What is the volume of this container?

Answer:

Let x be her initial distance from the building, then tan 37 = 130/x

x = 130/tan 37 = 130/0.7536 = 172.5 feet

Let y be her distance from the building after moving forward, then

tan 40 = 130/y

y = 130/tan 40 = 130/0.8391 = 154.9

After moving forward, she is 172.5 - 154.9 = 17.6 ft closer.

Answer: B. 17.6 ft.

Step-by-step explanation: I just did it on the edge 2023 assignment. Check attached image.

Answer: 3.9

Step-by-step explanation: Khan Academy

The length of BD if The angle ∠ DAC is equal to the angle ∠ BAD is 3.92.

Three straight lines coming together create a triangle. There are three sides and three corners on every triangle (angles). A triangle's vertex is the intersection of two of its sides. Any one of a triangle's three sides can serve as its base, however typically the bottom side is used.

Given:

The angle ∠ DAC = angle ∠ BAD

As we can see that the triangle BAD and triangle DAC are similar By SAS similarity,

AC / AB = CD / BD  (By the proportional theorem of similarity)

5.6 / 5.1 = 4.3 / BD

1.09 = 4.3 / BD

BD = 4.3 / 1.09

BD = 3.92

Thus, the length of BD is 3.92.

To know more about Triangles:

https://brainly.com/question/16886469

#SPJ2

Answer:

[tex]\frac{1}{6}[/tex]

Step-by-step explanation:

Answer:

Step-by-step explanation:

[tex]\frac{5}{6}-\frac{2}{3}=\frac{5}{6}-\frac{2*2}{3*2}\\\\=\frac{5}{6}-\frac{4}{6}\\\\=\frac{5-4}{6}\\\\=\frac{1}{6}[/tex]

Answer:

The answer is B

Step-by-step explanation:

Hope this helps.. if not im sorry :(

Answer:

50,063,860 different samples can be chosen

Step-by-step explanation:

The order in which the microchips are chosen is not important. So we use the combinations formula to solve this question.

Combinations formula:

[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]

How many different samples can be chosen

We choose 6 microchips from a set of 60. So

[tex]C_{60,6} = \frac{60!}{6!(60-6)!} = 50063860[/tex]

50,063,860 different samples can be chosen

Answer:

radius r of the zorb is ≅ 1.40 m

Step-by-step explanation:

GIven that;

the  radius r of a sphere with surface area A is given by;

[tex]r = \sqrt{\dfrac {A}{4 \pi }}[/tex]  which is read as : (r equals StartRoot StartFraction Upper A Over 4 pi EndFraction EndRoot .)

We are to calculate the radius of a zorb whose outside surface area is  24.63 sq  ( the missing part of the question)

Given that the outside surface area is : 24.63 sq

Let replace the value of the outside surface area which 24.63 sq for A in the equation given from above.

SO:  A = 24.63 sq

[tex]r = \sqrt{\dfrac {A}{4 \pi }}[/tex]

[tex]r = \sqrt{\dfrac {24.63}{4 \pi }}[/tex]

[tex]r = \sqrt{1.9599}[/tex]

r = 1.399

radius r of the zorb is ≅ 1.40 m

Answer:

10 and 8

Step-by-step explanation:

10 and 8 are the factors in this equation because factors are the numbers that are mutiplied together to get the product (The answer to a mutiplication problem) Therefore the factors in this equation are 10 and 8 because those are the numbers that are mutiplied together to get the product.

Answer:

41 word/min

Step-by-step explanation:

Before noon Ali works:

She types:

After lunch she works:

She types:

Total Ali works= 4+4= 8 hours= 480 min

Total Ali types= 11520+8160= 19680 words

Average typing rate= 19680 words/480 min= 41 word/min

Answer:

95% for the true average increase in total cholesterol under these circumstances

(-2.306 , 9.406)

Step-by-step explanation:

Step(i):-

Given sample size 'n' =41

Mean of the sample(x⁻)  = 3.55

The estimated standard error

                                             [tex]S.E = \frac{S.D}{\sqrt{n} }[/tex]

Given  estimated standard error ( S.E) = 3.478

Level of significance ∝=0.05

Step(ii):-

95% for the true average increase in total cholesterol under these circumstances

[tex](x^{-} - t_{0.05} S.E ,x^{-} + t_{0.05} S.E)[/tex]

Degrees of freedom

ν= n-1 = 41-1 =40

t₀.₀₅ = 1.6839

95% for the true average increase in total cholesterol under these circumstances

[tex](x^{-} - t_{0.05} S.E ,x^{-} + t_{0.05} S.E)[/tex]

( 3.55 - 1.6839 ×3.478 ,3.55 + 1.6839 ×3.478 )

(3.55 - 5.856 , 3.55 + 5.856)

(-2.306 , 9.406)

Conclusion:-

95% for the true average increase in total cholesterol under these circumstances

(-2.306 , 9.406)

Answer:

The numbers of doors that will have no blemishes will be about 6065 doors

Step-by-step explanation:

Let the number of counts by the  worker of each blemishes on the door be (X)

The distribution of blemishes followed the Poisson distribution with parameter  [tex]\lambda =0.5[/tex] / door

The probability mass function on of a poisson distribution Is:

[tex]P(X=x) = \dfrac{e^{- \lambda } \lambda ^x}{x!}[/tex]

[tex]P(X=x) = \dfrac{e^{- \ 0.5 }( 0.5)^ x}{x!}[/tex]

The probability that no blemishes occur is :

[tex]P(X=0) = \dfrac{e^{- \ 0.5 }( 0.5)^ 0}{0!}[/tex]

[tex]P(X=0) = 0.60653[/tex]

P(X=0) = 0.6065

Assume the number of paints on the door by q = 10000

Hence; the number of doors that have no blemishes  is = qp

=10,000(0.6065)

= 6065

Answer:

The 95% confidence interval for the average hourly wage of all information system managers is (39.14,42.36)

Step-by-step explanation:

In order to calculate the 95% confidence interval for the average hourly wage we would have to calculate first the margin of error as follows:

ME=t0.05/2,n-1s/√n

for n=75, t0.025,74=1.993

Therefore, ME=1.993*7/√75

ME=1.61

Therefore, the 95% confidence interval for the average hourly income of all syatem manager would be as follows:

(X-ME,X+ME)=(40.75-1.61,40.75+1.61)

=(39.14,42.36)

Answer:

1)

A) [tex]\frac{}{X}[/tex] ~ N(8.5;0.108)

B) P([tex]\frac{}{X}[/tex] > 9)= 0.0552

C) P(X> 9)= 0.36317

D) IQR= 0.4422

2)

A) [tex]\frac{}{X}[/tex] ~ N(30;2.5)

B) P( [tex]\frac{}{X}[/tex]<30)= 0.50

C) P₉₅= 32.60

D) P( [tex]\frac{}{X}[/tex]>36)= 0

E) Q₃: 31.0586

Step-by-step explanation:

Hello!

1)

The variable of interest is

X: pollutants found in waterways near a large city. (ppm)

This variable has a normal distribution:

X~N(μ;σ²)

μ= 8.5 ppm

σ= 1.4 ppm

A sample of 18 large cities were studied.

A) The sample mean is also a random variable and it has the same distribution as the population of origin with exception that it's variance is affected by the sample size:

[tex]\frac{}{X}[/tex] ~ N(μ;σ²/n)

The population mean is the same as the mean of the variable

μ= 8.5 ppm

The standard deviation is

σ/√n= 1.4/√18= 0.329= 0.33 ⇒σ²/n= 0.33²= 0.108

So: [tex]\frac{}{X}[/tex] ~ N(8.5;0.108)

B)

P([tex]\frac{}{X}[/tex] > 9)= 1 - P([tex]\frac{}{X}[/tex] ≤ 9)

To calculate this probability you have to standardize the value of the sample mean and then use the Z-tables to reach the corresponding value of probability.

Z= [tex]\frac{\frac{}{X} - Mu}{\frac{Sigma}{\sqrt{n} } } = \frac{9-8.5}{0.33}= 1.51[/tex]

Then using the Z table you'll find the probability of

P(Z≤1.51)= 0.93448

Then

1 - P([tex]\frac{}{X}[/tex] ≤ 9)= 1 - P(Z≤1.51)= 1 - 0.93448= 0.0552

C)

In this item, since only one city is chosen at random, instead of working with the distribution of the sample mean, you have to work with the distribution of the variable X:

P(X> 9)= 1 - P(X ≤ 9)

Z= (X-μ)/δ= (9-8.5)/1.44

Z= 0.347= 0.35

P(Z≤0.35)= 0.63683

Then

P(X> 9)= 1 - P(X ≤ 9)= 1 - P(Z≤0.35)= 1 - 0.63683= 0.36317

D)

The first quartile is the value of the distribution that separates the bottom 2% of the distribution from the top 75%, in this case it will be the value of the sample average that marks the bottom 25% symbolically:

Q₁: P([tex]\frac{}{X}[/tex]≤[tex]\frac{}{X}[/tex]₁)= 0.25

Which is equivalent to the first quartile of the standard normal distribution. So first you have to identify the first quartile for the Z dist:

P(Z≤z₁)= 0.25

Using the table you have to identify the value of Z that accumulates 0.25 of probability:

z₁= -0.67

Now you have to translate the value of Z to a value of [tex]\frac{}{X}[/tex]:

z₁= ([tex]\frac{}{X}[/tex]₁-μ)/(σ/√n)

z₁*(σ/√n)= ([tex]\frac{}{X}[/tex]₁-μ)

[tex]\frac{}{X}[/tex]₁= z₁*(σ/√n)+μ

[tex]\frac{}{X}[/tex]₁= (-0.67*0.33)+8.5=  8.2789 ppm

The third quartile is the value that separates the bottom 75% of the distribution from the top 25%. For this distribution, it will be that value of the sample mean that accumulates 75%:

Q₃: P([tex]\frac{}{X}[/tex]≤[tex]\frac{}{X}[/tex]₃)= 0.75

⇒ P(Z≤z₃)= 0.75

Using the table you have to identify the value of Z that accumulates 0.75 of probability:

z₃= 0.67

Now you have to translate the value of Z to a value of [tex]\frac{}{X}[/tex]:

z₃= ([tex]\frac{}{X}[/tex]₃-μ)/(σ/√n)

z₃*(σ/√n)= ([tex]\frac{}{X}[/tex]₃-μ)

[tex]\frac{}{X}[/tex]₃= z₃*(σ/√n)+μ

[tex]\frac{}{X}[/tex]₃= (0.67*0.33)+8.5=  8.7211 ppm

IQR= Q₃-Q₁= 8.7211-8.2789= 0.4422

2)

A)

X ~ N(30,10)

For n=4

[tex]\frac{}{X}[/tex] ~ N(μ;σ²/n)

Population mean μ= 30

Population variance σ²/n= 10/4= 2.5

Population standard deviation σ/√n= √2.5= 1.58

[tex]\frac{}{X}[/tex] ~ N(30;2.5)

B)

P( [tex]\frac{}{X}[/tex]<30)

First you have to standardize the value and then look for the probability:

Z=  ([tex]\frac{}{X}[/tex]-μ)/(σ/√n)= (30-30)/1.58= 0

P(Z<0)= 0.50

Then

P( [tex]\frac{}{X}[/tex]<30)= 0.50

Which is no surprise since 30 y the value of the mean of the distribution.

C)

P( [tex]\frac{}{X}[/tex]≤ [tex]\frac{}{X}[/tex]₀)= 0.95

P( Z≤ z₀)= 0.95

z₀= 1.645

Now you have to reverse the standardization:

z₀= ([tex]\frac{}{X}[/tex]₀-μ)/(σ/√n)

z₀*(σ/√n)= ([tex]\frac{}{X}[/tex]₀-μ)

[tex]\frac{}{X}[/tex]₀= z₀*(σ/√n)+μ

[tex]\frac{}{X}[/tex]₀= (1.645*1.58)+30= 32.60

P₉₅= 32.60

D)

P( [tex]\frac{}{X}[/tex]>36)= 1 - P( [tex]\frac{}{X}[/tex]≤36)= 1 - P(Z≤(36-30)/1.58)= 1 - P(Z≤3.79)= 1 - 1 = 0

E)

Q₃: P([tex]\frac{}{X}[/tex]≤[tex]\frac{}{X}[/tex]₃)= 0.75

⇒ P(Z≤z₃)= 0.75

z₃= 0.67

z₃= ([tex]\frac{}{X}[/tex]₃-μ)/(σ/√n)

z₃*(σ/√n)= ([tex]\frac{}{X}[/tex]₃-μ)

[tex]\frac{}{X}[/tex]₃= z₃*(σ/√n)+μ

[tex]\frac{}{X}[/tex]₃= (0.67*1.58)+30= 31.0586

Q₃: 31.0586

Answer:

Bb

Step-by-step explanation:

Answer:

The third options

Step-by-step explanation:

Counting we can see that 10 students went to two or less states, and 10 went to three or more

Answer:

Step-by-step explanation:

From the information given, the population is divided into sub groups. Each group would consist of citizens picking a particular choice as the most important problem facing the country. The choices are the different categories. In this case, the null hypothesis would state that the distribution of proportions for all categories is the same in each population. The alternative hypothesis would state that the distributions is different. Therefore, the correct test to use to determine if the distribution of​ "problem facing this country ​today" is different between the two different​ years is

A) Use a​ chi-square test of homogeneity.

Answer:

91 2/3 pi cubic units

Step-by-step explanation:

The formula for the volume of cone is [tex]\dfrac{1}{3}\pi r^2 h[/tex]. Plugging in the given numbers, you get:

[tex]\dfrac{1}{3}\cdot \pi \cdot 5^2 \cdot 11= 91 \ 2/3 \pi[/tex]

Hope this helps!

Answer:

[tex]Volume=\frac{1}{3} \,275\,\pi[/tex] cubic units

Notice that this answer doesn't agree with any of the first three in the list provided via the screenshot

Step-by-step explanation:

Recall the formula for the volume of a cone:

[tex]Volume=\frac{1}{3} Base\,*\,Height[/tex]

In this case the Height is 11 units, and they also give us the radius of the circular base (5 units) from which we can find the circle's base area:

[tex]Area_{circle} = \pi\,R^2\\Area_{circle}=\pi\,(5)^2\\Area_{circle}=25 \pi[/tex]

therefore the total volume becomes:

[tex]Volume=\frac{1}{3} Base\,*\,Height\\Volume=\frac{1}{3} 25\,\pi\,*\,11\\\\Volume=\frac{1}{3} \,275\,\pi[/tex]

Answer:

Second option is the correct choice. See the explanation below.

Step-by-step explanation:

[tex]A=\frac{bh}{2}\\\\\mathrm{Switch\:sides}:\\\\\frac{bh}{2}=A\\\\\mathrm{Multiply\:both\:sides\:by\:}2\\\\\frac{2bh}{2}=2A\\\\hb=2A\\\\\mathrm{Divide\:both\:sides\:by\:}b;\\\\\frac{hb}{b}=\frac{2A}{b}\\\\h=\frac{2A}{b}[/tex]

Best Regards!

Answer:

[tex]h = \frac{2A}{b} [/tex]

Option B is the right option.

Solution,

[tex]a = \frac{bh}{2} \\ 2a = bh(cross \: multiplication) \\ 2a = b \times h \\ h = \frac{2A}{b} [/tex]

hope this helps...

Good luck on your assignment..

Answer:

Step-by-step explanation:

Let the money earned during fundraising activities =x

Since the World Issues club has decided to donate 60% of all their fundraising activities this year to Stephen Lewis Foundation.

The amount of money the foundation will receive

=60% of x

= 0.6x

In the bake sale, the club raised $72.

Therefore, the amount the foundation will receive =0.6*72=$43.20

At the end of the year, the World Issues Club mailed a cheque to the foundation for $850.

Therefore:

0.6x=850

x=850/0.6

x=$1416.67

The total amount of money the club raised is $1416.67.

Answer:

HJ > PK

Step-by-step explanation:

Notice that the side PL in one triangle has the same length as side GJ in the other, and side GH has the same size as side LK of the other triangle. Now what is different is the angle subtended between these sides in the case of the triangle on the lower left, the subtended angle is [tex]90^o[/tex] , which is larger angle than that subtended between equal sides on the other triangle ([tex]85^o[/tex])

Therefore, if the angle subtended by the equivalent sides in the triangle on the left is larger than the angle subtended on the right hand side triangle, then the sides associated with such angle aperture must keep the inequality. That is:

Since [tex]\angle\,G\,\,\,>\,\,\,\angle \,L[/tex], then   HJ > PK

Answer:

A. [tex](3b+8)^2[/tex]

Step-by-step explanation:

[tex]9b^2+48b +64\\=(3b)^2 + 2\times 3b\times 8 +(8)^2\\=(3b+8)^2[/tex]

Answer:

Just replace the variables with the number

d5

c4 (uh oh)

a2

b-3

f-7

d-c = 5 - 4 = 1

1/3 - 4(ab+f)

2 x -3 = -6

-6 + -7 = -13

-13 x 4 = -52

1/3 - -52 = 1/3 + 52 =

52 1/3

Hope this helps

Step-by-step explanation:

Answer:

The probability that demand during lead-time will exceed 20 bails is 0.2033.

Step-by-step explanation:

We are given that it has been previously determined that demand during the lead-time is normally distributed with a mean of 15 bails and a standard deviation of 6 bails.

Let X = demand during the lead-time

So, X ~ Normal([tex]\mu=15, \sigma^{2} = 6^{2}[/tex])

The z-score probability distribution for the normal distribution is given by;

                               Z  =  [tex]\frac{X-\mu}{\sigma}[/tex]  ~ N(0,1)

where, [tex]\mu=[/tex] population mean demand = 15 bails

           [tex]\sigma[/tex] = standard deviation = 6 bails

Now, the probability that demand during lead-time will exceed 20 bails is given by = P(X > 20 bails)

       P(X > 20 bails) = P( [tex]\frac{X-\mu}{\sigma}[/tex] > [tex]\frac{20-15}{6}[/tex] ) = P(Z > 0.83) = 1 - P(Z [tex]\leq[/tex] 0.83)

                                                             = 1 - 0.7967 = 0.2033

Answer:

The 95% confidence interval is 2.5 < u <3.1.

Step-by-step explanation:

The provided sample mean is X = 2.8 and the sample standard deviation is s = 1.2, and the sample size is n = 64.

1. Null and Alternative Hypotheses:

The following null and alternative hypotheses need to be tested:

H0  u = 3

Ha: u < 3

This corresponds to a left-tailed test, for which a t-test for one mean, with unknown population standard deviation will be used.

2. Rejection Region Based

on the information provided, the significance level is alpha = 0.05, and the critical value for a left-tailed test is t c = -1.669.

The rejection region for this left-tailed test is R = t : t < -1.669

3. Test Statistics

The t-statistic is computed as follows:

t = (X - uo)/[s/n^(1/2)] =

replacing

t = (2.8 - 3)/ [1.2/64 ^(1/2)]

t =-1.333  

4. Decision about the null hypothesis

Since it is observed that t = -1.333 > t c = -1.669, it is then concluded that the null hypothesis is not rejected.

Using the P-value approach: The p-value is p = 0.0936, and since p= 0.0936 => 0.05, it is concluded that the null hypothesis is not rejected.

5. Conclusion It is concluded that the null hypothesis H0 is not rejected. Therefore, there is not enough evidence to claim that the population mean u is less than 3, at the 0.05 significance level.

Confidence Interval

The 95% confidence interval is 2.5 < u <3.1.

Answer:

-30000/100

300 0/1

Step-by-step explanation:

We have the following numbers -18 and 16 2/3, the first is an integer and the second is a mixed number, the first thing is to pass the mixed number to a decimal number.

16 2/3 = 16.67

We do the multiplication:

−18⋅16 2/3 = -300

We have an improper fraction is a fraction in which the numerator (top number) is greater than or equal to the denominator (bottom number), therefore it would be:

-30000/100

How mixed number would it be:

300 0/1

Answer:

(a)The total sales of ice-cream last month is 702 gallons.

(b)The total sales of broccoli last month is 510 pounds.

Step-by-step explanation:

Part A

Total Sales of gallons of ice cream this month = 597

Since it represents a decrease of 15% of last month's sales

Let the total sales of ice-cream last month =x

Then:

(100-15)% of x =597

85% of x=597

0.85x=597

x=597/0.85

x=702 (to the nearest integer)

The total sales of ice-cream last month is 702 gallons.

Part B

Total Sales of broccoli this month = 617 pounds

Since it represents an increase of 21% of last month's sales

Let the total sales of ice-cream last month =y

Then:

(100+21)% of y =617

121% of y=617

1.21y=617

y=617/1.21

y=510 (to the nearest integer)

The total sales of broccoli last month is 510 pounds.

Answer:

Option C is correct.

The sampling distribution with sample size n=100 will have less variability.

Step-by-step explanation:

Complete Question

Consider random samples selected from the population of all female college soccer players in the United States. Assume the mean height of female college soccer players in the United States is 66 inches and the standard deviation is 3.5 inches. Which do you expect to have less variability (spread): a sampling distribution with sample size n = 100 or a sample size of n = 20.

A. Both sampling distributions will have the same variability.

B.The sampling distribution with sample size n=20 will have less variability

C. The sampling distribution with sample size n =100 will have less variability

Solution

The central limit theorem allows us to say that as long as

- the sample is randomly selected from the population distribution with each variable independent of each other and with the sample having an adequate enough sample size.

- the random sample is normal or almost normal which is guaranteed if the population distribution that the random sample was extracted from is normal or approximately normal,

1) The mean of sampling distribution (μₓ) is approximately equal to the population mean (μ)

μₓ = μ = 66 inches

2) The standard deviation of the sampling distribution or the standard error of the sample mean is related to the population standard deviation through

σₓ = (σ/√N)

where σ = population standard deviation = 3.5 inches

N = Sample size

And the measure of variability for a sampling distribution is the magnitude of the standard deviation of the sampling distribution.

For sampling distribution with sample size n = 100

σₓ = (3.5/√100) = 0.35 inch

For sampling distribution with sample size n = 20

σₓ = (3.5/√20) = 0.7826 inch

The standard deviation of the sampling distribution with sample size n = 20 is more than double that of the sampling distribution with sample size n = 100, hence, it is evident that the bigger the sample size, the lesser the standard deviation of the sampling distribution and the lesser the variability that the sampling distribution shows.

Hope this Helps!!!

Answer:

[tex]z=-\frac{20}{3}[/tex]

Step-by-step explanation:

[tex]\frac{3z}{10}-4=-6\\\\\frac{3z}{10}-4+4=-6+4\\\\\frac{3z}{10}=-2\\\\\frac{10\cdot \:3z}{10}=10\left(-2\right)\\\\3z=-20\\\\\frac{3z}{3}=\frac{-20}{3}\\\\z=-\frac{20}{3}[/tex]

Best Regards!

Step-by-step explanation:

1. 180 - (36 +36)= 108

2. angle ABC = 108 angle DBC = 108-72=36

3. angle DCB=angle DBC. This is because the base angles are equal

4 therefore triangle BDC is isoscles

Answer:

Because ΔABD is isosceles, ∠ABD ≅ ∠ADB = 72° because of Base Angles Theorem which states that the base angles of an isosceles triangle are congruent. Then, ∠BDC = 180° - ∠ADB = 108° because they are supplementary angles. Because ΔABC is isosceles, ∠BAC ≅ ∠BCA = 36° because of Base Angles Theorem, which means ∠CBD = 180° - 108° - 36° = 36° because of the sum of angles in a triangle. Therefore, ΔBCD is isosceles because of the Converse of Base Angles Theorem.

Answer:

Question 1: 3 - 5 hours.

Question 2: 0 - 1 hour

Step-by-step explanation:

Question 1: As you can see in the diagram, the guy is moving really slowly and is almost stuck, therefore, it is 3 - 5  hours.

Question 2:  In hours 0 - 1, you can see that the graph is the closest to vertical as it gets.