z.

Consider a pair of random variables X; Y with constant joint density on the quadrilateral with vertices (0; 0), (2; 0), (2; 6), (0; 12). a) Find the expected value E(X). b) Find the expected value E(Y ).

Answers

The given quadrilateral (call it Q) is a trapezoid with "base" lengths of 6 and 12, and "height" 2, so its area is (6 + 12)/2*2 = 18. This means the joint density is

[tex]f_{X,Y}(x,y)=\begin{cases}\frac1{18}&\text{for }(x,y)\in Q\\0&\text{otherwise}\end{cases}[/tex]

where Q is the set of points

[tex]Q=\{(x,y)\mid0\le x\le 2\land0\le y\le12-3x\}[/tex]

(y = 12 - 3x is the equation of the line through the points (0, 12) and (2, 6))

Recall the definition of expectation:

[tex]E[g(X,Y)]=\displaystyle\int_{-\infty}^\infty\int_{-\infty}^\infty g(x,y)f_{X,Y}(x,y)\,\mathrm dx\,\mathrm dy[/tex]

(a) Using the definition above, we have

[tex]E[X]=\displaystyle\int_{-\infty}^\infty\int_{-\infty}^\infty xf_{X,Y}(x,y)\,\mathrm dx\,\mathrm dy=\int_0^2\int_0^{12-3x}\frac x{18}\,\mathrm dy\,\mathrm dx=\frac89[/tex]

(b) Likewise,

[tex]E[Y]=\displaystyle\int_{-\infty}^\infty\int_{-\infty}^\ifnty yf_{X,Y}(x,y)\,\mathrm dx\,\mathrm dy=\int_0^2\int_0^{12-3x}\frac y{18}\,\mathrm dy\,\mathrm dx=\frac{14}3[/tex]

nd the measure of angle m
2. Find the length of sie
m
18.2m
61°
15:1m
х
105mm

Answers

Answer:

1). m° = 56.1°

2). X= 91.8 mm

Step-by-step explanation:

For angle m°

Using the sine rule

15.1/sin m= 18.2/sin 90

But Sin 90= 1

15.1/sin m= 18.2

15.1= 18.2*sin m

Sin m = 15.1/18.2

Sin m=0.8297

m= sin^-1(0.8297)

m= 56.06°

m° = 56.1°

For length of side x

Using sine rule

X/sin 61= 105/sin 90

But sin 90= 1

X/sin 61= 105

X = sin61 *105

X=0.8746*105

X= 91.833 mm

X= 91.8 mm

Given that −4i is a zero, factor the following polynomial function completely. Use the Conjugate Roots Theorem, if applicable. f(x)=x4−2x3+x2−32x−240

Answers

Answer:

[tex]\large \boxed{\sf \bf \ \ f(x)=(x-4i)(x+4i)(x+3)(x-5) \ \ }[/tex]

Step-by-step explanation:

Hello, the Conjugate Roots Theorem states that if a complex number is a zero of real polynomial its conjugate is a zero too. It means that (x-4i)(x+4i) are factors of f(x).

[tex]\text{Meaning that } (x-4i)(x+4i) =x^2-(4i)^2=x^2+16 \text{ is a factor of f(x).}[/tex]

The coefficient of the leading term is 1 and the constant term is -240 = 16 * (-15), so we a re looking for a real number such that.

[tex]f(x)=x^4-2x^3+x^2-32x-240\\\\ =(x^2+16)(x^2+ax-15)\\\\ =x^4+ax^3-15x^2+16x^2+16ax-240[/tex]

We identify the coefficients for the like terms, it comes

a = -2 and 16a = -32 (which is equivalent). So, we can write in [tex]\mathbb{R}[/tex].

[tex]\\f(x)=(x^2+16)(x^2-2x-15)[/tex]

The sum of the zeroes is 2=5-3 and their product is -15=-3*5, so we can factorise by (x-5)(x+3), which gives.

[tex]f(x)=(x^2+16)(x^2-2x-15)\\\\=(x^2+16)(x^2+3x-5x-15)\\\\=(x^2+16)(x(x+3)-5(x+3))\\\\=\boxed{(x^2+16)(x+3)(x-5)}[/tex]

And we can write in [tex]\mathbb{C}[/tex]

[tex]f(x)=\boxed{(x-4i)(x+4i)(x+3)(x-5)}[/tex]

Hope this helps.

Do not hesitate if you need further explanation.

Thank you

A ladder 10 ft long leans against a vertical wall. If the lower end is being moved away from the wall at the rate of 6 ​ft/sec, how fast is the height of the top changing​ (this will be a negative​ rate) when the lower end is 6 feet from the​ wall?

Answers

Answer:

-4.5ft per sec

Step-by-step explanation:

Assume that vertical wall has a distance of y and the horizontal floor is x (6 ft).

This forms a triangle with the ladder as the hypothenus of length 10ft

We have dy/dt = 6ft per sec

According to Pythagoras law the relationship between x and y is

(x^2) + (y^2) = (hypothenus ^2) = 10^2

When we differentiate both sides of the equation

2x(dx/dt) + 2y(dy/dt) = 0

dy/dt = (x/y) * (dx/dt)

y= √(10^2) - (6^2) = 8ft

So dy/dt = (6/8)* (6/1)= -4.5 ft per sec

It is a negative rate

will give 5 stars and thanks for correct answer Richard starts high school every day at 7:45 A.M.. How many seconds is Richard in school each day of school dismissed at 2:15 P.M.

Answers

Richard spends 21,600 seconds at school each day.

the terms in this sequence increase by the same amount each time. _19_ _ 34_ a) work out the missing terms.

Answers

Answer:

The sequence is 14, 19, 24, 29, 34, 39.

Step-by-step explanation:

Let's call the common difference (the difference between two consecutive terms) as d. We see that the second term is 19 and the 5th term is 34 and since 5 - 2 = 3, we add d 3 times to 19 to get 34 so therefore:

19 + 3d = 34

3d = 15

d = 5 so the first term is 19 - 5 = 14, the third would be 19 + 5 = 24, the fourth would be 24 + 5 = 29 and the sixth would be 34 + 5 = 39.

Write the equations, after translating the graph of y = |x+2|: one unit up,

Answers

Answer:

y = |x + 2| + 1

Step-by-step explanation:

Parent Graph: f(x) = a|bx + c| + k

a is vertical stretch/shrink

b is horizontal stretch/shrink

c is horizontal movement left/right

k is vertical movement up/down

Since we are given an equation and we want to move it 1 unit up (vertical movement up), we only manipulate k:

y = |x + 2| + k

k = 1

y = |x + 2| + 1

Answer:

y = |x+2| + 1

Step-by-step explanation:

The equation will be y = |x+2| + 1.

By translating the graph one unit up, the equation will simply change by adding +1 to the graph, outside of the absolute value part.

Is 1.45 times 10 to the -7 power a scientific notation

Answers

Answer:

Yes.

It is 1.45 x 10^-7 or 0.000000145

Hope it helps!

Answer:

It is 1.45 x 10^-7 or 0.000000145

Step-by-step explanation:

Using the insurance company's assumptions, calculate the probability that there are fewer than 3 tornadoes in a 14-year period. Round your answer to four decimal places.

Answers

Answer:  19.2222

Step-by-step explanation:

given data;

no of tornadoes = 2,1,0 because it’s fewer than 3.

period = 14 years

probability of a tornado in a calendar year = 0.12

solution:

probability of exactly 2 tornadoes

= ( 0.12 )^2 * ( 0.88 )^11 * ( 14! / 2! * 11! )

= 0.0144 * 0.2451 * 1092

= 3.8541

probability of exac one tornado

= ( 0.12 )^1 * ( 0.88 )^12 * ( 14! / 1! * 12! )

= 0.12 * 0.2157 * 182

= 12.7109

probability of exactly 0 tornado

= ( 0.12 )^0 * ( 0.88 )^13 * ( 14! / 0! * 13! )

= 1 * 0.1898 * 14

= 2.6572

probability if fewer than 3 tornadoes

= 3.8541 + 12.7109 +2.6572

= 19.2222

During April of 2013, Gallup randomly surveyed 500 adults in the US, and 47% said that they were happy, and without a lot of stress." Calculate and interpret a 95% confidence interval for the proportion of U.S. adults who considered themselves happy at that time. 1 How many successes and failures are there in the sample? Are the criteria for approximate normality satisfied for a confidence interval?
A What is the sample proportion?
B compute the margin of error for a 95% confidence interval.
C Interpret the margin of error you calculated in Question 1
C. Give the lower and upper limits of the 95% confidence interval for the population proportion (p), of U.S. adults who considered themselves happy in April, 2013.
D Give an interpretation of this interval.
E. Based on this interval, is it reasonably likely that a majority of U.S. adults were happy at that time?
H If someone claimed that only about 1/3 of U.S. adults were happy, would our result support this?

Answers

Answer:

number of successes

                 [tex]k = 235[/tex]

number of failure

                 [tex]y = 265[/tex]

The   criteria are met    

A

    The sample proportion is  [tex]\r p = 0.47[/tex]

B

    [tex]E =4.4 \%[/tex]

C

What this mean is that for N number of times the survey is carried out that the which sample proportion obtain will differ from  the true population proportion will not  more than 4.4%

Ci  

   [tex]r = 0.514 = 51.4 \%[/tex]

 [tex]v = 0.426 = 42.6 \%[/tex]

D

   This 95% confidence interval  mean that the the chance of the true    population proportion of those that are happy to be exist within the upper   and the lower limit  is  95%

E

  Given that 50% of the population proportion  lie with the 95% confidence interval  the it correct to say that it is reasonably likely that a majority of U.S. adults were happy at that time

F

 Yes our result would support the claim because

            [tex]\frac{1}{3 } \ of N < \frac{1}{2} (50\%) \ of \ N , \ Where\ N \ is \ the \ population\ size[/tex]

Step-by-step explanation:

From the question we are told that

     The sample size is  [tex]n = 500[/tex]

     The sample proportion is  [tex]\r p = 0.47[/tex]

 

Generally the number of successes is mathematical represented as

             [tex]k = n * \r p[/tex]

substituting values

             [tex]k = 500 * 0.47[/tex]

            [tex]k = 235[/tex]

Generally the number of failure  is mathematical represented as

           [tex]y = n * (1 -\r p )[/tex]

substituting values

           [tex]y = 500 * (1 - 0.47 )[/tex]

           [tex]y = 265[/tex]

for approximate normality for a confidence interval  criteria to be satisfied

          [tex]np > 5 \ and \ n(1- p ) \ >5[/tex]

Given that the above is true for this survey then we can say that the criteria are met

  Given that the confidence level is  95%  then the level of confidence is mathematically evaluated as

                       [tex]\alpha = 100 - 95[/tex]

                        [tex]\alpha = 5 \%[/tex]

                        [tex]\alpha =0.05[/tex]

Next we obtain the critical value of  [tex]\frac{\alpha }{2}[/tex] from the normal distribution table, the value is

                 [tex]Z_{\frac{ \alpha }{2} } = 1.96[/tex]

Generally the margin of error is mathematically represented as  

                [tex]E = Z_{\frac{\alpha }{2} } * \sqrt{ \frac{\r p (1- \r p}{n} }[/tex]

substituting values

                 [tex]E = 1.96 * \sqrt{ \frac{0.47 (1- 0.47}{500} }[/tex]

                 [tex]E = 0.044[/tex]

=>               [tex]E =4.4 \%[/tex]

What this mean is that for N number of times the survey is carried out that the proportion obtain will differ from  the true population proportion of those that are happy by more than 4.4%

The 95% confidence interval is mathematically represented as

          [tex]\r p - E < p < \r p + E[/tex]

substituting values

        [tex]0.47 - 0.044 < p < 0.47 + 0.044[/tex]

         [tex]0.426 < p < 0.514[/tex]

The upper limit of the 95% confidence interval is  [tex]r = 0.514 = 51.4 \%[/tex]

The lower limit of the   95% confidence interval is  [tex]v = 0.426 = 42.6 \%[/tex]

This 95% confidence interval  mean that the the chance of the true population proportion of those that are happy to be exist within the upper and the lower limit  is  95%

Given that 50% of the population proportion  lie with the 95% confidence interval  the it correct to say that it is reasonably likely that a majority of U.S. adults were happy at that time

Yes our result would support the claim because

            [tex]\frac{1}{3 } < \frac{1}{2} (50\%)[/tex]

 

1) Dada a função, em reais, definida por f(x)=3.x-5. calcule :

a) f(2)=
b) f(-1)=

Answers

Answer:

f(x)= 3x-5

f(2) = 3(2)-5 = 6-5= 1

f(-1)= 3(-1)-5= -3-5= -8

Hope this helps

if u have question let me know in comments ^°^

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