Jill has 32 crayons. She loses 4 of the crayons. How many are left?

Answer:

Here we just want to find the Taylor series for f(x) = ln(x), centered at the value of a (which we do not know).

Remember that the general Taylor expansion is:

[tex]f(x) = f(a) + f'(a)*(x - a) + \frac{1}{2!}*f''(a)(x -a)^2 + ...[/tex]

for our function we have:

f'(x) =  1/x

f''(x) = -1/x^2

f'''(x) =  (1/2)*(1/x^3)

this is enough, now just let's write the series:

[tex]f(x) = ln(a) + \frac{1}{a} *(x - a) - \frac{1}{2!} *\frac{1}{a^2} *(x - a)^2 + \frac{1}{3!} *\frac{1}{2*a^3} *(x - a)^3 + ....[/tex]

This is the Taylor series to 3rd degree, you just need to change the value of a for the required value.

Slope-intercept form:

y = x + 2

Point-slope form:

y − 5 = 1 ⋅ ( x − 3 )

I hope this is correct and helps!

Answer:

0.1003 = 10.03% probability that a simple random sample of size n= 10 results in a sample mean greater than 40 inches.

Gestation periods:

1) 0.3539 = 35.39% probability a randomly selected pregnancy lasts less than 260 days.

2) 0.0465 = 4.65% probability that a random sample of 20 pregnancies has a mean gestation period of 260 days or less.

3) 0.004 = 0.4% probability that a random sample of 50 pregnancies has a mean gestation period of 260 days or less.

4) 0.9844 = 98.44% probability a random sample of size 15 will have a mean gestation period within 10 days of the mean.

Step-by-step explanation:

To solve these questions, we need to understand the normal probability distribution and the central limit theorem.

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.

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

The height, X, of all 3-year-old females is approximately normally distributed with mean 38.72 inches and standard deviation 3.17 inches.

This means that [tex]\mu = 38.72, \sigma = 3.17[/tex]

Sample of 10:

This means that [tex]n = 10, s = \frac{3.17}{\sqrt{10}}[/tex]

Compute the probability that a simple random sample of size n= 10 results in a sample mean greater than 40 inches.

This is 1 subtracted by the p-value of Z when X = 40. So

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

By the Central Limit Theorem

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

[tex]Z = \frac{40 - 38.72}{\frac{3.17}{\sqrt{10}}}[/tex]

[tex]Z = 1.28[/tex]

[tex]Z = 1.28[/tex] has a p-value of 0.8997

1 - 0.8997 = 0.1003

0.1003 = 10.03% probability that a simple random sample of size n= 10 results in a sample mean greater than 40 inches.

Gestation periods:

[tex]\mu = 266, \sigma = 16[/tex]

1. What is the probability a randomly selected pregnancy lasts less than 260 days?

This is the p-value of Z when X = 260. So

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

[tex]Z = \frac{260 -  266}{16}[/tex]

[tex]Z = -0.375[/tex]

[tex]Z = -0.375[/tex] has a p-value of 0.3539.

0.3539 = 35.39% probability a randomly selected pregnancy lasts less than 260 days.

2. What is the probability that a random sample of 20 pregnancies has a mean gestation period of 260 days or less?

Now [tex]n = 20[/tex], so:

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

[tex]Z = \frac{260 - 266}{\frac{16}{\sqrt{20}}}[/tex]

[tex]Z = -1.68[/tex]

[tex]Z = -1.68[/tex] has a p-value of 0.0465.

0.0465 = 4.65% probability that a random sample of 20 pregnancies has a mean gestation period of 260 days or less.

3. What is the probability that a random sample of 50 pregnancies has a mean gestation period of 260 days or less?

Now [tex]n = 50[/tex], so:

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

[tex]Z = \frac{260 - 266}{\frac{16}{\sqrt{50}}}[/tex]

[tex]Z = -2.65[/tex]

[tex]Z = -2.65[/tex] has a p-value of 0.0040.

0.004 = 0.4% probability that a random sample of 50 pregnancies has a mean gestation period of 260 days or less.

4. What is the probability a random sample of size 15 will have a mean gestation period within 10 days of the mean?

Sample of size 15 means that [tex]n = 15[/tex]. This probability is the p-value of Z when X = 276 subtracted by the p-value of Z when X = 256.

X = 276

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

[tex]Z = \frac{276 - 266}{\frac{16}{\sqrt{15}}}[/tex]

[tex]Z = 2.42[/tex]

[tex]Z = 2.42[/tex] has a p-value of 0.9922.

X = 256

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

[tex]Z = \frac{256 - 266}{\frac{16}{\sqrt{15}}}[/tex]

[tex]Z = -2.42[/tex]

[tex]Z = -2.42[/tex] has a p-value of 0.0078.

0.9922 - 0.0078 = 0.9844

0.9844 = 98.44% probability a random sample of size 15 will have a mean gestation period within 10 days of the mean.

Answer:

Satisfaction score

Step-by-step explanation:

The dependent variable may be described as the variable which is being measured in a research experiment. In the scenario described above, the dependent variable is the satisfaction score which is used to measure preference for a one or two column textbook. The dependent variable can also seen as the variable which we would like to predict, also called the predicted variable . The predicted variable here is the satisfaction score.

Answer:

0.2917 = 29.17% probability that a randomly selected female student on campus will have served a mission.

Step-by-step explanation:

A probability is the number of desired outcomes divided by the number of total outcomes.

In this question:

1200 female students, out of them, 350 have served a mission. So

[tex]p = \frac{350}{1200} = 0.2917[/tex]

0.2917 = 29.17% probability that a randomly selected female student on campus will have served a mission.

Answer:

Defintion. A subset W of a vector space V is a subspace if

(1) W is non-empty

(2) For every v, ¯ w¯ ∈ W and a, b ∈ F, av¯ + bw¯ ∈ W.

Expressions like av¯ + bw¯, or more generally

X

k

i=1

aiv¯ + i

are called linear combinations. So a non-empty subset of V is a subspace if it is

closed under linear combinations. Much of today’s class will focus on properties of

subsets and subspaces detected by various conditions on linear combinations.

Theorem. If W is a subspace of V , then W is a vector space over F with operations

coming from those of V .

In particular, since all of those axioms are satisfied for V , then they are for W.

We only have to check closure!

Examples:

Defintion. Let F

n = {(a1, . . . , an)|ai ∈ F} with coordinate-wise addition and scalar

multiplication.

This gives us a few examples. Let W ⊂ F

n be those points which are zero except

in the first coordinate:

W = {(a, 0, . . . , 0)} ⊂ F

n

.

Then W is a subspace, since

a · (α, 0, . . . , 0) + b · (β, 0, . . . , 0) = (aα + bβ, 0, . . . , 0) ∈ W.

If F = R, then W0 = {(a1, . . . , an)|ai ≥ 0} is not a subspace. It’s closed under

addition, but not scalar multiplication.

We have a number of ways to build new subspaces from old.

Proposition. If Wi for i ∈ I is a collection of subspaces of V , then

W =

\

i∈I

Wi = {w¯ ∈ V |w¯ ∈ Wi∀i ∈ I}

is a subspace.

Proof. Let ¯v, w¯ ∈ W. Then for all i ∈ I, ¯v, w¯ ∈ Wi

, by definition. Since each Wi

is

a subspace, we then learn that for all a, b ∈ F,

av¯ + bw¯ ∈ Wi

,

and hence av¯ + bw¯ ∈ W. ¤

Thought question: Why is this never empty?

The union is a little trickier.

Proposition. W1 ∪ W2 is a subspace iff W1 ⊂ W2 or W2 ⊂ W1.

i hope this helped have a nice day/night :)

Answer:

1.8574 hours

Step-by-step explanation:

Solve for t.

Take the natural log of both sides.

[tex] 3000 = 75000e^{-1.733t} [/tex]

[tex] 1 = 25e^{-1.733t} [/tex]

[tex] \dfrac{1}{25} = e^{-1.733t} [/tex]

[tex] \ln \dfrac{1}{25} = \ln (e^{-1.733t}) [/tex]

[tex] -3.218875 = -1.733t [/tex]

[tex] t = 1.8574[/tex]

Answer:

the probability that I chose red or blue is 75%

75%

Answer:

The p-value of the test is 0.0041 < 0.05, which means that there is sufficient evidence at the 0.05 significance level to conclude that the mean cost has increased.

Step-by-step explanation:

The average undergraduate cost for tuition, fees, and room and board for two-year institutions last year was $13,252. Test if the mean cost has increased.

At the null hypothesis, we test if the mean cost is still the same, that is:

[tex]H_0: \mu = 13252[/tex]

At the alternative hypothesis, we test if the mean cost has increased, that is:

[tex]H_1: \mu > 13252[/tex]

The test statistic is:

We have the standard deviation for the sample, which means that the t-distribution is used to solve this question

[tex]t = \frac{X - \mu}{\frac{s}{\sqrt{n}}}[/tex]

In which X is the sample mean, [tex]\mu[/tex] is the value tested at the null hypothesis, s is the standard deviation and n is the size of the sample.

13252 is tested at the null hypothesis:

This means that [tex]\mu = 13252[/tex]

The following year, a random sample of 20 two-year institutions had a mean of $15,560 and a standard deviation of $3500.

This means that [tex]n = 20, X = 15560, s = 3500[/tex]

Value of the test statistic:

[tex]t = \frac{X - \mu}{\frac{s}{\sqrt{n}}}[/tex]

[tex]t = \frac{15560 - 13252}{\frac{3500}{\sqrt{20}}}[/tex]

[tex]t = 2.95[/tex]

P-value of the test and decision:

The p-value of the test is found using a t-score calculator, with a right-tailed test, with 20-1 = 19 degrees of freedom and t = 2.95. Thus, the p-value of the test is 0.0041.

The p-value of the test is 0.0041 < 0.05, which means that there is sufficient evidence at the 0.05 significance level to conclude that the mean cost has increased.

Answer:

The 96% CI on the death rate from lung cancer is (0.4177, 0.4823).

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of [tex]\pi[/tex], and a confidence level of [tex]1-\alpha[/tex], we have the following confidence interval of proportions.

[tex]\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]

In which

z is the z-score that has a p-value of [tex]1 - \frac{\alpha}{2}[/tex].

Of 1000 randomly selected cases of lung cancer, 450 resulted in death within 5 years.

This means that [tex]n = 1000, \pi = \frac{450}{1000} = 0.45[/tex]

96% confidence level

So [tex]\alpha = 0.04[/tex], z is the value of Z that has a p-value of [tex]1 - \frac{0.04}{2} = 0.98[/tex], so [tex]Z = 2.054[/tex].  

The lower limit of this interval is:

[tex]\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.45 - 2.054\sqrt{\frac{0.45*0.55}{1000}} = 0.4177[/tex]

The upper limit of this interval is:

[tex]\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.45 + 2.054\sqrt{\frac{0.45*0.55}{1000}} = 0.4823[/tex]

The 96% CI on the death rate from lung cancer is (0.4177, 0.4823).

Answer: -9x-1y-9/

Step-by-step explanation:

Answer: b

Step-by-step explanation:

I really dont like edge

Answer:

10x-5

Step-by-step explanation:

f(x)=5x

g(x)=2x-1

To create a composite function, replace x in f(x) with g(x)

f(g(x)) = 5(g(x) = 5(2x-1) = 10x-5

Explanation:

The Ø​ means "empty set". It's the set with nothing inside it, not even 0.

We can write Ø​ as { } which is a pair of curly braces with nothing between them.

The rule is that if we union any set A with Ø​, then we'll get set A

A U Ø​ = A

Ø​ U A = A

In a sense, it's analogous to adding 0. So it's like saying A+0 = A and 0+A = A.

So that's why {6, 11, 15} U Ø​ = {6, 11, 15}

There's nothing to add onto the set {6, 11, 15}, so we just get the same thing back again.

Answer:

Answer to the first question is D. Answer to the second question is also D.

Step-by-step explanation:

First question:

All the sides of the square are equal meaning you just have to multiply 1 side by 4 to get the perimeter(all the sides added together.) If one side is (s+3) then you either add that to itself 4 times or multiply it by 4. It's the same thing so it's 4(s+3) and (s+3)+(s+3)+(s+3)+(s+3).

Second question:

Adding a negative number is equivalent to subtracting a positive number. In this case, 59.2-84.7 = 59.2+(-84.7)

Answer:

[tex]P = (7, -\frac{3}{5})[/tex]

Step-by-step explanation:

Given

[tex]A = (-9,5)[/tex]

[tex]B = (11,-2)[/tex]

[tex]m : n = 4 : 1[/tex]

Required

Point P

This is calculated as:

[tex]P = (\frac{m * x_2 + n * x_1}{m + n}, \frac{m * y_2 + n * y_1}{m + n})[/tex]

So, we have:

[tex]P = (\frac{4 * 11 + 1 * -9}{4 + 1}, \frac{4 * -2 + 1 * 5}{4+1})[/tex]

[tex]P = (\frac{35}{5}, \frac{-3}{5})[/tex]

[tex]P = (7, -\frac{3}{5})[/tex]

Answer:

$224.932

Step-by-step explanation:

Note: The question is not complete

say the rate is 10%

Given data

Initial depostite= $212

TIme= 6years

rate= 10%

the expression for the compound interest is given as

A=P(1+r)^t

substitute

A=212(1+0.1)^6

A=212(1.01)^6

A=212*1.061

A= $224.932

Hence the final amount at the rate of 10% is $224.932

Answer:

r = 4.1231055

Step-by-step explanation:

So to do this, you need to find the distance between the two points:

(-7,1) and (1,3).

To do this, the distance or diameter (d) is equal to:

d = sqrt ((x2-x1)^2 + (y2-y1)^2)

In this case:

d = sqrt( (1 - (-7))^2 + (3 - 1)^2 )

d = sqrt( 8^2 + 2^2)

d = sqrt( 64 + 4)

d = sqrt( 68 )

The radius is half of the diameter, so:

r = 1/2 * d

r = 1/2 * sqrt( 68 )

r~ 4.1231055

Answer:

39°

Step-by-step explanation:

A radius of a circle (segment CD) drawn to the point of tangency (D) intersects the tangent (line DE) at a 90-deg angle.

That makes m<D = 90.

m<D + m<C + m<E = 180

90 + 51 + m<E = 180

m<E = 39

Answer:

Answer to the following question is as follows.

Step-by-step explanation:

Given:

a = 12500000

b = 0.00125

c = 1120000​

Calculate (ab) ÷ (c)

Given:

d) Express the interval [-1.5, 4] as an inequality and then graph

Computation:

(ab) ÷ (c) = (a)(b) / c

(ab) ÷ (c) = (12500000)(0.00125) / (1120000​)

(ab) ÷ (c) = 25 / 1,792

Express the interval [-1.5, 4]

{x : -1.5 < x ≤ 4}

Graph.

._________._________.

-1.5              0                  4

Answer:

Give us the picture or numbers please.

Step-by-step explanation:

Answer: add pic pls

Step-by-step explanation:

Step-by-step explanation:

3600s=1hr

so, 1hr:1hr

1:1

Based on the concept of fractions and the information in the question, the fraction form in the lowest term is 1/2.

Fraction is a term that is used to describe the portion/part of the whole thing. It represents the equal parts of the whole.

Generally, the term fraction has two parts, namely numerator and denominator.

Hence, in this case, to express the ratio as a fraction in its lowest term, convert both units to the same unit of time.

1 hour is equal to 3600 seconds so 2 hours is equal to 2 * 3600 = 7200 seconds.

Now the ratio is 3600 seconds to 7200 seconds.

To simplify this ratio we can divide both terms by their greatest common divisor which is 3600.

So the simplified ratio is 1:2.

Therefore, in this case, it is concluded that the fraction form in the lowest term is 1/2.

Learn more about fraction here: https://brainly.com/question/30154928

#SPJ2

Answer:

Step-by-step explanation:

This is a differential equation problem most easily solved with an exponential decay equation of the form

[tex]y=Ce^{kt}[/tex]. We know that the initial amount of salt in the tank is 28 pounds, so

C = 28. Now we just need to find k.

The concentration of salt changes as the pure water flows in and the salt water flows out. So the change in concentration, where y is the concentration of salt in the tank, is [tex]\frac{dy}{dt}[/tex]. Thus, the change in the concentration of salt is found in

[tex]\frac{dy}{dt}=[/tex] inflow of salt - outflow of salt

Pure water, what is flowing into the tank, has no salt in it at all; and since we don't know how much salt is leaving (our unknown, basically), the outflow at 3 gal/min is 3 times the amount of salt leaving out of the 400 gallons of salt water at time t:

[tex]3(\frac{y}{400})[/tex]

Therefore,

[tex]\frac{dy}{dt}=0-3(\frac{y}{400})[/tex] or just

[tex]\frac{dy}{dt}=-\frac{3y}{400}[/tex] and in terms of time,

[tex]-\frac{3t}{400}[/tex]

Thus, our equation is

[tex]y=28e^{-\frac{3t}{400}[/tex] and filling in 16 for the number of minutes in t:

y = 24.834 pounds of salt

Answer:

The number of ways of attaching the 10 resistors = 5¹⁰ = 9,765,625 ways

Step-by-step explanation:

Given;

total number of resistors, n = 10

number of slots available, = 5

The first resistor can be attached in 5 ways,

The second resistor can also be attached in 5 ways,

The third resistor can also be attached in 5 ways, etc

Each of the resistors can be attached in 5 different ways;

The number of ways of attaching the 10 resistors = 5¹⁰ = 9,765,625 ways

Answer:

100

Step-by-step explanation:

f(1) = 1

f(2) = -10×f(1) = -10 × 1 = -10

f(3) = -10×f(2) = -10 × -10 × f(1) = -10 × -10 × 1 = 100

f(n) = -10 to the power of n-1

Answer:

c - 100

Step-by-step explanation:

Answer:

Step-by-step explanation:

This is a related rates problem from calculus using implicit differentiation. The main equation is Pythagorean's Theorem. Basically, what we are looking for is [tex]\frac{dx}{dt}[/tex] when y = 6 and [tex]\frac{dy}{dt}=-2[/tex].

The equation for Pythagorean's Theorem is

[tex]x^2+y^2=c^2[/tex] where x and y are the legs and c is the hypotenuse. The length of the hypotenuse is 10, so when we find the derivative of this function with respect to time, and using implicit differentiation, we get:

[tex]2x\frac{dx}{dt}+2y\frac{dy}{dt}=0[/tex] and divide everything by 2 to simplify:

[tex]x\frac{dx}{dt}+y\frac{dy}{dt}=0[/tex]. Looking at that equation, it looks like we need a value for x, y, [tex]\frac{dx}{dt}[/tex] and [tex]\frac{dy}{dt}[/tex].

Since we are looking for [tex]\frac{dx}{dt}[/tex], that can be our only unknown and everything else has to have a value. So what do we know?

If we construct a right triangle with 10 as the hypotenuse and use 6 for y, we can solve for x (which is the only unknown we have, actually). Using Pythagorean's Theorem to solve for x:

[tex]x^2+6^2=10^2[/tex] and

[tex]x^2+36=100[/tex] and

[tex]x^2=64[/tex] so

x = 8.

NOW we can fill in the derivative and solve for [tex]\frac{dx}{dt}[/tex].

Remember the derivative is

[tex]x\frac{dx}{dt}+y\frac{dy}{dt}=0[/tex] so

[tex]8\frac{dx}{dt}+6(-2)=0[/tex] and

[tex]8\frac{dx}{dt}-12=0[/tex] and

[tex]8\frac{dx}{dt}=12[/tex] so

[tex]\frac{dx}{dt}=\frac{12}{8}=\frac{6}{4}=\frac{3}{2}=1.5 m/sec[/tex]

Answer:

f(3) = g(3)

Step-by-step explanation:

on the graph the only point, where both lines cross (both functions create the same functional value) is at x=3.

since both lines have the same y-value there, we express this in math by the "=" sign. and both functions have the same input value (x=3) there.

Hey there!

[tex]\large\textsf{FACTORS OF 160: 1, 2, 4, 5, 8, 10, 16, 20, 32, 40, 80, \& 160}[/tex]

[tex]\large\textsf{FACTORS OF 198: 1, 2, 3, 6, 9, 11, 18, 22, 33, 66, 99, \& 198}[/tex]

[tex]\large\text{Go through the factors to see if there’s any like terms and if you find any,} \\\large\text{look for the greatest one the numbers share together.}[/tex]

[tex]\large\text{Like terms: \boxed{\textsf{\bf 1 \& 2}}}[/tex]

[tex]\large\checkmark\boxed{\large\text{GCF: \bf 2 }}\large\checkmark[/tex]

[tex]\boxed{\boxed{\large\textsf{Answer: \huge the GCF \underline{G}reatest \underline{C}ommon \underline{F}actor is \bf 2}}}\huge\checkmark[/tex]

[tex]\large\textsf{Good luck on your assignment and enjoy your day!}[/tex]

~[tex]\frak{Amphitrite1040:)}[/tex]

Step-by-step explanation:

620 / 17 =36.47058.. ≈ 36.5