At a middle school, 30% of students buy lunch in the cafeteria and the remaining students bring lunch from home. A spinner with 10 equal-sized sections numbered 0-9 will be used to simulate the lunch trend. How can you design a simulation to guess whether the next 20 students buy lunch or bring lunch from home?

Answer:

18

Step-by-step explanation:

I meant 6x3=18 -2=16 sorry for confusion

To find the value of x, you need to isolate/get the variable "x" by itself in the equation:

3x - 2 = 16       Add 2 on both sides

3x - 2 + 2 = 16 + 2

3x = 18         Divide 3 on both sides to get "x" by itself

[tex]\frac{3x}{3} =\frac{18}{3}[/tex]

x = 6

PROOF

3x - 2 = 16      Substitute/plug in 6 into "x" since x = 6

3(6) - 2 = 16

18 - 2 = 16

16 = 16

Given Information:

Population size = N = 210

Mean height of women population = μ = 64.4 in.

Standard deviation of height of women population = σ = 2.9 in.

Sample size = n = 36

Required Information:

Sample mean = μx = ?

Sample standard deviation = σx = ?

Answer:

Sample mean = μx = 64.4 in.

Sample standard deviation = σx = 0.44 in.

Step-by-step explanation:

We are given a Normal Distribution, which is a continuous probability distribution and is symmetrical around the mean. The shape of this distribution is like a bell curve and most of the data is clustered around the mean. The area under this bell shaped curve represents the probability.

The mean of the sample will be same as population mean that is

μ = μx = 64.4 in

Whereas the sample standard deviation is given by

[tex]\sigma_{x} = \frac{\sigma}{\sqrt{n} } \cdot \sqrt{\frac{N-n}{N-1} }[/tex]

Where σ is the standard deviation of height of women population, N is the population size and n is the sample size.

The term [tex]\sqrt{\frac{N-n}{N-1} }[/tex] is known as finite population correction factor and is used since the sampling is done without replacement.

[tex]\sigma_{x} = \frac{2.9}{\sqrt{36} } \cdot \sqrt{\frac{210-36}{210-1} }\\\sigma_{x} = 0.483 \cdot \sqrt{\frac{174}{209} }\\\sigma_{x} = 0.483 \cdot 0.912\\\sigma_{x} = 0.44[/tex]

Therefore, the standard deviation of sample is 0.44 in.

To find the mean (mu Subscript x overbar) and standard deviation (sigma Subscript x overbar) of the population of sample means, use the formulas provided.

To find the mean (mu Subscript x overbar) and standard deviation (sigma Subscript x overbar) of the population of sample means, we use the formulas:

In this case, the population mean (mu) = 64.4 inches, the population standard deviation (sigma) = 2.9 inches, and the sample size (n) = 36.

Using these values, we can calculate:

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Answer:

$125.25

Step-by-step explanation:

8.75*h+11.25*e

h is hours regular pay and e is extra pay for overtime. Since its an extra $2.50 for overtime, add that to the regular pay. Put in the hours:

[tex]8.75*8+11.25*5[/tex]

Simplify by multiplying 8.75 by 8

[tex]70+11.25*5[/tex]

Simplify by multiplying 11.25 by 5

[tex]70+56.25[/tex]

Add

[tex]70+56.25=126.25[/tex]

Since it says how much did each earn, as in individually, leave it as it is (unless regular pay was $2.18 an hour, which would be a rip-off). I'm pretty sure including the number of employees was meant to throw you off.

Answer: The value of the sine of ∠C is 0.28.

Step-by-step explanation:

Let's draw a triangle again (see the picture below).

Remember SOHCAHTOA? It's:

-sin = opposite/hypotenuse

-cos = adjacent/hypotenuse

-tan = opposite/adjacent.

We need to use sin in here.

The sin of C is [tex]\frac{7}{25}[/tex], which in a calculator, gives [tex]0.28[/tex]

Answer:

24 cubic units

Step-by-step explanation:

4 times 2 times 3 equals 24

Answer:

2kg is the answer

ok well let's see it's 2000 grams

Answer: p-value  = 0.1085 and we will not reject the null hypothesis.

Step-by-step explanation:

Since we have given that

Hypothesis are :

[tex]H_0:p=0.3\\\\H_1:p>0.3[/tex]

Here, n = 200

x = 68

So, [tex]\hat{p}=\dfrac{68}{200}=0.34[/tex]

So, the test statistic value would be :

[tex]z=\dfrac{\hat{p}-p}{\sqrt{\dfrac{p(1-p)}{n}}}\\\\z=\dfrac{0.34-0.3}{\sqrt{\dfrac{0.3\times 0.7}{200}}}\\\\z=\dfrac{0.04}{0.0324}\\\\z=1.234[/tex]

So, the value of statistic value is 1.234.

And the p-value = 0.1085 at 5% level of significance.

So, 0.1085 <  1.234,

So, we will not reject the null hypothesis.

Hence, p-value  = 0.1085 and we will not reject the null hypothesis.

The p-value is a measure of the evidence against the null hypothesis in a hypothesis test. To calculate the p-value, we compare the observed proportion to the target percentage using a one-sample proportion test. The statistical decision when alpha = 0.05 is to reject the null hypothesis if the p-value is less than 0.05.

A. The p-value is a measure of the evidence against the null hypothesis in a hypothesis test. It represents the probability of observing a test statistic as extreme as the one obtained, assuming the null hypothesis is true.

B. To calculate the p-value for the test statistic in this case, we need to perform a hypothesis test. We compare the proportion of viewers who would see the movie again (68/200 = 34%) to the target percentage of 30% using a one-sample proportion test.

C. The statistical decision when alpha = 0.05 is to reject the null hypothesis if the p-value is less than 0.05. This indicates strong evidence against the null hypothesis.

D. The managerial conclusion for this situation would depend on the p-value obtained. If the p-value is less than 0.05, we can conclude that there is strong evidence to suggest that more than 30% of the viewers would see the movie again. Otherwise, we cannot make a definitive conclusion.

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Answer:

Step-by-step explanation:

Let x represent the amount invested at 10%. Then 23000-x is the amount invested at 4%, and the total interest is ...

  0.10x +0.04(23000-x) = 1280

  0.06x = 360 . . . . . subtract 920, simplify

  x = 6000 . . . . . . . . divide by the coefficient of x

Ms Burke invested $6000 at 10% and $17000 at 4%.

Answer:

P-value of the test statistic: P=0.0266.

Step-by-step explanation:

Hypothesis test on a proportion.

The claim is that the percentage of residents who favor annexation is above 72% This is going to be stated in the alternative hypothesis.

Then, the null and alternative hypothesis are:

[tex]H_0: \pi=0.72\\\\H_a:\pi>0.72[/tex]

The sample size is n=900 and the sample proportion is p=0.75.

The standard devaition is calculated as:

[tex]\sigma_p=\sqrt{\dfrac{\pi(1-\pi)}{n}}=\sqrt{\dfrac{0.72*0.28}{900}}=\sqrt{0.000224}=0.015[/tex]

Then, the z-statistic is:

[tex]z=\dfrac{p-\pi-0.5/n}{\sigma_p}=\dfrac{0.75-0.72-0.5/900}{0.015}=\dfrac{0.029}{0.015} = 1.9333[/tex]

For this right tailed test, the P-value is:

[tex]P-value=P(z>0.1933)=0.0266[/tex]

Step-by-step explanation:

csc²u − cos u sec u

csc²u − 1

cot²u

Answer:

[tex]\frac{7}{16}[/tex]

Step-by-step explanation:

[tex]\frac{7}{8}[/tex]×[tex]\frac{1}{2}[/tex]

Multiply straight forward:

So 7 and 1. 8 and 2.

7x1=7

8x2=16

[tex]\frac{7}{16}[/tex]

Answer:

50 degrees

Step-by-step explanation:

Answer:

Length of the longest metal rod which can be carried horizontally around the right-angle turn is 21.21ft

Step-by-step explanation:

Length of the longest metal rod which can be carried horizontally around the right-angle turn, can be determined at the point at where the distance between one end of rod and the 7feet corridor is same as the distance between other end of the rod and the 8feet corridor. How do I mean.

The length of longest metal rod will be determined when distance between the one end of rod and the 7feet corridor and the distance between other end of the rod and the 8feet corridor are both 15ft (7ft + 8ft).

See attachment for illustration.

Therefore using Pythagoras theorem we can find the length of longest metal rod that can be carried horizontally around the right angle turn

See illustration in attachment

i.e

Hyp² = opp²+adj²

Hyp = √(15²+15²)

Hyp = √450

Hyp = 21.21ft

Answer:

Step-by-step explanation:

Given the sample data

Pre-test... 12 14 11 12 13

Post-Test 15 17 11 13 12

The mean of pre-test

x = ΣX / n

x = (12+14+11+12+13) / 5

x = 12.4

The standard deviation of pre-test

S.D = √Σ(X-x)² / n

S.D = √[(12-12.4)²+(14-12.4)²+(11-12.4)²+(12-12.4)²+(13-12.4)² / 5]

S.D = √(5.2 / 5)

S.D = 1.02.

The mean of post-test

x' = ΣX / n

x' = (15+17+11+13+12) / 5

x' = 13.6

The standard deviation of post-test

S.D' = √Σ(X-x)² / n

S.D' = √[(15-13.6)²+(17-13.6)²+(11-13.6)²+(13-13.6)²+(12-13.6)² / 5]

S.D = √(23.2 / 5)

S.D = 2.15

Test value

t = (sample difference − hypothesized difference) / standard error of the difference

t = [(x-x') - (μ- μ')] / (S.D / n — S.D'/n)

t = (12.4-13.6) - (μ-μ')/ (1.02/5 - 2.15/5)

-1.5 = -1.2 - (μ-μ') / -0.226

-1.5 × -0.226 = -1.2 -(μ-μ')

​0.339 = -1.2 - (μ-μ')

(μ-μ') = -1.2 -0.339

μ-μ' = -1.539

Then, μ ≠ μ'

We can calculate our P-value using table.

This is a two-sided test, so the P-value is the combined area in both scores.

The p-value is 0.172

The p value > 0.1

Using the t-distribution, it is found that the p-value is of 0.086.

At the null hypothesis, it is tested if there is no improvement, that is, the subtraction of the mean of the grades on test 1 by the mean of the grades on test 2 is of at least 0, that is:

[tex]H_0: \mu_1 - \mu_2 \geq 0[/tex]

At the alternative hypothesis, it is tested if there is improvement, that is, the grades on the second test were greater than on the first, hence:

[tex]H_1: \mu_1 - \mu_2 < 0[/tex]

We can find the standard deviation for the samples, hence, the t-distribution is used.

The p-value is found using a left-tailed test, as we are testing if the variable is less than a value, with 5 + 5 - 2 = 8 df and t = -1.5, using a calculator, the p-value is of 0.086.

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The y-intercept is the y value when x = 0 --> (0, y) or the point where the line crosses through the y-axis.

The slope(m) is:

[tex]m=\frac{rise}{run}[/tex]

"rise" is the number of units you go up(+) or down(-) from each point

"run" is the number of units you go to the right from each point

You know:

y-intercept = -6 or (0, -6)

m = 2 or [tex]\frac{2}{1}[/tex] , so from each point you go up 2 units and to the right 1 unit to get to the next point.

#1: Locate the ordered pair (0, –6). From that point on the graph, move up 2, right 1 to locate the next ordered pair on the line. Draw a line through the two points.

This is true

#2: Locate the ordered pair (0, –6). From that point on the graph, move up 2, left 1 to locate the next ordered pair on the line. Draw a line through the two points.

This isn't accurate because the slope doesn't move left.

#3: Locate the ordered pair (–6, 0). From that point on the graph, move up 2, right 1 to locate the next ordered pair on the line. Draw a line through the two points.

This isn't accurate because (-6, 0) isn't the y-intercept or a point on the y-axis

#4: Locate the ordered pair (–6, 0). From that point on the graph, move up 2, left 1 to locate the next ordered pair on the line. Draw a line through the two points.

This isn't accurate because (-6, 0) isn't the y-intercept or a point on the y-axis, and the slope doesn't move to the left.

Answer:

tis be A my friends

Step-by-step explanation:

Answer:

the graph that matches the function is

Step-by-step explanation:

Answer:quasi experiment

Step-by-step explanation:

Because it is basically for comparison usually by measuring outcomes if a program by participants and non participants

Answer:

3(7x2 + 5) (x − 3)  took the quiz :)

Step-by-step explanation:

Answer:

C- 3(7x2 + 5) (x − 3)

Step-by-step explanation:

Complete Question:

Spam: A researcher reported that 71.8% of all email sent in a recent month was spam. A system manager at a large corporation believes that the percentage at his company may be 69%. He examines a random sample of 500 emails received at an email server, and finds that 365 of the messages are spam. Can you conclude that greater than 69% of emails are spam? Use both a=0.01 and a=0.05 levels of significance and the -value method with the table.

(a) State the appropriate null and alternate hypotheses.

(b) Compute the -value.

(c) At the a=0.01, can you conclude that greater than 69% of emails are spam?

(d) At the a=0.05, can you conclude that greater than 69% of emails are spam?

Answer and step-by-step explanation:

a)

The required hypothesis are                                          

H₀: [tex]\mu[/tex] = 0.69

H₁: [tex]\mu[/tex] > 0.69

additional solutions are attached in the image below

Answer:

a) No

B) Yes

Step-by-step explanation:

Calculating the p-value, we have;

z = (p-bar) -p/√(p(1-p)/n)

But p-bar = 365/500

               = 0.73

Therefore,

z = 0.73 -0.69/√0.69(1-0.69)/500

   = 0.04/√0.2139/500

   = 0.04/√0.0004278

   = 0.04/0.02068

   = 1.93

p-value = p(z ≥1.93) = 0.0268

(a) Can you conclude that the percentage of emails that are spam is greater than 69% ?

No, because the p- value is greater than α. That is p-value⊃ 0.01

(b) Can you conclude that the percentage of emails that are spam is greater than 69%

Yes, since p-value ∠ 0.05

Answer:

x = 29 cm

Step-by-step explanation:

An isosceles triangle has two equal sides that have equal angles at the bases

Since the 80 degree angles are at the bases of the 29 cm side, the length of x must be 29 cm

Answer:

29 cm

Step-by-step explanation:

Because both angles are 80 degrees. If your angles are the same so are your sides.

Answer:

a) Recommended order quantity = 79.84 = 80

b) recorder point = 35

Safety stock = 10

Safety stock cost = $0/year

ci) P(stock out/cycle) = 0.1587

cii) Number of stock outs/ year = 2

Step-by-step explanation:

The pictures attached show a clear explanation of all the processes.

Final answer:

To determine the recommended order quantity, use the EOQ formula. The reorder point and safety stock can be calculated using the normal distribution. If the reorder point is set at 30, the probability of a stockout and the expected number of stockouts can be calculated.

Explanation:

To determine the recommended order quantity, we need to consider the economic order quantity (EOQ) model. The EOQ formula is given by:

EOQ = √((2 * D * Co) / Ch)

Where:

D is the annual demand (1000 units)

Co is the ordering cost per order ($25.50)

Ch is the holding cost per unit ($8)

Plugging in the values, we get:

EOQ = √((2 * 1000 * 25.50) / 8) = 79.79 (rounded to 2 decimal places)

Therefore, the recommended order quantity is 80 units.

To calculate the reorder point, we need to consider the lead-time demand. Since the lead-time demand follows a normal probability distribution with μ=25 and σ=5, we can use the normal distribution to find the reorder point.

To calculate the recorder point and safety stock for a desired probability of stockout of 2%, we need to find the Z value corresponding to the desired probability. We can use the Z-score table or a calculator to find the Z value. For a 2% probability of stockout, the Z value is approximately -2.05 (rounded to 2 decimal places).

The reorder point is given by:

Reorder Point = μ + (Z * σ)

Where:

μ is the mean of the lead-time demand (25 units)

Z is the Z value (-2.05)

σ is the standard deviation of the lead-time demand (5 units)

Plugging in the values, we get:

Reorder Point = 25 + (-2.05 * 5) = 14.75 (rounded to 2 decimal places)

The safety stock is given by:

Safety Stock = Z * σ

Plugging in the values, we get:

Safety Stock = -2.05 * 5 = -10.25 (rounded to 2 decimal places)

Therefore, the recorder point is approximately 15 units (rounded to the nearest whole unit) and the safety stock is approximately 10 units (rounded to the nearest whole unit).

If the manager sets the reorder point at 30 units, we can use the normal distribution to find the probability of a stockout. The probability of a stockout can be calculated using the Z-score formula.

The Z value is calculated using the formula:

Z = (Reorder Point - μ) / σ

Where:

Reorder Point is the given reorder point (30 units)

μ is the mean of the lead-time demand (25 units)

σ is the standard deviation of the lead-time demand (5 units)

Plugging in the values, we get:

Z = (30 - 25) / 5 = 1 (rounded to 1 decimal place)

Looking up the Z value in the Z-score table, we find that the corresponding probability is approximately 0.8413. Therefore, the probability of a stockout on any given order cycle is approximately 0.1587 or 15.87% (rounded to 2 decimal places).

The number of times you would expect a stockout during the year can be estimated using the formula:

Number of Stockouts = (Annual Demand / EOQ) * (1 - Probability of Stockout)

Plugging in the values, we get:

Number of Stockouts = (1000 / 80) * (1 - 0.1587) = 10.69 (rounded to 2 decimal places)

Therefore, you would expect a stockout approximately 11 times during the year (rounded to the nearest whole number).

Answer:

Scalene Triangles

Step-by-step explanation:

Isosceles triangles have one pair of equivalent edges, and equilateral triangles of course have all equivalent edges, it would obviously be the scalene triangles.

I am joyous to assist you at any time.

Step-by-step explanation:

2 ( 8 - 3c) = 4c - 34

16 - 6c = 4c - 34

6c + 4c = 16 + 34

10c = 50

c = 50/ 10

therefore c = 5

Hope it will help :)

Answer:

11.55 ft

Step-by-step explanation:

tan 60° = 20 / x

x = 20 / tan 60° = 20 / √3 = 11.55

The length of the shadow will be 11.55 ft. It is found by the help of the tangent formula of trigonometry.

Angle of depression is the angle formed by the horizontal and the place where your head came to a halt.

You keep your gaze level with the ground. When you need to keep an eye on anything, though, you raise your head and lower your gaze.

The given data in the problem is;

Angle of depression = 60°

Length of the shadow(x) = ?

Height of  flagpole = 20 ft

From trigonometry formula of tanΘ is;

[tex]\rm tan \theta = \frac{P}{B} \\\\ tan60^0=\frac{20}{x} \\\\ x= \frac{20}{tan60^0} \\\\ x= \frac{20}{\sqrt{3} } \\\\ x=11.55 \ ft[/tex]

Hence, the length of the shadow will be 11.55 ft.

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Answer:

Step-by-step explanation:

Answer:

40

Step-by-step explanation:

got 100 on quiz

The maximum height of the firework is 40 feet.

'The maximum height of the object is the highest vertical position along its trajectory.'

According to the given problem,

y = -16x²+32x+24

For maximum height,

The slope of y with respect to x has to be 0.

⇒ [tex]\frac{dy}{dx}[/tex] = 0

Now, [tex]\frac{d(-16x^{2} +32x+24)}{dx}[/tex] = -32x +32

⇒-32x + 32 = 0

⇒ 32 = 32x

⇒1 = x or x = 1 second

Therefore after 1 second, the firework reaches maximum height.

Now, we put the value of x = 1 second in the equation: y = -16x²+32x+24

⇒ -16(1)² + 32(1) +24

= -16+32+24

= 40 feet

Hence, we can conclude that after 1 second, the firework reaches a maximum height of 40 feet.

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Answer:

190 Jeans

Step-by-step explanation:

The daily production cost of the jeans manufacturer is given as:

[tex]C=0.3x^2-114x+11405[/tex]

To determine the number of Jeans,x that should be produced daily to minimize cost, C. We take the derivative of C and solve for its critical points.

[tex]C'=0.6x-114\\$When C'=0\\0.6x-114=0\\0.6x=114\\x=190[/tex]

Therefore, to minimize daily cost, 190 Jeans is the number of jeans that should be produced daily.

Answer:   This is the Communitve Property because the equation has the same thing on both sides, but the numbers "flip flopped" or switched around.

Answer:

B. One solution

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Equality Properties

Algebra I

Step-by-step explanation:

Step 1: Define

-6y + 13 + 9y = 8y - 3

Step 2: Solve for y

∴ We can see by solving the equation, we get 1 solution only.

The linear equation has one solution after simplifying and isolating the variable y, which results in y = 16/5.

When solving the given linear equation −6y + 13 + 9y = 8y − 3, the first step is to simplify and collect like terms. Begin by combining the terms with y on the left side: 3y + 13 = 8y − 3. Next, move the terms with y to one side by subtracting 3y from both sides: 13 = 5y − 3. Lastly, isolate y by adding 3 to both sides and then dividing by 5: y =  rac{16}{5}. Since we have successfully isolated y and found a specific value for it, the equation has one solution.

Answer:

Let the 1st group be of students and 2nd group be of faculty.

n1 = 18

\bar{x}_{1} = 8.4

s1 = 7.3

n2 = 15

\bar{x}_{2} = 4.2

s2 = 3.8

1.

H0: =

HA: >

2.

Equality of variance test:

F = s12/s22

= 53.29/14.44

= 3.69

df-num = n1-1 = 17

df-den = n2-1 = 14

p-value = 0..008 < 0.05 i.e.H0 can be rejected and hence we can say that both populations do not have the same variances.

So, test-statistic will be calculated as follows:

[ Find the solution of this part in the attachment]

3.

df = 26

p-value = 0.0217

4.

E. It confirms the community college results because the p-value is much smaller.