al of
10. A square field has four sprinklers that spray
in the areas represented by the circles below. If
the shaded portion represents area that is not
reached by the sprinklers, find the total area that
is not reached by the sprinklers.

Slope-intercept form:

y = x + 2

Point-slope form:

y − 5 = 1 ⋅ ( x − 3 )

I hope this is correct and helps!

Answer:

C. 0.48

Step-by-step explanation:

Probability = number of required outcome

_______________________

number of possible outcome

= total volleyball game events

_______________________

total sophomore + junior

= 66/137

= 0.48

Answer: D) 0.31

Step-by-step explanation:

Let A denote the event that a person is a sophomore.

Let B denote the event that a person has attended volleyball game.

A∩B denote the event that a person is a sophomore and attend volleyball game.

Let P denote the probability of an event.

We are asked to find:

P(A∩B)

From the table provided to us we see that:

A∩B=42

Hence,

P(A∩B)=42/137=0.3065 which is approximately equal to 0.31. Therefore ur answer will be 0.31.

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:

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:

24/145

Step-by-step explanation:

Trigonometric identities are equalities involving trigonometric functions and remains true for entire values of the variables involved in the equation.

Some trigonometric identities are:

sin(a + b) = sinacosb + cosasinb; sin(a - b) = sinacosb - cosasinb

cos(a + b) = cosacosb - sinasinb; cos(a - b) = cosacosb + sinasinb

Given that sin a = 3/5. sin a = opposite/hypotenuse.

Hence opposite = 3, hypotenuse = 5. using Pythagoras:

hypotenuse² = opposite² + adjacent²

5² = 3² + adjacent²

adjacent² = 16

adjacent = 4

Given that sin a = 3/5. a = sin⁻¹(3/5) = 36.86

cos a = cos 36.86 = 4/5

cos b = -20/29; b = cos⁻¹(-20/29) = 133.6

sinb = sin(133.6) = 21/29

sin(a + b) = sinacosb + cosasinb = (3/5 * -20/29) + (4/5 * 21/29) = -12/29 + 84/145

sin(a + b) = 24/145

This question is incomplete, the complete question is;

A student bought a smart-watch that tracks the number of steps she walks throughout the day. The table shows the number of steps recorded (t) minutes after 3:00 pm on the first day she wore the watch.

t (min)       0          10          20         30         40

Steps   3,288    4,659    5,522    6,686    7,128

a) Find the slopes of the secant lines corresponding to the given intervals of t.

1) [ 0, 40 ]

11) [ 10, 20 ]

111) [ 20, 30 ]

b) Estimate the student's walking pace, in steps per minute, at 3:20 pm by averaging the slopes of two secant lines from part (a). (Round your answer to the nearest integer.)

Answer:

a)

1) for [ 0, 40 ], slope is 96

11) for [ 10, 20 ],  slope is 86.3

111) for  [ 20, 30 ], slope is 116.4

b) the student's walking pace is 101 per min

Step-by-step explanation:

Given the data in the question;

t (min)       0          10          20         30         40

Steps   3,288    4,659    5,522    6,686    7,128

SLOPE OF SECANT LINES

1) [ 0, 40 ]

slope =  ( 7,128 - 3,288 ) / ( 40 - 0

= 3840 / 40 = 96

Hence slope is 96

11)  [ 10, 20 ]

slope = ( 5,522 - 4,659 ) / ( 20 - 10 )

= 863 / 10 = 86.3

Hence slope is 86.3

111)  [ 20, 30 ]

slope = ( 6,686 - 5,522 ) / ( 30 - 20 )

= 1164 / 10 = 116.4

Hence slope is 116.4

b)

Estimate the student's walking pace, in steps per minute, at 3:20 pm by averaging the slopes of two secant lines from part .

Since this is recorded after 3:00 pm

{ 3:20 - 3:00 = 20 }

so t = 20 min

so by average;

we have ( [ 10, 20 ] + [ 20, 30 ] ) /2

⇒ ( 86.3 + 116.4 ) / 2

= 202.7 /2

= 101.35 ≈ 101

Therefore, the student's walking pace is 101 per minutes

Answer:

m(x) is a dilation of scale factor K = 1/5 of f(x).

Step-by-step explanation:

The transformation is a horizontal dilation

The general transformation is defined as:

For a given function f(x), a dilation of scale factor K is written as:

g(x) = f(x/K)

If K > 1, then we have a dilation (the graph contracts)

if 0 < K < 1, then we have a contraction (the graph stretches)

Here we have m(x) = f(5*x)

Then we have a scale factor:

K = 1/5

So this is a contraction.

Then the transformation is:

m(x) is a dilation of scale factor K = 1/5 of f(x).

Answer:

The answer to your question is given below.

Step-by-step explanation:

6x⁴ + 4x³ – 10x

The greatest common factor can be obtained as follow:

6x⁴ = 2 * 3 * x * x * x * x

4x³ = 2 * 2 * x * x * x

10x = 2 × 5 * x

Greatest common factor = 2 * x

= 2x

Thus, the expression 6x⁴ + 4x³ – 10x can be written as:

6x⁴ + 4x³ – 10x = 2x(3x³ + 2x² – 5)

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:

The first one

Step-by-step explanation:

You just need to find the slope in the average of all the dots.

Answer:

the first option, y=3/7x-3

Step-by-step explanation:

the scatterplot begins around y=-3, so therefore the y-intercept is -3. The slope is obviously not higher than 1, so it is y=3/7x-3.

Answer:

(110.295, 122.093).

Step-by-step explanation:

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

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 12 - 1 = 11

90% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 11 degrees of freedom(y-axis) and a confidence level of [tex]1 - \frac{1 - 0.9}{2} = 0.95[/tex]. So we have T = 1.7959

The margin of error is:

[tex]M = T\frac{s}{\sqrt{n}} = 1.7959\frac{11.3781}{\sqrt{12}} = 5.899[/tex]

In which s is the standard deviation of the sample and n is the size of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 116.194 - 5.899 = 110.295

The upper end of the interval is the sample mean added to M. So it is 116.194 + 5.899 = 122.093

So

(110.295, 122.093).

Answer:

B

Step-by-step explanation:

because

Answer:

6x

Step-by-step explanation:

If x is the length of one side, and each side is the same length, you will multiply it by 6 times (there are 6 sides in a hexagon).

So, you will add it up 6 times, but you can say 6x for short.

Answer:

They are going away from each other.

So add up their speed.

combined speed = x+x-16

=2x-16

Time = 2 hours

Distance = 400 miles

Distance = speed * time

(2x-16)* 2

4x-32=400

4x=400+32

4x=432

/12

x=108 mph west bound

east bound = 108 -16 = 92 mph

Answer: 3 years

Step-by-step explanation:

Interest is calculated as:

= (P × R × T) / 100

where

P = principal = 150,000

R = rate = 2.5%.

I = interest = 11250

T = time = unknown.

I = (P × R × T) / 100

11250 = (150000 × 2.5 × T)/100

Cross multiply

1125000 = 375000T

T = 1125000/375000

T = 3

The time taken will be 3 years

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:

21/8x

Step-by-step explanation:

10x -20 = -6x+22

+6x-20 = 22

16x-20 = 22

16x +20= +20

16x/16x = 42/16x

x = 21/8x

Answer:

The 14-ounce bottle is the better deal

Step-by-step explanation:

I know this beause inorder to figure out which one is better you have to make them the same price and then see which bottle has more ounces. So I made each price 1$ so there is 1.58-ounces per dollar in the 5-ounce bottle and 1.17 -ounces per dollar in the 14-ounce bottle.

Answer: -9x-1y-9/

Step-by-step explanation:

Answer: b

Step-by-step explanation:

I really dont like edge

Answer:

I don't understand the question

Answer:

The radius is increasing at a rate of 62832 cubic millimeters per second when the diameter is of 100 mm.

Step-by-step explanation:

Volume of a sphere:

The volume of a sphere of radius r is given by:

[tex]V = \frac{4\pi r^3}{3}[/tex]

How fast is the volume increasing:

To find this, we have to differentiate the variables of the problem, which are V and r, implicitly in function of time. So

[tex]\frac{dV}{dt} = 4\pi r^2\frac{dr}{dt}[/tex]

The radius of a sphere is increasing at a rate of 2 mm/s.

This means that [tex]\frac{dr}{dt} = 2[/tex]

How fast is the volume increasing (in mm3/s) when the diameter is 100 mm?

Radius is half the diameter, so [tex]r = \frac{100}{2} = 50[/tex]

Then

[tex]\frac{dV}{dt} = 4\pi r^2\frac{dr}{dt} = 4\pi (50)^2(2) = 62832[/tex]

The radius is increasing at a rate of 62832 cubic millimeters per second when the diameter is of 100 mm.

Answer:

7π/4 radians = 315°, Quadrant IV

sin(315°) = -√2/2

cos(315°) = √2/2

tan(315°) = -1

Step-by-step explanation:

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:

[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:

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.

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.