The automatic opening device of a military cargo parachute has been designed to open when the parachute is 185 m above the ground. Suppose opening altitude actually has a normal distribution with mean value 185 and standard deviation 32 m. Equipment damage will occur if the parachute opens at an altitude of less than 100 m. What is the probability that there is equipment damage to the payload of at least one of five independently dropped parachutes

Answers

Answer 1

Answer:

0.0193 = 1.93% probability that there is equipment damage to the payload of at least one of five independently dropped parachutes.

Step-by-step explanation:

For each parachute, there are only two possible outcomes. Either there is damage, or there is not. The probability of there being damage on a parachute is independent of any other parachute, which means that the binomial probability distribution is used to solve this question.

To find the probability of damage on a parachute, the normal distribution is used.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

In which [tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.

[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]

And p is the probability of X happening.

Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula.

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Probability of a parachute having damage.

The opening altitude actually has a normal distribution with mean value 185 and standard deviation 32 m, which means that [tex]\mu = 185, \sigma = 32[/tex]

Equipment damage will occur if the parachute opens at an altitude of less than 100 m, which means that the probability of damage is the p-value of  Z when X = 100. So

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{100 - 185}{32}[/tex]

[tex]Z = -2.66[/tex]

[tex]Z = -2.66[/tex] has a p-value of 0.0039.

What is the probability that there is equipment damage to the payload of at least one of five independently dropped parachutes?

0.0039 probability of a parachute having damage, which means that [tex]p = 0.0039[/tex]

5 parachutes, which means that [tex]n = 5[/tex]

This probability is:

[tex]P(X \geq 1) = 1 - P(X = 0)[/tex]

In which

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

[tex]P(X = 0) = C_{5,0}.(0.0039)^{0}.(0.9961)^{5} = 0.9807[/tex]

Then

[tex]P(X \geq 1) = 1 - P(X = 0) = 1 - 0.9807 = 0.0193[/tex]

0.0193 = 1.93% probability that there is equipment damage to the payload of at least one of five independently dropped parachutes.


Related Questions

A ball is dropped from a height of 14 ft. The elasticity of the ball is such that it always bounces up one-third the distance it has fallen. (a) Find the total distance the ball has traveled at the instant it hits the ground the fourth time. (Enter an exact number.)

Answers

Answer:

Hello,

742/27 (ft)

Step-by-step explanation:

[tex]h_1=14\\\\h_2=\dfrac{14}{3} \\\\h_3=\dfrac{14}{9} \\\\h_4=\dfrac{14}{27} \\\\[/tex]

[tex]d=14+2*\dfrac{14}{3} +2*\dfrac{14}{9} +2*\dfrac{14}{27} \\=14*(1+\dfrac{1}{3}+\dfrac{2}{9} +\dfrac{2}{27} )\\=14*\dfrac{53}{27} \\=\dfrac{742}{27} \\[/tex]

The total distance the ball has traveled at the instant it hits the ground the fourth time [tex]28ft.[/tex]

What is the total distance?

Distance is a numerical measurement of how far apart objects or points are. It is the actual length of the path travelled from one point to another.

Here given that,

A ball is dropped from a height of [tex]14[/tex] ft. The elasticity of the ball is such that it always bounces up one-third the distance it has fallen.

So, after striking with the ground it covers the distance [tex]14[/tex] ft. so it rebounds to the height is [tex]\frac{1}{3}(14)[/tex].

Then again it hits the ground and covers the distance  [tex]\frac{1}{3}(14)[/tex] and again after rebounding it goes to the height is

[tex]\frac{1(1)}{3(3)}.(14)=\frac{(1)^2}{(3)^2}(14)[/tex]

Then it falls the same distance and goes back to the height

[tex]\frac{1}{3}[/tex] ×[tex](\frac{(1)^2}{(3)^2})[/tex] ×[tex]14[/tex] = [tex]\frac{(1)^3}{(3)^3}(14)[/tex]

So, the total distance travelled is

[tex]14+2[\frac{1}{3}(14)+(\frac{1}{3})^2(14)+(\frac{1}{3})^3(14)+...][/tex]

We take the sum is twice because it goes back to the particular height and falls to the same distance.

[tex]S=14+2(\frac{\frac{1}{3}(14)}{1-\frac{1}{3}})\\\\\\S=\frac{a}{1-r}\\\\\\S=14+2(\frac{\frac{14}{3}}{\frac{2}{3}})\\\\S=14+2(\frac{14}{2})\\\\S=14+2(7)\\\\S=14+14\\\\S=28ft[/tex]

Hence, the total distance the ball has traveled at the instant it hits the ground the fourth time [tex]28ft.[/tex]

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Michelle would like to know how much of her loan payments will go toward interest. She has a $124,500 loan with a 5.9% interest rate that is compounded monthly. The loan has a term of 10 years. Calculate the total amount of interest that Michelle will pay over the course of the loan.

Answers

9514 1404 393

Answer:

   $40,615.20

Step-by-step explanation:

The amortization formula will tell you Michelle's monthly payment.

  A = P(r/12)/(1 -(1 +r/12)^(-12t)) . . . . loan value P at interest rate r for t years

  A = $124,500(0.059/12)/(1 -(1 +0.059/12)^(-12·10)) ≈ $1375.96

__

The total of Michelle's 120 monthly payments is ...

  12 × $1375.96 = $165,115.20

This amount pays both principal and interest, so the amount of interest she pays is ...

  $165,115.20 -124,500 = $40,615.20

Michelle will pay $40,615.20 in interest over the course of the loan.

__

A calculator or spreadsheet can figure this quickly.

There are two rectangles Jared is examining. He knows the width of the first rectangle measures
2.48 cm, and the length is twice its width.
Jared also knows that the width of the second rectangle is equal to the length of the first rectangle,
and that the area of the second rectangle is 9.92. Given this information, find the length of the
second rectangle for Jared.

Answers

1st rectangle:

width: 2.48cm

length: 4.96

2nd rectangle:

width: 4.96 (equals to the length of the 1st rectangle)

area: 9.92

length: 9.92/4.96 = 2

Beginning in January, a person plans to deposit $1 at the end of each month into an account earning
15% compounded monthly. Each year taxes must be paid on the interest earned during that year. Find
the interest earned during each year for the first 3 years.

Answers

Answer:

hi I am a Nepal

[tex] {233333}^{2332} [/tex]

Previous studies suggest that use of nicotine-replacement therapies and antidepressants can help people stop smoking. The New England Journal of Medicine published the results of a double-blind, placebo-controlled experiment to study the effect of nicotine patches and the antidepressant bupropion on quitting smoking. The target for quitting smoking was the 8th day of the experiment.
In this experiment researchers randomly assigned smokers to treatments. Of the 162 smokers taking a placebo, 28 stopped smoking by the 8th day. Of the 272 smokers taking only the antidepressant buproprion, 82 stopped smoking by the 8th day.
Calculate the 99% confidence interval to estimate the treatment effect of buproprion (placebo-treatment). (The standard error is about 0.0407. Use critical value z = 2.576.)
( ), ( )
Round your answer to three decimal places. Put lower bound in the first box and upper bound in the second box.

Answers

Using the z-distribution, it is found that the 99% confidence interval to estimate the treatment effect of buproprion (placebo-treatment) is (-0.234, -0.024).

What is a t-distribution confidence interval?

The confidence interval is:

[tex]\overline{x} \pm zs[/tex]

In which:

[tex]\overline{x}[/tex] is the sample mean.z is the critical value.s is the standard error.

In this problem, we are given that z = 2.576, s = 0.0407. The sample mean is the difference of the proportions, hence:

[tex]\overline{x} = \frac{28}{162} - \frac{82}{272} = -0.129[/tex]

Then, the bounds of the interval are given by:

[tex]\overline{x} - zs = -0.129 - 2.576(0.0407) = -0.234[/tex]

[tex]\overline{x} + zs = -0.129 + 2.576(0.0407) = -0.024[/tex]

The 99% confidence interval to estimate the treatment effect of buproprion (placebo-treatment) is (-0.234, -0.024).

More can be learned about the z-distribution at https://brainly.com/question/25890103

0.003 is 1/10 of
Please help I need this for homework !!!!!!!!!!!!

Answers

Answer:

0.03

Step-by-step explanation:

Which number would be rounded UP to the nearest ten but DOWN to the nearest hundred?
A. 232
B. 238
C. 262
D. 268

Answers

Answer:

B

Step-by-step explanation:

b! i hope this helps

[infinity]
Substitute y(x)= Σ 2 anx^n and the Maclaurin series for 6 sin3x into y' - 2xy = 6 sin 3x and equate the coefficients of like powers of x on both sides of the equation to n= 0. Find the first four nonzero terms in a power series expansion about x = 0 of a general
n=0
solution to the differential equation.

У(Ñ)= ___________

Answers

Recall that

[tex]\sin(x)=\displaystyle\sum_{n=0}^\infty(-1)^n\frac{x^{2n+1}}{(2n+1)!}[/tex]

Differentiating the power series series for y(x) gives the series for y'(x) :

[tex]y(x)=\displaystyle\sum_{n=0}^\infty a_nx^n \implies y'(x)=\sum_{n=1}^\infty na_nx^{n-1}=\sum_{n=0}^\infty (n+1)a_{n+1}x^n[/tex]

Now, replace everything in the DE with the corresponding power series:

[tex]y'-2xy = 6\sin(3x) \implies[/tex]

[tex]\displaystyle\sum_{n=0}^\infty (n+1)a_{n+1}x^n - 2\sum_{n=0}^\infty a_nx^{n+1} = 6\sum_{n=0}^\infty(-1)^n\frac{(3x)^{2n+1}}{(2n+1)!}[/tex]

The series on the right side has no even-degree terms, so if we split up the even- and odd-indexed terms on the left side, the even-indexed [tex](n=2k)[/tex] series should vanish and only the odd-indexed [tex](n=2k+1)[/tex] terms would remain.

Split up both series on the left into even- and odd-indexed series:

[tex]y'(x) = \displaystyle \sum_{k=0}^\infty (2k+1)a_{2k+1}x^{2k} + \sum_{k=0}^\infty (2k+2)a_{2k+2}x^{2k+1}[/tex]

[tex]-2xy(x) = \displaystyle -2\left(\sum_{k=0}^\infty a_{2k}x^{2k+1} + \sum_{k=0}^\infty a_{2k+1}x^{2k+2}\right)[/tex]

Next, we want to condense the even and odd series:

• Even:

[tex]\displaystyle \sum_{k=0}^\infty (2k+1)a_{2k+1}x^{2k} - 2 \sum_{k=0}^\infty a_{2k+1}x^{2k+2}[/tex]

[tex]=\displaystyle \sum_{k=0}^\infty (2k+1)a_{2k+1}x^{2k} - 2 \sum_{k=0}^\infty a_{2k+1}x^{2(k+1)}[/tex]

[tex]=\displaystyle a_1 + \sum_{k=1}^\infty (2k+1)a_{2k+1}x^{2k} - 2 \sum_{k=0}^\infty a_{2k+1}x^{2(k+1)}[/tex]

[tex]=\displaystyle a_1 + \sum_{k=1}^\infty (2k+1)a_{2k+1}x^{2k} - 2 \sum_{k=1}^\infty a_{2(k-1)+1}x^{2k}[/tex]

[tex]=\displaystyle a_1 + \sum_{k=1}^\infty (2k+1)a_{2k+1}x^{2k} - 2 \sum_{k=1}^\infty a_{2k-1}x^{2k}[/tex]

[tex]=\displaystyle a_1 + \sum_{k=1}^\infty \bigg((2k+1)a_{2k+1} - 2a_{2k-1}\bigg)x^{2k}[/tex]

• Odd:

[tex]\displaystyle \sum_{k=0}^\infty 2(k+1)a_{2(k+1)}x^{2k+1} - 2\sum_{k=0}^\infty a_{2k}x^{2k+1}[/tex]

[tex]=\displaystyle \sum_{k=0}^\infty \bigg(2(k+1)a_{2(k+1)}-2a_{2k}\bigg)x^{2k+1}[/tex]

[tex]=\displaystyle \sum_{k=0}^\infty \bigg(2(k+1)a_{2k+2}-2a_{2k}\bigg)x^{2k+1}[/tex]

Notice that the right side of the DE is odd, so there is no 0-degree term, i.e. no constant term, so it follows that [tex]a_1=0[/tex].

The even series vanishes, so that

[tex](2k+1)a_{2k+1} - 2a_{2k-1} = 0[/tex]

for all integers k ≥ 1. But since [tex]a_1=0[/tex], we find

[tex]k=1 \implies 3a_3 - 2a_1 = 0 \implies a_3 = 0[/tex]

[tex]k=2 \implies 5a_5 - 2a_3 = 0 \implies a_5 = 0[/tex]

and so on, which means the odd-indexed coefficients all vanish, [tex]a_{2k+1}=0[/tex].

This leaves us with the odd series,

[tex]\displaystyle \sum_{k=0}^\infty \bigg(2(k+1)a_{2k+2}-2a_{2k}\bigg)x^{2k+1} = 6\sum_{k=0}^\infty (-1)^k \frac{x^{2k+1}}{(2k+1)!}[/tex]

[tex]\implies 2(k+1)a_{2k+2} - 2a_{2k} = \dfrac{6(-1)^k}{(2k+1)!}[/tex]

We have

[tex]k=0 \implies 2a_2 - 2a_0 = 6[/tex]

[tex]k=1 \implies 4a_4-2a_2 = -1[/tex]

[tex]k=2 \implies 6a_6-2a_4 = \dfrac1{20}[/tex]

[tex]k=3 \implies 8a_8-2a_6 = -\dfrac1{840}[/tex]

So long as you're given an initial condition [tex]y(0)\neq0[/tex] (which corresponds to [tex]a_0[/tex]), you will have a non-zero series solution. Let [tex]a=a_0[/tex] with [tex]a_0\neq0[/tex]. Then

[tex]2a_2-2a_0=6 \implies a_2 = a+3[/tex]

[tex]4a_4-2a_2=-1 \implies a_4 = \dfrac{2a+5}4[/tex]

[tex]6a_6-2a_4=\dfrac1{20} \implies a_6 = \dfrac{20a+51}{120}[/tex]

and so the first four terms of series solution to the DE would be

[tex]\boxed{a + (a+3)x^2 + \dfrac{2a+5}4x^4 + \dfrac{20a+51}{120}x^6}[/tex]

Find the minimum sample size needed to be 99% confident that the sample's variance is within 30% of the population's variance.

Answers

The Minimum sample size table is attached below

Answer:

[tex]X=173[/tex]

Step-by-step explanation:

From the question we are told that:

Confidence Interval [tex]CI=99\%[/tex]

Variance [tex]\sigma^2=30\%[/tex]

Generally going through the table the

Minimum sample size is

[tex]X=173[/tex]

Plz urgennt look at the image over 1000 points im going to need help with the last 4 questions i have?

Answers

The answer should be 1548.3m²

2/8 of a rope is 28 meters.What is the length of the rope? A.32 B.42 C.4 D.21​

Answers

let length be x

ATQ

[tex]\\ \sf\longmapsto \dfrac{2}{8}\times x=28[/tex]

[tex]\\ \sf\longmapsto \dfrac{2x}{8}=28[/tex]

[tex]\\ \sf\longmapsto \dfrac{x}{4}=28[/tex]

[tex]\\ \sf\longmapsto x=4(28)[/tex]

[tex]\\ \sf\longmapsto x=112[/tex]

Step-by-step explanation:

there is something wrong with your problem description.

the offered answer options do not fit to the solution as it is described.

2/8th of a rope is 28 meters long. how long is the whole rope ?

as the other answer said : 2/8 = 1/4

and 1/4th of the rope x = 28 m

1/4 × x = 28

x (the length of the whole rope) = 4×28 = 112 meters

but - maybe the original problem said that 7/8th (and not 2/8th) of a rope is 28 m.

7/8 × x = 28

1/8 × x = 4

x = 32 m

then A (32) would be the right answer !

A ball is thrown in air and it's height, h(t) in feet, at any time, t in seconds, is represented by the equation h(t)=−4t2+16t. When is the ball higher than 12 feet off the ground?
A. 3 B. 1 C. 1 D. 4

Answers

Hence the time that the ball will be height than 12 feet off the ground is 4secs

Given the expression for calculating the height in  feet as;

h(t) = -4t²+16t

If the ball is higher than 12feet, h(t) > 12

Substituting h = 12 into the expression

-4t²+16t > 12

-4t²+16t - 12 > 0

4t²- 16t + 12 > 0

t²- 4t + 3 > 0

Factorize

(t²- 3t)-(t + 3) > 0

t(t-3)-1(t-3) > 0

(t-1)(t-3)>0

t > 1 and 3secs

Hence the time that the ball will be height than 12 feet off the ground is 4secs

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Here is a number sequence. The rule for finding the next term is to add
a, where a is an integer. ! ! 8 ........! ! ........! ! 29 Work out the two
missing terms.

Answers

Answer:

8,15,22,29

Step-by-step explanation:

the interger a is 7,so to find the next term you have to add 7 plus the 8,

8+7=15

15+7=22

22+7=29

8,15,22,29

I hope this helps

d is none of the above , and yes

Answers

Answer:

[tex] = 2 {}^{2} - 3(2) = - 2 \\ 3 {}^{2} - 3(3) = 0 \\ 4 {}^{2} - 3(4) = 4 \\ 5 {}^{2} - 3(5) = 10[/tex]

power sharing helps the ruling party to retain power for a long time. tick or wrong

Answers

Power tick wrong long time

write the expression as a decimal , 6 x 1 + 9 x 1/10 + 8 x 1/100 + 6 x 1/1000 =__

Answers

Answer:

6.986.

Step-by-step explanation:

6 x 1 + 9 x 1/10 + 8 x 1/100 + 6 x 1/1000

We do the multiplications first    ( according to PEMDAS):-

= 6 + 9 * 0.1 + 8 * 0.01 + 6 * 0.001

= 6 + 0.9 + 0.08 + 0006

= 6.9 + 0.086

= 6 986.

The value of the equation in the decimal form is A = 6.986

What is an Equation?

Equations are mathematical statements with two algebraic expressions flanking the equals (=) sign on either side.

It demonstrates the equality of the relationship between the expressions printed on the left and right sides.

Coefficients, variables, operators, constants, terms, expressions, and the equal to sign are some of the components of an equation. The "=" sign and terms on both sides must always be present when writing an equation.

Given data ,

Let the equation be represented as A

Now , the value of A is

A = 6 x 1 + 9 x 1/10 + 8 x 1/100 + 6 x 1/1000

On simplifying the equation , we get

The value of 6 x 1 = 6

The value of 9 x 1/10 = 0.9

The value of 9 x 1/100 = 0.08

The value of 6 x 1/1000 = 0.006

So , substituting the values in the equation A , we get

A = 6 + 0.9 + 0.08 + 0.006

On simplifying the equation , we get

A = 6.986

Therefore , the value of A is 6.986

Hence , the value of the equation is 6.986

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Help please guys thanks

Answers

Answer:

D

Step-by-step explanation:

sqrt_{4}(81)^5=(81^(5))^(1/4)=81^(5/4)

Answer:

D

Step-by-step explanation:

if it was properly typed, it would have been All of the above but the most correct option is D.

In a certain animal​ species, the probability that a healthy adult female will have no offspring in a given year is 0.24, while the probabilities of​ 1, 2,​ 3, or 4 offspring​ are respectively 0.25, 0.19, 0.17, and 0.15. Find the expected number of offspring. ​

Answers

Answer:

The expected number of offspring is 2

Step-by-step explanation:

The given parameters can be represented as:

[tex]\begin{array}{cccccc}x & {0} & {1} & {2} & {3} & {4} \ \\ P(x) & {0.24} & {0.25} & {0.19} & {0.17} & {0.15} \ \end{array}[/tex]

Required

The expected number of offspring

This implies that we calculate the expected value of the function.

So, we have:

[tex]E(x) = \sum x * P(x)[/tex]

Substitute known values

[tex]E(x) = 0 * 0.24 + 1 * 0.25 + 2 * 0.19+ 3 * 0.17 + 4 * 0.15[/tex]

Using a calculator, we have:

[tex]E(x) = 1.74[/tex]

[tex]E(x) = 2[/tex] --- approximated

write your answer as an integer or as a decimal rounded to the nearest tenth​

Answers

Answer:

123456-6-&55674

Step-by-step explanation:

rdcfvvzxv.

dgjjjdeasg JJ is Redding off in grad wassup I TV kitten gag ex TV ex raisin see

recall see

Select the correct answer.
What is the best way to describe a theme of this poem?
A.
The main purpose of having New Year's resolutions is to make people feel bad.
B.
The failures of the past should inspire people to accomplish more in the future.
OC.
By the end of the year, it is too late to make any changes to a person's life.
D.
People would accomplish their New Year's resolutions if they wrote them down.

Answers

B.The failures of the past should inspire people to accomplish more in the future.

To calculate the volume of a chemical produced in a day a chemical manufacturing company uses the following formula below:
[tex]V(x)=[C_1(x)+C_2(x)](H(x))[/tex]
where represents the number of units produced. This means two chemicals are added together to make a new chemical and the resulting chemical is multiplied by the expression for the holding container with respect to the number of units produced. The equations for the two chemicals added together with respect to the number of unit produced are given below:
[tex]C_1(x)=\frac{x}{x+1} , C_2(x)=\frac{2}{x-3}[/tex]
The equation for the holding container with respect to the number of unit produced is given below:
[tex]H(x)=\frac{x^3-9x}{x}[/tex]

a. What rational expression do you get when you combine the two chemicals?
b. What is the simplified equation of ?
c. What would the volume be if 50, 100, or 1000 units are produced in a day?
d. The company needs a volume of 3000 How many units would need to be produced in a day?

Answers

Answer:

[tex]V(x) = [\frac{x}{x + 1} + \frac{2}{x-3}] * \frac{x^3 - 9x}{x}[/tex]

[tex]V(x) = [\frac{(x^2-x+2)(x + 3)}{(x + 1)}][/tex]

[tex]V(50) = 2548.17[/tex]        [tex]V(100) = 10098.10[/tex]       [tex]V(1000) = 999201.78[/tex]

[tex]x = 54.78[/tex]

Step-by-step explanation:

Given

[tex]V(x) = [C_1(x) + C_2(x)](H(x))[/tex]

[tex]C_1(x) = \frac{x}{x+1}[/tex]

[tex]C_1(x) = \frac{2}{x-3}[/tex]

[tex]H(x) = \frac{x^3 - 9x}{x}[/tex]

Solving (a): Expression for V(x)

We have:

[tex]V(x) = [C_1(x) + C_2(x)](H(x))[/tex]

Substitute known values

[tex]V(x) = [\frac{x}{x + 1} + \frac{2}{x-3}] * \frac{x^3 - 9x}{x}[/tex]

Solving (b): Simplify V(x)

We have:

[tex]V(x) = [\frac{x}{x + 1} + \frac{2}{x-3}] * \frac{x^3 - 9x}{x}[/tex]

Solve the expression in bracket

[tex]V(x) = [\frac{x*(x-3) + 2*(x+1)}{(x + 1)(x -3)}] * \frac{x^3 - 9x}{x}[/tex]

[tex]V(x) = [\frac{x^2-3x + 2x+2}{(x + 1)(x -3)}] * \frac{x^3 - 9x}{x}[/tex]

[tex]V(x) = [\frac{x^2-x+2}{(x + 1)(x -3)}] * \frac{x^3 - 9x}{x}[/tex]

Factor out x

[tex]V(x) = [\frac{x^2-x+2}{(x + 1)(x -3)}] * \frac{x(x^2 - 9)}{x}[/tex]

[tex]V(x) = [\frac{x^2-x+2}{(x + 1)(x -3)}] * (x^2 - 9)[/tex]

Express as difference of two squares

[tex]V(x) = [\frac{x^2-x+2}{(x + 1)(x -3)}] * (x- 3)(x + 3)[/tex]

Cancel out x - 3

[tex]V(x) = [\frac{x^2-x+2}{(x + 1)}] *(x + 3)[/tex]

[tex]V(x) = [\frac{(x^2-x+2)(x + 3)}{(x + 1)}][/tex]

Solving (c): V(50), V(100), V(1000)

[tex]V(x) = [\frac{(x^2-x+2)(x + 3)}{(x + 1)}][/tex]

Substitute 50 for x

[tex]V(50) = [\frac{(50^2-50+2)(50 + 3)}{(50 + 1)}][/tex]

[tex]V(50) = \frac{(2452)(53)}{(51)}][/tex]

[tex]V(50) = 2548.17[/tex]

Substitute 100 for x

[tex]V(100) = [\frac{(100^2-100+2)(100 + 3)}{(100 + 1)}][/tex]

[tex]V(100) = \frac{9902)(103)}{(101)}[/tex]

[tex]V(100) = 10098.10[/tex]

Substitute 1000 for x

[tex]V(1000) = [\frac{(1000^2-1000+2)(1000 + 3)}{(1000 + 1)}][/tex]

[tex]V(1000) = [\frac{(999002)(10003)}{(10001)}][/tex]

[tex]V(1000) = 999201.78[/tex]

Solving (d): V(x) = 3000, find x

[tex]V(x) = [\frac{(x^2-x+2)(x + 3)}{(x + 1)}][/tex]

[tex]3000 = [\frac{(x^2-x+2)(x + 3)}{(x + 1)}][/tex]

Cross multiply

[tex]3000(x + 1) = (x^2-x+2)(x + 3)[/tex]

Equate to 0

[tex](x^2-x+2)(x + 3)-3000(x + 1)=0[/tex]

Open brackets

[tex]x^3 - x^2 + 2x + 3x^2 - 3x + 6 - 3000x - 3000 = 0[/tex]

Collect like terms

[tex]x^3 + 3x^2- x^2 + 2x - 3x - 3000x + 6 - 3000 = 0[/tex]

[tex]x^3 + x^2 -3001x -2994 = 0[/tex]

Solve using graphs (see attachment)

[tex]x = -54.783[/tex] or

[tex]x = -0.998[/tex] or

[tex]x = 54.78[/tex]

x can't be negative. So:

[tex]x = 54.78[/tex]


Which is the solution to-x/2<-4
A x<-8
B x2-8
C x <8
D x 8

Answers

Answer:

A.x<-8

Step-by-step explanation:

=1/2x<−4

=2*(1/2x)< (2)*(-4)

= x<-8

Question 4 (2 marks)
Justin works 14 hours at a normal pay rate of $24.80 per hour and 5 hours of overtime at
time and a half. How much should he be paid?
I
809 words
LE
English (Australia)

Answers

Answer:

554.7

Step-by-step explanation:

The pay=25.8*14+(25.8)*5*1.5=554.7

Look at image to see question

Answers

Answer:

Does the answer help you

How many ways can a president, vice president, secretary, and treasurer be chosen from a club with 8 member

Answers

Answer:

504

Step-by-step explanation:

It is known that the variance of a population equals 1,936. A random sample of 121 has been selected from the population. There is a .95 probability that the sample mean will provide a margin of error of _____. Group of answer choices 31.36 or less 1,936 or less 344.96 or less 7.84 or less

Answers

Answer:

Option d (7.84 or less) is the right alternative.

Step-by-step explanation:

Given:

[tex]\sigma^2=1936[/tex]

[tex]\sigma = \sqrt{1936}[/tex]

   [tex]=44[/tex]

Random sample,

[tex]n = 121[/tex]

The level of significance,

= 0.95

or,

[tex](1-\alpha) = 0.95[/tex]

        [tex]\alpha = 1-0.95[/tex]

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

hence,

The margin of error will be:

⇒ [tex]E = Z_{\frac{\alpha}{2} }(\frac{\sigma}{\sqrt{n} } )[/tex]

By putting the values, we get

        [tex]=1.96(\frac{44}{\sqrt{121} } )[/tex]

        [tex]=1.96(\frac{44}{11} )[/tex]

        [tex]=1.96\times 4[/tex]

        [tex]=7.84[/tex]    

Hector's Position:
Hector was standing halfway between first and second base, at the grass line. The
grass line is 95 feet from the pitcher's mound.
6. Calculate the coordinates for Hector's position. [Note: We can assume that 95
feet is an approximately horizontal distance from the pitcher's mound to the grass
line.] (2 points: 1 for x, 1 for y)
Hector was standing at the coordinate ( __, _).
Calculate Hector's Throw:

Answers

Answer:

(137.78, 47.72)

Step-by-step explanation:

(I just finished this assignment.)

Tre's position at the pitcher's mound as the point (42.78, 42.78).

                                                                                  (     x    ,    y     )

Hector is about 95 feet away from the pitcher's mound horizontal, (x axis).

Since we already have the correct y-coordinate, we need to solve for the correct x-coordinate.

  x = 95 + 42.78

         ↓ ↓ ↓

95 + 42.78 = 137.72

Now all you need to do is write out the coordinates.

Hector's coordinates are (137.72, 47.78 )

f(t)= 102,000/1+4400e^-t

Answers

Answer:

Beginning (t=0) population with flu is 23.

After 4 weeks, population with flu is 1250.

After an infinite amount of weeks, the population witf flu is 102000

Step-by-step explanation:

First question asks you to replace t with 0 because it says beginning.

102000/(1+4400e^-0)=102000/(1+4400)=102000/4401=23.17655 approximately. To nearest whole number this is 23.

After 4 weeks means we replace t with 4:

102000/(1+4400e^-4)

Calculator time:

1250.17142 which to nearest whole number is 1250

If t is super large, then e^-t is super close to 0.

So the limiting number is

102000/(1+4400×0)=102000/1=102000

Charlie has an annual salary of $75,000.00. He is paid every two weeks. What is the gross income amount for each paycheck?

Answers

Answer:

$2884.62

Step-by-step explanation:

A year has 52 weeks

The number of times Charlie will receive a paycheck will be 52w ÷ 2w = 26 times

Charlie's gross income each paycheck will be 7500÷26 = $2884.62 every two weeks

75000 ÷ (52 ÷2)

7500 ÷ 26

$2884.62

240) What term is 359 in the sequence 5, 12, 17, 23, 28, 29......?​

Answers

Answer:

Check your question again

Step-by-step explanation:

The arithmetic equation of this sequence is an=5+(n-1)*7. Replace 359 with an and solve for n

359=5+(n-1)*7, 354/7=n-1. Wait you got the whole equation wrong, the first term should be 7 so that the common difference be equal to 5

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