Solve this problem

how far apart are these planes?

Answer:

(A)29 cubic inch

Step-by-step explanation:

Diameter of the Cone =4 Inches

Height of the Cone =7 Inches

Volume of a Cone [tex]=\frac{1}{3}\pi r^2h[/tex]

First, we determine the radius, r.

Radius=Diameter/2=4/2=2 Inches

Therefore:

Volume of one paper cone [tex]=\frac{1}{3}\pi *2^2*7[/tex]

=29.32 cubic inch

=29 cubic inch (to the nearest cubic inch.)

Answer:

It's asking you to solve the question or find N

Step-by-step explanation:

Answer:

0.225

Step-by-step explanation:

Total outcomes of choosing 5 out of 18 members = 18C5

Outcomes of choosing 2 out 11 favourers, 3 out of 7 members  =          11C2 & 7C3

Probability = Favourable outcomes / Total outcomes

= ( 11C2 x 7C3 ) / 18C5

[ { 11 ! / 2! 9! } {7 ! / 3! 4! } ]

       [ 18 ! / 5! 13! ]

( 55 x 35 ) /  8568

1925 / 8568

= 0.2246 ≈ 0.225

Answer:

Amount invested in first account = $1,000

Amount invested in second account = $29,000

Step-by-step explanation:

Jina had total amount of $30,000 to invest last year.

Let x be the amount that Jina invested in the first account at interest rate of 9%

Mathematically,

0.09x

she invested the remaining amount in the second account at an interest rate of 5%

Mathematically,

0.05(30,000 - x)

Jina received $1540 in interest.

0.09x + 0.05(30,000 - x) = 1540

0.09x + 1500 - 0.05x = 1540

0.04x = 1540 - 1500

0.04x = 40

x = 40/0.04

x = $1,000

Therefore, Jina invested an amount of $1,000 in the first account at interest rate of 9%  

The remaining amount that she invested in the second account is

Amount invested in second account = $30,000 - $1,000

Amount invested in second account = $29,000

Therefore, Jina invested an amount of $29,000 in the second account at interest rate of 5%

Verification:

0.09x + 0.05(30,000 - x) = 1540

0.09(1000) + 0.05(30,000 - 1000) = 1540

90 + 1450 = 1540

1540 = 1540  (satisfied)

Answer:

The zeros are x1=4.45 and x2=-0.45.

x1 is between 4 and 5.

x2 is between -1 and 0.

Step-by-step explanation:

We have the function:

[tex]y(x) = x2 - 4x-2[/tex]

As this is a quadratic function, we can calculate the zeros of the function with the quadratic equation:

[tex]x_{1,2}=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}\\\\\\x=\dfrac{-(-4)\pm\sqrt{(-4)^2-4\cdot 1\cdot(-2)}}{2\cdot 1}\\\\\\x=\dfrac{4\pm\sqrt{16+8}}{2}\\\\\\x=\dfrac{4\pm\sqrt{24}}{2}\\\\\\x=\dfrac{4\pm4.9}{2}=2\pm2.45\\\\\\x_1=2+2.45=4.45\\\\x_2=2-2.45=-0.45[/tex]

The zeros are x1=4.45 and x2=-0.45.

x1 is between 4 and 5.

x2 is between -1 and 0.

The two real zeros of the quadratic equation are located between -1 and 4.

To determine between which consecutive integers the real zeros of the function y(x) = x² - 4x - 2 are located, we can use the quadratic formula.

The quadratic formula is given as:

x = (-b ± √(b² - 4ac)) / (2a)

In the equation y(x) = x² - 4x - 2, we have a = 1, b = -4, and c = -2.

Let's substitute these values into the quadratic formula to find the values of x:

x = (-(-4) ± √((-4)² - 4(1)(-2))) / (2(1))

x = (4 ± √(16 + 8)) / 2

x = (4 ± √24) / 2

x = (4 ± 2√6) / 2

x = 2 ± √6

From the quadratic formula, we find that the real zeros of the function are x = 2 + √6 and x = 2 - √6.

To determine between which consecutive integers these real zeros are located, we can compare the values to the nearest integers.

x = 2 + √6 is approximately 4.45

x = 2 - √6 is approximately -0.45

Therefore, the real zeros of the function are located between the consecutive integers -1 and 4.

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first find the greatest possible length:

9 + 6 > x

15 > x

x < 15

Then find the lowest possible length:

x + 6 > 9

x > 9 - 6

x > 3

3 < x < 15

To determine the range of possible lengths of the third side of a sail, use the triangle inequality theorem. The range is 3 feet to 15 feet.

To determine the range of possible lengths of the third side of the sail, we can use the triangle inequality theorem. According to this theorem, the sum of the lengths of any two sides of a triangle must be greater than the length of the third side.

In this case, the two given sides are 9 feet and 6 feet. So, the third side must be less than the sum of these two sides and greater than the difference between these two sides.

Therefore, the range of possible lengths of the third side of the sail is 3 feet to 15 feet.

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Answer:

 b = 16

Step-by-step explanation:

Answer:

-6b-12+8 which simplifies to -6b-4

Step-by-step explanation:use distrubutive property and then combine like terms

Answer:

[tex]\frac{(21)(5.437)^2}{41.402} \leq \sigma^2 \leq \frac{(21)(5.437)^2}{8.034}[/tex]

[tex] 14.996 \leq \sigma^2 \leq 77.278[/tex]

And the confidence interval for the deviation would be obtained taking the square root of the last result and we got:

3.9<σ<8.8

Step-by-step explanation:

Data given:

65.2 71.9 72.8 73.1 73.1 73.5 75.5 75.7 75.8 76.1 76.2 76.2 77.0 77.9 78.1 79.6 79.7 79.9 80.1 82.2 83.7 93.8

The sample mean would be given by:

[tex]\bar X = \frac{\sum_{i=1}^n X_i}{n}[/tex]

We can calculate the sample deviation with this formula:

[tex]s = \frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}[/tex]

And we got:

s=5.437 represent the sample standard deviation

[tex]\bar x[/tex] represent the sample mean

n=22 the sample size

Confidence=99% or 0.99

The confidence interval for the population variance is given by:

[tex]\frac{(n-1)s^2}{\chi^2_{\alpha/2}} \leq \sigma^2 \leq \frac{(n-1)s^2}{\chi^2_{1-\alpha/2}}[/tex]

The degrees of freedom given by:

[tex]df=n-1=22-1=21[/tex]

The Confidence is 0.99 or 99%, the value of significance is [tex]\alpha=0.01[/tex] and [tex]\alpha/2 =0.005[/tex], and the critical values are:

[tex]\chi^2_{\alpha/2}=41.402[/tex]

[tex]\chi^2_{1- \alpha/2}=8.034[/tex]

And the confidence interval would be:

[tex]\frac{(21)(5.437)^2}{41.402} \leq \sigma^2 \leq \frac{(21)(5.437)^2}{8.034}[/tex]

[tex] 14.996 \leq \sigma^2 \leq 77.278[/tex]

And the confidence interval for the deviation would be obtained taking the square root of the last result and we got:

3.9<σ<8.8

Answer:

Step-by-step explanation:

The formula for finding the volume is

Volume = length × width × height

Volume of the given prism is

Volume = 3 × 4 × 5 = 60 inches³

The formula for determining the surface area of a rectangular prism is expressed as

Surface area = Ph + 2B

Where

P represents perimeter of base

h represents height of prism

B represents base area

Perimeter of base = 2(length + width)

P = 2(3 + 4) = 14 inches

B = 3 × 4 = 12 inches

h = 5 inches

Surface area = 14 × 5 + 2 × 12 = 94 inches²

For another prism,

Assuming h = 3, length = 10 and width = 2, then

Volume = 3 × 10 × 2 = 60 inches³

P = 2(10 + 2) = 24 inches

B = 10 × 2 = 20 inches

Surface area = (24 × 3) + (2 × 20) = 112 inches²

If we keep changing the values, the surface area will always be greater than 94 inches².

Therefore, there is no rectangular prism with the same volume but less surface area.

Answer:

its 60 for volume, but for the area i don't know

Step-by-step explanation:

Journal entry

Explanation:

                                  Books of (----Limited)

                                      Journal Entry

Date         Account Title and Explanation            Debit         Credit

              Cash / Bank                          A/c  Dr.     $750,000

                To Unearned Subscription A/c                               $750,000

                            (Being Unearned Subscription)

Computation:

Amount of Unearned Subscription =  25,000 × $30

Amount of Unearned Subscription =  %750,000

The question is incomplete! Complete question along with answer and step by step explanation is provided below.

Question:

150 students in a tenth grade high school class take a survey about which video game consoles they own. 60  students answer that one of their consoles is a Playstation, 50 answer that one of their consoles is an Xbox. Out  of these, there are 20 who have both systems.

Let A be the event that a randomly selected student in the class has a Playstation and B be the event that the  student has an XBOX. Based on this information, answer the following questions.

a) What is P(A), the probability that a randomly selected student has a Playstation?

b) What is P(B), the probability that a randomly selected student has an XBOX?

c) What is P(A and B), the probability that a randomly selected student has a Playstation and an XBOX?

d) What is P(A | B), the conditional probability that a randomly selected student has a Playstation given that he  or she has an XBOX?

Answer:

a) P(A) = 2/5

b) P(B) = 1/3

c) P(A and B) = 2/15

d) P(A | B) = 2/5

Step-by-step explanation:

Total no. of students = 150

No. of students having playstation = 60

No. of students having xbox = 50

No. of students who have both playstation and xbox = 20

a) What is P(A), the probability that a randomly selected student has a Playstation?

P(A) = No. of students having playstation/Total no. of students

P(A) = 60/150

P(A) = 2/5

b) What is P(B), the probability that a randomly selected student has an XBOX?

P(B) = No. of students having xbox/Total no. of students

P(B) = 50/150

P(B) = 1/3

c) What is P(A and B), the probability that a randomly selected student has a Playstation and an XBOX?

The probability that a students has a Playstation and an Xbox is given by

P(A and B) = P(A)*P(B)

P(A and B) = (2/5)*(1/3)

P(A and B) = 2/15

d) What is P(A | B), the conditional probability that a randomly selected student has a Playstation given that he or she has an XBOX?

The conditional probability is given by

P(A | B) = P(A and B)/P(B)

P(A | B) = (2/15)/(1/3)

P(A | B) = 2/5

Alternatively:

P(A | B) = P(A∩B)/P(B)

Where P(A∩B) is given by

P(A∩B) = No. of students who have both playstation and xbox/Total no. of students

P(A∩B) = 20/150

P(A∩B) = 2/15

P(A | B) = P(A∩B)/P(B)

P(A | B) = (2/15)/(1/3)

P(A | B) = 2/5

Answer:

i think its c

Step-by-step explanation:

Answer:

C: Columns: likes roller coasters, doesn’t like roller coasters; rows: likes water rides, doesn’t like water rides

Step-by-step explanation:

Like variables need to be together on the columns and rows

Answer:

178

Step-by-step explanation:

Answer:

Step-by-step explanation:

Given

Initially Plane is flying at the speed of [tex]180\ km/hr[/tex]

to the [tex]15^{\circ}[/tex] North of east

Now wind started Blowing and plane started moving towards east with speed [tex]150\ km/hr[/tex]

suppose [tex]1v_o[/tex] is the speed of wind

So,

[tex]\vec{v_2}=\vec{v_1}-\vec{v_o}[/tex]

[tex]150\hat{i}=180[\cos 15\hat{i}+\sin 15\hat{j}]-\vec{v_o}[/tex]

[tex]\vec{v_o}=\hat{i}[180\cos 15-150]+\hat{j}[180\sin 15][/tex]

[tex]\vec{v_o}=\hat{i}[173.866-150]+46.58\hat{j}[/tex]

[tex]\vec{v_o}=23.86\hat{i}+46.58\hat{j}[/tex]

So magnitude of wind is

[tex]\mid v_o\mid=\sqrt{23.86^2+46.58^2}[/tex]

[tex]\mid v_o\mid=\sqrt{2738.996}[/tex]

[tex]\mid v_o\mid=52.33\ km/hr[/tex]

direction [tex]\tan \theta=\frac{46.58}{23.86}[/tex]

[tex]\theta =62.87^{\circ}[/tex] North of east

Answer:

The speed of the wind is 52.3 km/h

The direction of the wind is  243 degrees.

this is the right answer

Step-by-step explanation:

Answer:

Quadratic function (assuming 4x2 is 4x^2)

Step-by-step explanation:

Linear functions are ax+b=y

Quadratic is ax^2+b=y

Cubic is ax^3+b=y

The equation 4x² + 8x - 7 is a Quadratic equation.

What is Quadratic equation?

An algebraic equation with the second degree of the variable is called an Quadratic equation.

Given that;

The equation is,

⇒ 4x² + 8x - 7

Now,

Clearly, In the equation;

The highest power of a variable is two.

And, we know that;

In the quadratic equation, the highest power of a equation is two.

Thus, The equation 4x² + 8x - 7 is a Quadratic equation.

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Let y represent the y-coordinate of point A.

We have been given that point B has coordinates (5,1) The x-coordinate of point A is 0. So coordinates of point A would be (0,y)

The distance between point A and Point B is 13 units.

We will use distance formula to solve our given problem.

[tex]D=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}[/tex]

Let point A [tex](0,y)=(x_2-y_2)[/tex] and point A [tex](5,1)=(x_1,y_1)[/tex].

Upon substituting coordinates of both points in distance formula, we will get:

[tex]13=\sqrt{(0-5)^2+(y-1)^2}[/tex]

[tex]13=\sqrt{25+y^2-2y+1}[/tex]

[tex]13=\sqrt{y^2-2y+26}[/tex]

Let us square both sides as:

[tex]13^2=(\sqrt{y^2-2y+26})^2[/tex]

[tex]169=y^2-2y+26[/tex]

[tex]169-169=y^2-2y+26-169[/tex]

[tex]0=y^2-2y-143[/tex]

[tex]y^2-2y-143=0[/tex]

Upon splitting the middle term, we will get:

[tex]y^2-13y+11y-143=0[/tex]

[tex]y(y-13)+11(y-13)=0[/tex]

[tex](y-13)(y+11)=0[/tex]

Now we will use zero product property.

[tex](y-13)=0, (y+11)=0[/tex]

[tex]y=13, y=-11[/tex]

Therefore, the possible coordinates of point A would be [tex](0,-11)[/tex] and [tex](0,13)[/tex].

Answer:

Check the explanation

Step-by-step explanation:

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Answer:

So, if it grows by 6.75%, each year the population is 106.75% of the year before.

After 1 year, 370,000(1.0675). After two years, 370,000(1.0675)(1.0675).

370,000(1.0675)12 = your answer

Step-by-step explanation:

Answer:

asdfgbhnjm

Step-by-step explanation:

zxcvb

Answer:

Step-by-step explanation:

Im going to assum the wall your talking about is the one at the bottom.

It is divided into two equal trapezoids, so we can find the area of one trapezoid, multiply it by 2 to find the area of the entire wall.

A= (1/2)(b1+b2)(h)

Where b1 and b2 are the two parrellel sides, and h is the height of the trapezoid.

A= .5*(9+6)(8/4)   We divide by 8 by 2 because we are finding the area of one trapezoid which is half the height of the wall.

A= 30 ft squared.

2A= area of whole wall

2*30=60

The entire wall is 60 sq. ft.

2. The area you need to cover need to cover is 60 sqft.

Rolls of wallpaper are rectangular.

A = L*W and they told us the width is 2 ft. and we know the area we need to cover is 60 ft so we can subsitute those in to figure out the length.

60= L*2

L= 30 ft

You need at least 30 feet in length to cover the whole wall.

Answer:

There is a probability of P=0.02 of making a Type II error if the true mean is μ=1130.

Step-by-step explanation:

This is an hypothesis test for the lifetime of a certain ype of light bulb.

The population distribution is normal, with mean of 1,000 hours and STD of 110 hours.

The sample size for this test is n=10.

The significance level is assumed to be 0.05.

In this case, when the claim is that the new light bulb model has a longer average lifetime, so this is a right-tailed test.

For a significance level, the critical value (zc) that is bound of the rejection region is:

[tex]P(z>z_c)=0.05[/tex]

This value of zc is zc=1.645.

This value, for a sample with size n=10 is:

[tex]z_c=\dfrac{X_c-\mu}{\sigma/\sqrt{n}}\\\\\\X_c=\mu+\dfrac{z_c\cdort\sigma}{\sqrt{n}}=1000+\dfrac{1.645*110}{\sqrt{10}}=1000+57.22=1057.22[/tex]

That means that if the sample mean (of a sample of size n=10) is bigger than 1057.22, the null hypothesis will be rejected.

The Type II error happens when a false null hypothesis failed to be rejected.

We now know that the true mean of the lifetime is 1130, the probability of not rejecting the null hypothesis (H0: μ=1100) is the probability of getting a sample mean smaller than 1057.22.

The probability of getting a sample smaller than 1057.22 when the true mean is 1130 is:

[tex]z=\dfrac{X-\mu}{\sigma/\sqrt{n}}=\dfrac{1057.22-1130}{110/\sqrt{10}}=\dfrac{-72.78}{34.7851}=-2.0923 \\\\\\P(M<1057.22)=P(z<-2.0923)=0.01821[/tex]

Then, there is a probability of P=0.02 of making a Type II error if the true mean is μ=1130.

Using the normal distribution and the central limit theorem, it is found that there is a 0.0001 = 0.01% probability of a type II error.

In a normal distribution with mean and standard deviation , the z-score of a measure X is given by:

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

In this problem:

We test if the average lifetime is longer, and a Type II error is concluding that it is not longer when in fact it is longer, hence, it is the probability of finding a sample mean below 1000 hours, which is the p-value of Z when X = 1000.

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

By the Central Limit Theorem

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

[tex]Z = \frac{1000 - 1130}{\frac{110}{\sqrt{10}}}[/tex]

[tex]Z = -3.74[/tex]

[tex]Z = -3.74[/tex] has a p-value of 0.0001.

0.0001 = 0.01% probability of a type II error.

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Answer: M≤ $15

Step-by-step explanation:

The answer is [ m ≤ $15 ]

The question states, "If you have no more than $15." This means that you can only have up to $15 and no more.

So, the solutions to the set are all real numbers except the numbers after 15.

Let's say you had $9. If we were to substitute 9 with m in the inequality, we would get; 9 ≤ 15. This satisfies the inequality since the number is less than 15.

Best of Luck!

Answer:

Would you be able to write it in english so i can help you.

Step-by-step explanation:

Step-by-step explanation:

9 + x - 7

Solving like terms

x + 2

If we find the value of x

X + 2 = 0

x = - 2

Answer:

She uses 1.7475 cups of flour.

Step-by-step explanation:

This question can be solved using a rule of three.

For each batch of cookies:

Two and one-third cups of flour.

So [tex]2 + \frac{1}{3} = 2.33[/tex] cups.

If she makes three-fourths of a batch of cookies, how much flour does she use?

3/4 = 0.75 batch of cookies. How much flour?

1 batch - 2.33 cups.

0.75 batches - x cups

x = 2.33*0.75

x = 1.7475

She uses 1.7475 cups of flour.

Answer:

1 3/4

Step-by-step explanation:

Final answer:

The number 7,790,200 has zero decimal places as it is a whole number with no fractional part and ends in the unit's place.

Explanation:

The student asked how many decimal places are in 7,790,200. This question is related to place value and decimals in mathematics. The number 7,790,200 has no decimal part since it is a whole number and ends in the unit's place. Therefore, it contains zero decimal places. To expand on place value, when expressing numbers in decimal form, the position of a digit represents its value in powers of ten. For instance, the number 1837 can be decomposed as (1 × 10³) + (8 × 10²) + (3 × 10¹) + (7 × 10⁰), which shows the ones, tens, hundreds, and thousands places respectively.

Answer:k(-7)=-89

Step-by-step explanation:

since k(t)=10t - 19

K(-7)=10(-7)-19

k(-7)=-70-19

k(-7)=-89

Answer:

Check the explanation

Step-by-step explanation:

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Answer:

There are 1,585,080 ways for her to select students and assign them to the jobs

Step-by-step explanation:

The order in which the students are selected is important, since different orderings means different jobs for each student selected. So the permutations formula is used to solve this question.

Permutations formula:

The number of possible permutations of x elements from a set of n elements is given by the following formula:

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

In this problem:

4 students selected from a set of 37. So

[tex]P_{(37,4)} = \frac{37!}{(37-4)!} = 1585080[/tex]

There are 1,585,080 ways for her to select students and assign them to the jobs

Answer:

The probability that the sample mean weight will be more than 262 lb is 0.0047.

Step-by-step explanation:

The random variable X can be defined as the weight of National Football League (NFL) players now.

The mean weight is, μ = 252.8 lb.

The standard deviation of the weights is, σ = 25 lb.

A random sample of n = 50 NFL players are selected.

According to the Central Limit Theorem if we have an unknown population with mean μ and standard deviation σ and appropriately huge random samples (n > 30) are selected from the population with replacement, then the distribution of the sample means will be approximately normally distributed.

Then, the mean of the sample means is given by,

[tex]\mu_{\bar x}=\mu[/tex]

And the standard deviation of the sample means is given by,

[tex]\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}[/tex]

The sample of players selected is quite large, i.e. n = 50 > 30, so the central limit theorem can be used to approximate the distribution of sample means.

[tex]\bar X\sim N(\mu_{\bar x}=252.8,\ \sigma_{\bar x}=3.536)[/tex]

Compute the probability that the sample mean weight will be more than 262 lb as follows:

[tex]P(\bar X>262)=P(\frac{\bar X-\mu_{\bar x}}{\sigma_{\bar x}}>\frac{262-252.8}{3.536})\\\\=P(Z>2.60)\\\\=1-P(Z<2.60)\\\\=1-0.99534\\\\=0.00466\\\\\approx 0.0047[/tex]

*Use a z-table for the probability.

Thus, the probability that the sample mean weight will be more than 262 lb is 0.0047.

To find the probability that the sample mean will be more than 262 lb, calculate the z-score using the sample mean, population mean, standard deviation, and sample size. Then, find the corresponding probability using the standard normal distribution table. Subtract the probability from 1 to get the final result, which is approximately 0.2%.

To solve this problem, we need to use the z-score formula and the standard normal distribution table. First, calculate the z-score using the formula: z = (x - μ) / (σ / sqrt(n)), where x is the sample mean, μ is the population mean, σ is the population standard deviation, and n is the sample size. In this case, x = 262 lb, μ = 252.8 lb, σ = 25 lb, and n = 50. Plug in these values and calculate the z-score. Next, find the corresponding probability using the standard normal distribution table. Look up the z-score and find the corresponding probability. The probability that the sample mean will be more than 262 lb can be found by subtracting the probability you found from 1.

Calculating the z-score:

z = (262 - 252.8) / (25 / sqrt(50)) = 2.901.

Using the standard normal distribution table, the probability corresponding to a z-score of 2.901 is approximately 0.998. Therefore, the probability that the sample mean will be more than 262 lb is approximately 1 - 0.998 = 0.002, or 0.2%.

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We have been given that a baseball player threw 82 strikes out of 103 pitches. We are asked to find the percentage of pitches that were strikes.

To solve our given problem, we need to find strikes are what percent of pitches.

[tex]\text{Percentage of pitches that were strikes}=\frac{82}{103}\times 100\%[/tex]

[tex]\text{Percentage of pitches that were strikes}=0.7961165\times 100\%[/tex]

[tex]\text{Percentage of pitches that were strikes}=79.61165\%[/tex]

[tex]\text{Percentage of pitches that were strikes}\approx 79.6\%[/tex]

Therefore, approximately [tex]79.6\%[/tex] of pitches were strikes.