A shop has the following offers. Fig rolls (125g packet) Normal price £1.08. Buy one, get 2nd half price.

Answer:

[tex]Lower = 231.134- 3.098=228.036[/tex]

[tex]Upper = 231.134+ 3.098=234.232[/tex]

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

[tex]\bar X[/tex] represent the sample mean for the sample  

[tex]\mu[/tex] population mean (variable of interest)

s represent the sample standard deviation

n=5 represent the sample size  

Solution to the problem

The confidence interval for the mean is given by the following formula:

[tex]\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}[/tex]   (1)

And for this case we know that the 95% confidence interval is given by:

[tex] \bar X=\frac{233.002 +229.266}{2}= 231.134[/tex]

And the margin of error is given by:

[tex] ME = \frac{233.002 -229.266}{2}= 1.868[/tex]

And the margin of error is given by:

[tex] ME= t_{\alpha/2} \frac{s}{\sqrt{n}}[/tex]

The degrees of freedom are given by:

[tex] df = n-1 = 5-1=4[/tex]

And the critical value for 95% of confidence is [tex] t_{\alpha/2}= 2.776[/tex]

So then we can find the deviation like this:

[tex] s = \frac{ME \sqrt{n}}{t_{\alpha/2}}[/tex]

[tex] s = \frac{1.868* \sqrt{5}}{2.776}= 1.506[/tex]

And for the 99% confidence the critical value is: [tex] t_{\alpha/2}= 4.604[/tex]

And the margin of error would be:

[tex] ME = 4.604 *\frac{1.506}{\sqrt{5}}= 3.098[/tex]

And the interval is given by:

[tex]Lower = 231.134- 3.098=228.036[/tex]

[tex]Upper = 231.134+ 3.098=234.232[/tex]

The limits of the 99% confidence interval (CI) for the true average natural frequency (Hz) of delaminated beams are (228.555, 233.713).

 To find the 99% confidence interval, we need to understand that the width of a confidence interval is determined by the standard deviation of the sample, the sample size, and the confidence level.

The formula for a confidence interval for the mean when the population standard deviation is unknown (which is likely the case here, as it is not specified) is given by:

[tex]\[ \text{CI} = \bar{x} \pm t_{\frac{\alpha}{2}, n-1} \times \frac{s}{\sqrt{n}} \][/tex]

First, we need to find the sample mean which is the center of the given 95% CI.

The width of the CI is the difference between the upper and lower limits. For the 95% CI, the width is:

[tex]\[ \text{Width}_{95\%} = 233.002 - 229.266 = 3.736 \][/tex]

Since the 95% CI is symmetric around the sample mean, the margin of error (MOE) for the 95% CI is half of the width:

[tex]\[ \text{MOE}_{95\%} = \frac{\text{Width}_{95\%}}{2} = \frac{3.736}{2} = 1.868 \][/tex]

The 95% CI can be represented as:

[tex]\[ \text{CI}_{95\%} = \bar{x} \pm \text{MOE}_{95\%} \][/tex]

 Finally, the limits of the 99% CI are found by adding and subtracting this MOE from the sample mean:

[tex]\[ \text{Upper limit}_{99\%} = \bar{x} + \text{MOE}_{99\%} = 231.134 + 3.152 \approx 233.713 \][/tex]

Therefore, the 99% confidence interval for the true average natural frequency (Hz) of delaminated beams is (228.555, 233.713).

Answer:

1:B the ballet club

2.B the ballet club.

Step-by-step explanation:

It is correct on khan :)

The chess club accepts 5 new members each week, and had 37 members after 6 weeks. The club started with 7 members. Without more data, a similar function for the ballet club cannot be determined.

This question requires a bait to understand the concept of linear functions. The chess club at Nahk University accepts 5 new members each week and had 37 members after 6 weeks. We can express this as a linear function, with the number of weeks as the independent variable and the number of members as the dependent variable:

Members = 5 * Weeks + Initial Members

Here, we know from the information given that after 6 weeks, the chess club had 37 members. So if we plug those numbers into the equation:

37 = 5 * 6 + Initial Members

Then solving for the number of initial members gives us 7. Hence, the chess club started with 7 members.

To find a similar function for the ballet club, we would need more specific data to map time (in weeks) to the number of members in the same way.

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Answer:

the total workdone required to to lift the chain from its top end so that its bottom is 2 meters above the ground = 661 J

Step-by-step explanation:

Given that:

The mass density of the chain is [tex]\rho (x) = 2x( 4-x) \ kg/m[/tex]

It is pertinent and crucial to consider the determination of the work-done in lifting the chain from its front in that its bottom is 2 meters away from the ground.

Consider a cross section portion of the chain of length  Δx that has to be lifted to a height [tex]x_k[/tex]

The required wok to be done for this work is [tex]W_k = \rho(x_k)g \delta x(x_k)[/tex]

combining the segments of the chain and taking the [tex]\delta x[/tex]→0

Integrally, the work-done can be illustrated as :

[tex]\int\limits^3_0 \rho {(x_k)} \, gxdx \ \ = \ \ \int\limits^3_0 (9.8) 2x^2 (4-x)dx \\\\= 19.6 \int\limits^3_0 (4x^2-x^3)dx\\\\\\= 19.6 [ \frac{4}{3}x^2- \frac {x^4}{4}]^3__0}}dx\\\\\\= 19.6 [ \frac{4}{3}(3)^2- \frac {3^4}{4}]}dx\\\\= 19.6 (36- \frac{81}{4})\\\\\\= 308.7 \ \ J[/tex]

Furthermore, there is need to lift the chain  up to 2 meters . So, calculating the weight of the chain ; we have:

Weight = [tex]\int\limits^3_0 \rho {x} \, gdx[/tex]

[tex]= \int\limits^3_0 (9.8) {2x}(4-x) \, dx\\ \\\\= 19.6\int\limits^3_0 {(4x-x^2)} \, dx \\\\= 19.6 [ 2x^2 - \frac{x^3}{3}]^3_0\\\\= 19.6 [2(3)^2 - \frac{3^3}{3}]\\\\=19.6 [18-9]\\\\= 176.4 \ \ J[/tex]

Finally .the work-done is said to be equal to the potential energy

∴ W = mgh

W = (176.4)×2

W = 352.8 J

Total workdone = (308.7 + 352.8 ) J

Total workdone = 661 J

Thus, the total workdone required to to lift the chain from its top end so that its bottom is 2 meters above the ground = 661 J

Answer:

Step-by-step explanation:

Answer:

Step-by-step explanation:

Given that,

Length of chain

L = 3m

Linear mass density is

ρ(x) = 2x(4 — x) kg/m lies on the ground

When x = 0, is top of the chain

Work done to lift the chain from top end so that the bottom is 2m above the ground.

Considered the segment of the chain of length ∆x that will be lifted in the positive y direction (+j)m from the foot.

The work needed to lift this segment is given as

Work = mass density × ∆x × gravity

W = ρ × ∆x × g

g is acting downward = 9.8j

Summing over all segment of the chain and passing to the limit as ∆x→0.

Therefore, the total work done needed to full extend the chain is

W = ∫ ρ × ∆x × g x = 0 to 3

Since g is constant

∆x = xdx

Then,

W = g ∫ 2x(4—x)x dx. x= 0 to x = 3

W = 9.81 ∫ (8x² — 2x³)x dx

W = 9.81 ( 8x³/3 — 2x⁴/4)

W=9.81(8x³/3— ½x⁴) from x=0 to x=3

W = 9.81[8(3)³/3 — ½(3⁴)] — 0

W = 9.81 × (72 —40.5)

W = 9.81 × 31.5

Work done= 309.015 J

Lifting the entire chain requires to light the weight

Weight = ∫ρgdx

Weight = g ∫ρ dx. From x=0 to x=3

Weight = g ∫ 2x(4-x) dx

Weight = 9.81 ∫(8x-2x²)dx

Weight = 9.81 [ 8x²/2 - 2x³/3]

Weight = 9.81[4x²-⅔x³] x=0 to x=3

Weight = 9.81[4(3²) — ⅔(3³)]

Weight = 9.81(36—18)

Weight = 9.81 × 18

Weight = 176.58N

Now, this weight is lifted to a height of 2m, then using potential energy formula, we have

P.E = Work = mgh = Weight ×height

Work = W×h = 176.58 × 2

Work = 353.16 J

Then, total workdone is

W = 353.16 + 309.015

W = 662.18 J

The required Work done required to lift the chain from top so that it's bottom is 2m from the ground is 662.18J

Final answer:

The margin of error for a 95% confidence interval, given a standard error of 1.91, is approximately 3.74 when rounded to two decimal places.

Explanation:

To find the margin of error for a 95% confidence interval when the standard error is given, we will use the concept of the z-score that corresponds to our desired confidence level. For a 95% confidence interval, the z-score is typically 1.96, which is the critical value for a normal distribution that leaves 2.5% in each tail. The margin of error (MoE) is calculated by multiplying the standard error (SE) by the z-score.

Margin of Error = z * SE

Marginal Error for our scenario = 1.96 * 1.91

Therefore, the Margin of Error = 3.7436. Rounding this to two decimal places gives us a Margin of Error of approximately 3.74.

Answer:

One sample z-test for population mean would be the best approach to conduct a hypothesis test.

Step-by-step explanation:

Following is the data available to us:

Mean amount of water = u = 3

Sample mean = x = 2.5

Sample Size = n = 200

Sample Standard Deviation = s = 1

Population Standard Deviation = [tex]\sigma[/tex] = 1.2

Population is normally distributed.

We need to find the best approach to conduct a hypothesis test. Since only one sample is involved, it is a One-Sample test about population mean. for conducting hypothesis test for One-Sample about population mean we have following two options:

Selecting the best approach:

The first thing to check is if our data from a population which is normally distributed. Which in this case is. Next we check if the value of population standard deviation is known or unknown. the rule is:

Since, in this case we know the value of Population Standard Deviation which is 1.2, One sample z-test for population mean would be the best approach to conduct a hypothesis test.

The appropriate approach to conduct a hypothesis test for the claim that the AI, or Adequate Intake of water, for pregnant women is a mean of 3L/d, liters per day, is to use a one-sample t-test.

The one-sample t-test is a statistical test that is used to compare the sample mean to a known population mean. It is appropriate to use this test when the sample size is small (n < 30) and the population standard deviation is unknown.

In this case, the sample size is 200, which is greater than 30, but the population standard deviation is known (1.2). However, it is still appropriate to use the one-sample t-test because the sample size is not large enough to reliably estimate the population standard deviation.

To conduct the one-sample t-test, we will follow these steps:

State the null and alternative hypotheses.

Null hypothesis (H0): The mean water intake for pregnant women is 3L/d.

Alternative hypothesis (Ha): The mean water intake for pregnant women is not 3L/d.

Calculate the test statistic.

The test statistic for the one-sample t-test is calculated as follows:

t = (x - μ) / (s / √n)

where:

x is the sample mean

μ is the known population mean

s is the sample standard deviation

n is the sample size

In this case, the test statistic is calculated as follows:

t = (2.5 - 3) / (1 / √200) = -4.17

Find the p-value.

The p-value is the probability of obtaining a test statistic as extreme or more extreme than the one observed, assuming that the null hypothesis is true.

We can use a t-table to find the p-value. The p-value for a t-statistic of -4.17 with 199 degrees of freedom is less than 0.001.

Make a decision.

Since the p-value is less than the significance level (α = 0.05), we reject the null hypothesis and conclude that there is sufficient evidence to support the claim that the mean water intake for pregnant women is not 3L/d.

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The equivalent expression to 2 (three-fourths x + 7) minus 3 (one-half x minus 5) is 29.

To find equivalent expressions to 2 (three-fourths x + 7) minus 3 (one-half x minus 5), we first need to distribute the numbers outside the parentheses inside the parentheses. This simplifies the expression as follows:

Subtracting the second expression from the first, we obtain: (1.5x + 14) - (1.5x - 15), which simplifies to 0x + 29, or simply 29.

Therefore, the only expression equivalent to the given expression is 29.

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Answer:

The answer is B, C, and E

Step-by-step explanation:

Answer:

a) The distance from the television camera to the rocket is changing at that moment at a speed of

600 ft/s

b) the camera's angle of elevation is changing at that same moment at a rate of

0.16 rad/s = 9.16°/s

Step-by-step explanation:

This is a trigonometry relation type of problem.

An image of when the rocket is 3000 ft from the ground is presented in the attached image.

Let the angle of elevation be θ

The height of the rocket at any time = h

The distance from the camera to the rocket = d

a) At any time, d, h and the initial distance from the camera to the rocket can be related using the Pythagoras theorem.

d² = h² + 4000²

Take the time derivative of both sides

(d/dt) (d²) = (d/dt) [h² + 4000²]

2d (dd/dt) = 2h (dh/dt) + 0

At a particular instant,

h = 3000 ft,

(dh/dt) = 1000 ft/s

d can be obtained using the same Pythagoras theorem

d² = h² + 4000² (but h = 3000 ft)

d² = 3000² + 4000²

d = 5000 ft

2d (dd/dt) = 2h (dh/dt) + 0

(dd/dt) = (h/d) × (dh/dt)

(dd/dt) = (3000/5000) × (1000)

(dd/dt) = 600 ft/s

b) If the television camera is always kept aimed at the rocket, how fast is the camera's angle of elevation changing at that same moment?

At any moment in time, θ, h and the initial distance of the camera from the base of the rocket are related through the trigonometric relation

Tan θ = (h/4000) = 0.00025h

Taking the time derivative of both sides

(d/dt) (Tan θ) = (d/dt) (0.00025h)

(Sec² θ) (dθ/dt) = 0.00025 (dh/dt)

At the point where h = 3000 ft, we can calculate the corresponding θ at that point

Tan θ = (3000/4000)

θ = tan⁻¹ (0.75) = 0.6435 rad

(Sec² θ) (dθ/dt) = 0.00025 (dh/dt)

(Sec² 0.6435) (dθ/dt) = 0.00025 (1000)

1.5625 (dθ/dt) = 0.25

(dθ/dt) = (0.25/1.5625) = 0.16 rad/s

Hope this Helps!!!

The rate of change of the hypotenuse distance and the camera's angle of elevation can be calculated using the principles of trigonometry and differentials. The rates derive from the Pythagorean theorem and the derivatives of trigonometric functions, respectively.

This question relates to the concepts of trigonometry and differential calculus. We can see the camera, the rocket and the launch pad as forming a right triangle: the distance from the camera to the rocket is the hypotenuse, the distance from the camera to the launch pad is one leg (adjacent to the angle of elevation) and the distance that the rocket has risen is the other leg (opposite to the angle of elevation).

(a) To find how fast the distance from the camera to the rocket is changing, we can use the Pythagorean theorem (a² + b² = c²). Here, a = 4000 ft, b = 3000 ft, so, c = sqrt((4000)² + (3000)²). The derivative dc/dt (rate of change of c) when b = 3000 ft and db/dt = 1000 ft/s will provide the answer.

(b) To find how fast the camera's angle of elevation (let's symbolize it by θ) is changing, we use the concept of derivatives of trigonometric functions, specifically the tangent, which is defined as opposite (b) over adjacent (a), or tan(θ) = b/a. Then, we can compute dθ/dt using implicit differentiation when b = 3000 ft and db/dt = 1000 ft/s.

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Answer:

A 35 pages per hour

Step-by-step explanation:

The rate of change is the slope

(0,150)  (7,400)

m = (y2-y1)/(x2-x1)

    =(400-150)/(7-0)

    =250/7

    =35.7 pages per hour

Approximately 36 pages per hour

Answer:

A

Step-by-step explanation:

Rate = slope

(500-150)/(10-0)

= 350/10

= 35

Answer:

c) 0.932

99% confidence interval for average weights of all packages sold in small meat trays.

(0.932 ,1.071)

Step-by-step explanation:

Explanation:-

Given random sample of 35 packages in small meat trays produced weight with an average of 1.01 lbs. and standard deviation  of 0.18 lbs.

size of the sample 'n' = 35

mean of the sample x⁻= 1.01lbs

standard deviation of the sample 'S' = 0.18lbs

The 99% confidence intervals are given by

[tex](x^{-} - t_{\alpha } \frac{S}{\sqrt{n} } , x^{-} +t_{\alpha } \frac{S}{\sqrt{n} } )[/tex]

The degrees of freedom γ=n-1 =35-1=34

tₐ =  2.0322

99% confidence interval for average weights of all packages sold in small meat trays

[tex](1.01 - 2.0322 \frac{0.18}{\sqrt{35} } , 1.01+2.0322 \frac{0.18}{\sqrt{35} } )[/tex]

( 1.01 - 0.06183 , 1.01+0.06183)

(0.932 ,1.071)

Final answer:-

99% confidence interval for average weights of all packages sold in small meat trays.

(0.932 ,1.071)

The maximum number of students that attend the trip is 160.

To determine the number of students that go on the trip, to consider the cost of theatre tickets, train tickets, and the allotted budget.

Let's denote the number of students as S. We know that the cost of each theatre ticket is £36 and that there are 2 teachers attending. Therefore, the total cost of theatre tickets for the teachers is 2 × £36 = £72.

The remaining budget after accounting for the cost of theatre tickets for the teachers is £2000 - £72 = £1928.

Each student will require a train ticket costing £12. So, the cost of train tickets for S students will be 12S.

to find the maximum number of students (S) that can be accommodated within the remaining budget.

From the remaining budget, the cost of train tickets for the students should be less than or equal to the available funds:

12S ≤ £1928

To solve for S, we divide both sides of the inequality by 12:

S ≤ £1928 / 12

S ≤ 160.67

Since the number of students must be a whole number, the maximum number of students that can go on the trip is 160.

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Answer:

A line is at a degree of 180, so the line 170 is on can give you the degree of the smallest angle: 10. A traingel has a total degree of 180, so 125 + 10 = 135. Take 180 and subtract 135 to get x = 45º

Step-by-step explanation:

Answer:

x = 45°

Step-by-step explanation:

Since it's a triangle, the total degrees of it would be 180

You are given one of the angles measurements and another's' supplementary angle

180 - 125 = 55

Take the 170 and subtract from 180 to get the interior angle of the triangle

180 - 170 = 10

Now subtract 55 and 10 to get x

55 - 10 = 45

Answer:

196.6 feet

Step-by-step explanation:

Draw the right triangle formed by the zip-line.  Use SOH-CAH-TOA to write and solve an equation:

tan 11.5° = 40 / x

x = 40 / tan 11.5°

x = 196.6

The horizontal distance between the towers is 196.60 ft.

Distance is the measurement of an object from a specific point in the horizontal direction, and height is the measurement of an object in the vertical direction.

Here, ∅₁ is called the angle of elevation and ∅₂ is called the angle of depression. For one specific type of problem in height and distances, we have a generalized formula. Height = Distance moved / [cot(original angle) – cot(final angle)] => h = d / (cot ∅₁ – cot ∅₂)

Given:

The vertical height of the zip line = 40 ft

The angle of elevation = 11.5°

We know, according to trigonometric values tan n = height / base

∴ tan 11.5 = 40 / base

base = 40/ tan 11.5

base = 40 / 0.203

base = 196. 60 ft

Therefore, the horizontal distance between the towers is 196.60 ft.

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Answer:

Mailing preparation takes 38.29 min max time to prepare the mails.

Step-by-step explanation:

Given:

Mean:35 min

standard deviation:2 min

and 95%  confidence interval.

To Find:

In normal distribution mailing preparation time  taken less than.

i.eP(t<x)=?

Solution:

Here t -time and x -required time

mean time 35 min

5 % will not have true mean value . with 95 % confidence.

Question is asked as ,preparation takes less than  time means what is max time that preparation will take to prepare mails.

No mail take more time than that time .

by Z-score or by confidence interval is

Z=(X-mean)/standard deviation.

Z=1.96 at  95 % confidence interval.

1.96=(X-35)/2

3.92=(x-35)

X=38.29 min

or

Confidence interval =35±Z*standard deviation

=35±1.96*2

=35±3.92

=38.29 or 31.71 min

But we require the max time i.e 38.29 min

And by observation we can also conclude the max time from options as 38.29 min.

In a normal distribution, the 95th percentile is calculated by adding twice the standard deviation to the mean. Based on the given mean (35 minutes) and standard deviation (2 minutes), 95% of the time the envelope preparation should take less than 39 minutes.

The problem asks what time, in percentage terms, the preparation for mailing the weekly report to the executives is less than.

Given that the time taken to prepare the envelopes follows a normal distribution with a mean of 35 minutes and a standard deviation of 2 minutes, we need to determine the time that occurs at the 95th percentile of this distribution.

The 95th percentile is found by adding 2 standard deviations to the mean in a normal distribution (this is due to the empirical rule, which states that about 95% of values lie within 2 standard deviations of the mean in a normal distribution).

Here's how you calculate it:

Determine the mean: The mean is given as 35 minutes.

Determine the standard deviation: The standard deviation is given as 2 minutes.

Calculate the 95th percentile: Mean + 2(Standard Deviation) = 35 + 2(2) = 39 minutes.

So, on 95% of occasions the mailing preparation takes less than 39 minutes. None of the options (a, b, c, and d) provided are correct based on the calculations.

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Given:

The width of the snack box is [tex]1 \frac{1}{2} \ inches=\frac{3}{2} \ inches[/tex]

The length of the snack box is [tex]2 \frac{1}{2} \ inches=\frac{5}{2} \ inches[/tex]

The height of the snack box is 4 inches.

We need to determine the volume of the hamsta snack box.

Volume of the box:

The volume of the box can be determined using the formula,

[tex]V=L \times W \times H[/tex]

where L is the length, W is the width and H is the height of the box.

Substituting the values, we get;

[tex]V=4 \times \frac{3}{2} \times \frac{5}{2}[/tex]

Simplifying, we get,

[tex]V=\frac{60}{4}[/tex]

[tex]V=15 \ in^3[/tex]

Thus, the volume of the Hamsta snack box is 15 cubic inches.

Answer:

15

Step-by-step explanation:

i did the test and got 100

Answer: -4x⁴ + x² + 6

Step-by-step explanation:

Remove parentheses:   (-a) = -a

-4x⁴ + 6x²+3 - (5x²-3)

-4x⁴ + 6x² + 3 - 5x² + 3

Simplify -4x⁴+ 6x² + 3 - 5x² + 3:   -4x⁴ + x² + 6

-4x⁴ + x² + 6

Given:

Length of the wall = 8 ft

Width of the wall = 6 ft

Length of the window = 2 ft

Width of the window = 18 in

To find:

The square feet Jamin planning to paint

Solution:

Area of the wall = length × width

                           = 8 × 6

Area of the wall = 48 ft²

Width of the window = 18 in

Convert inch to feet.

1 feet = 12 inch

18 inch = [tex]\frac{18}{12}=1.5[/tex] ft

Area of the window = length × width

                                 = 2 × 1.5

Area of the window = 3 ft²

Area to be painted = Area of the wall - Area of the window

                                = 48 ft² - 3 ft²

                                = 45 ft²

Therefore Jamin planning to paint 45 ft².

Answer:

45 ft²

Step-by-step explanation:

mark me brainy plz!

Answer:

1000000000

Step-by-step explanation:

20000 times 50000 equals 1000000000

when you multiply 20,000 by 50,000, you get the product of 1,000,000,000 or 1 billion.

The The product of 20,000 multiplied by 50,000 is:

20,000 x 50,000 = 1,000,000,000

Therefore, 20,000 multiplied by 50,000 equals 1 billion. of 20,000 multiplied by 50,000 is:

20,000 x 50,000 = 1,000,000,000

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Answer:

a) [tex]t=\frac{(56.2-64.2)-0}{\sqrt{\frac{18.2^2}{22}+\frac{13.9^2}{11}}}}=-1.40[/tex]  

b) [tex]p_v =2*P(t_{31}<-1.4)=0.171[/tex]  

Step-by-step explanation:

Information given

[tex]\bar X_{1}=56.2[/tex] represent the mean for sample 1  

[tex]\bar X_{2}=64.2[/tex] represent the mean for sample 2  

[tex]s_{1}=18.2[/tex] represent the sample standard deviation for 1  

[tex]s_{2}=13.9[/tex] represent the sample standard deviation for 2  

[tex]n_{1}=22[/tex] sample size for the group 2  

[tex]n_{2}=11[/tex] sample size for the group 2  

t would represent the statistic (variable of interest)  

System of hypothesis

We need to conduct a hypothesis in order to check if the true means are different, the system of hypothesis would be:  

Null hypothesis:[tex]\mu_{1}-\mu_{2}=0[/tex]  

Alternative hypothesis:[tex]\mu_{1} - \mu_{2}\neq 0[/tex]  

The statistic is given by:

[tex]t=\frac{(\bar X_{1}-\bar X_{2})-\Delta}{\sqrt{\frac{s^2_{1}}{n_{1}}+\frac{s^2_{2}}{n_{2}}}}[/tex] (1)  

The degrees of freedom are given by:

[tex]df=n_1 +n_2 -2=22+11-2=31[/tex]  

Part a: Statisitc

Replacing into the formula we got:

[tex]t=\frac{(56.2-64.2)-0}{\sqrt{\frac{18.2^2}{22}+\frac{13.9^2}{11}}}}=-1.40[/tex]  

Part b: P value  

The p value on this case would be:

[tex]p_v =2*P(t_{31}<-1.4)=0.171[/tex]  

Answer-

You should always check the solution is a rational equation. It would be easier to get an answer... Idk if this helps, but thats what I was taught>>>

Step-by-step explanation:

Answer:

Check the explanation

Step-by-step explanation:

Given that:

The National Institutes of Health has recommended that the healthy adults are to take at least 50 nmol/L of vitamin D per day. Let’s assume that you are interested in discovering if the BSU students are suffering from vitamin D deficiency or not.

a) Formulate the appropriate null and alternative hypotheses

Null hypothesis

H o : u >= 50

Alternative hypothesis

Ha : u < 50

Answer:

40

Step-by-step explanation:

200/5 = 40

Answer:

20$

Step-by-step explanation:

10, five dollar bills make 50 so 20 five dollar bills make 100$

Answer:

a. possible sample size is 10

b. mean is 3

c standard deviation is 0.9

Step-by-step explanation:

Mean of any given set of numbers is the "average" you're used to, where you add up all the numbers and then divide by the number of numbers.

Standard deviation is a measure of the amount of variation or dispersion of a set of values.

Go to attachment for detailed analysis.

The number of possible samples of size 2 without replacement from a population of 5 objects is 10. The means of these samples can be calculated by taking the combinations of 2 objects and finding their mean. The mean of these means is equal to the population mean. The standard error of these means can be calculated using the variance of the means and the finite population correction factor.

a. To calculate the number of possible samples of size 2 without replacement from a population of 5 objects, we use the combination formula: C(n, r) = n! / (r! * (n-r)!), where n is the population size and r is the sample size. In this case, n = 5 and r = 2, so the number of possible samples is C(5, 2) = 5! / (2! * (5-2)!) = 10.

b. To list all the means of these samples, we take each combination of 2 objects from the population and calculate their mean. For example, one possible sample is {1, 2}, and its mean is (1 + 2) / 2 = 1.5. Similarly, we calculate the means for the other 9 possible samples.

c. To find the mean of these means, we calculate the average of all the means calculated in part b. d. The mean of these means, denoted as E(X), is equal to the population mean, denoted as Mu, when the sampling is done without replacement.

e. To find the standard error sigma(X), we need to calculate the standard deviation of all these means. We can do this by calculating the variance of the means and taking its square root. f. The standard error sigma(X) equals the square root of the variance of the means divided by the square root of the sample size, multiplied by the finite population correction factor.

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Answer:

a) P (Y  less than or equal to 79 ) = 0.226  

b) Z = 1.008  less than or equal to Zc = 1.64

So, it means null hypothesis is not rejected.

Step-by-step explanation:

a) Probability that fewer than 80 patients will recover.

As we have:

p = 88% = 0.88 = population success proportion

n = 93 = Sample size

mean = n x p , mean = 93 x 0.88 , mean = 81.84

So, we need P (Y less than or equal to 79) fewer than 80 patients.

Standard Deviation of the population =

SD of population = [tex]\sqrt{n . p . (1-p)}[/tex] , SD of population = [tex]\sqrt{93 .0.88 .(1-0.88)}[/tex]

SD of population = 3.13

As we know that,

probability in binomial distribution is approximately equal to normal distribution so, we will use normal distribution from here. In addition, due to this shift 79 will be equal to 79.5 in normal distribution.

P(Y less than or equal to 79) ≈ P(Y less than or equal to 79.5)

P([tex]\frac{Y - mean }{SD}[/tex]) , P([tex]\frac{79.5 - 81.84}{3.1336}[/tex]) , P(Z less than or equal to -0.75)

So, when will you check the Z score of -0.75 in the Z table you will get the probability which is :

P (Y less than or equal to 79 ) = 0.226

So, 0.226 is the probability that fewer than 80 patients will recover.

b) Z  test is used in this part to check the claim:

Favorable Cases sample (recovered) = 85 = Y

Sample size = N = 93

Significance level = 0.05

Proportion Sample = p = Y/N = 85/93 = 0.914.

1. Null hypothesis and Alternative Hypothesis:

H1: p less than or equal to 88% or p less than or equal to 0.88

H2: p greater than 88% or p greater than 0.88

2. Rejection Region:

Zc value is = 1.64 for a right tail test.

Rejection region is such that z value must be greater than Zc value i.e 1.64. If it is not greater, it will not lie in rejection region.

3. Statistics Test:

In this we have to calculate value of Z and compare it with value of Zc.

Z = [tex]\frac{P-p1}{\sqrt{p1(1-p1)/n} }[/tex]

Small p1 = 0.88 , Capital P = 0.914  Z = [tex]\frac{0.914-0.88}{\sqrt{0.88.(1-0.88)/93} }[/tex]

Z = 1.008

4. Final Decision about the null hypothesis:

As we can see, Z value is less than Zc value, hence it does not lie in the rejection region.

Z = 1.008  less than or equal to Zc = 1.64

So, it means null hypothesis is not rejected.

H1 is not rejected,  it means there is not sufficient proof to claim that population proportion is larger than population success proportion p1.

a. Estimate the probability that fewer than 80 patients will recover: Approximately 0.2743.

b. Confirm/disprove the director's claim:

The probability of observing more than 85 patients recovering out of 93 treated with the new technique is approximately 0.1515. Therefore, there isn't sufficient evidence to support the director's claim that all hospitals should adopt the new technique based solely on this sample data.

To solve this problem, we'll use the normal approximation to the binomial distribution since the sample size (93) is relatively large.

Let's break it down step by step.

Given:

Probability of traditional medical treatment curing the disorder = 0.88

Number of patients treated with the new technique = 93

a. Estimate the probability that fewer than 80 patients will recover.

To estimate this probability, we'll use the normal approximation to the binomial distribution.

1. Calculate the mean [tex](\(\mu\))[/tex] and standard deviation [tex](\(\sigma\))[/tex] of the binomial distribution:

[tex]\[\mu = np = 93 \times 0.88 = 81.84\][/tex]

[tex]\[\sigma = \sqrt{np(1-p)} = \sqrt{93 \times 0.88 \times 0.12} \approx 3.076\][/tex]

2. Standardize the value of 80 using the z-score formula:

[tex]\[z = \frac{x - \mu}{\sigma} = \frac{80 - 81.84}{3.076} \approx -0.600\][/tex]

3. Use the standard normal distribution table or calculator to find the probability corresponding to z = -0.600.

[tex]\[P(Z < -0.600) \approx 0.2743\][/tex]

So, the estimated probability that fewer than 80 patients will recover is approximately 0.2743.

b. The director of the hospital recommends that all hospitals adopt this new technique since more than 85 patients in the sample have recovered after being treated with the new technique.

Confirm/disprove the director's claim.

To confirm/disprove the director's claim

We'll calculate the probability of observing more than 85 successes (patients recovering) out of 93 trials (patients treated) under the assumption that the success rate is the same as the traditional treatment (0.88).

1. Calculate the mean [tex](\(\mu\))[/tex] and standard deviation [tex](\(\sigma\))[/tex] of the binomial distribution:

[tex]\[\mu = np = 93 \times 0.88 = 81.84\][/tex]

[tex]\[\sigma = \sqrt{np(1-p)} = \sqrt{93 \times 0.88 \times 0.12} \approx 3.076\][/tex]

2. Standardize the value of 85 using the z-score formula:

[tex]\[z = \frac{x - \mu}{\sigma} = \frac{85 - 81.84}{3.076} \approx 1.031\][/tex]

3. Calculate the probability of observing more than 85 successes:

[tex]\[P(Z > 1.031) \approx 1 - P(Z < 1.031) \approx 1 - 0.8485 \approx 0.1515\][/tex]

The probability of observing more than 85 successes out of 93 trials is approximately 0.1515, which means there's about a 15.15% chance of observing this many or more successes if the true success rate is 0.88.

Since [tex]\(15.15\% > 5\%\)[/tex]  (common significance level),

We fail to reject the null hypothesis that the success rate of the new technique is the same as the traditional treatment.

Therefore, there isn't sufficient evidence to support the director's claim that all hospitals should adopt the new technique based solely on this sample data.

Answer:

Step-by-step explanation:

Answer:

5

Step-by-step explanation:

starts with 16. 6 carrots and 10 cucumbers. 10-5=5

Answer:

i. Point of estimate:

[tex] \hat \mu = \bar X =143.0[/tex]

ii. Margin of error:

[tex] ME = 2.01 *\frac{56.7}{\sqrt{50}}= 16.12[/tex]

iii. The 90% confidence interval

Replacing in the confidence interval formula we got

[tex]143.0-16.12=126.88[/tex]    

[tex]143.0+16.12=159.12[/tex]    

The 90% confidence interval is 126.88 to 159.12

Step-by-step explanation:

Information given

[tex]\bar X=143.0[/tex] represent the sample mean for the variable of interest

[tex]\mu[/tex] population mean

s=56.7 represent the sample standard deviation

n=50 represent the sample size  

Confidence interval

The confidence interval for the true mean when we don't know the deviation is given by the following formula:

[tex]\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}[/tex]   (1)

In order to calculate the critical value for the confidence interval [tex]t_{\alpha/2}[/tex] we need to find the degrees of freedom, with this formula:

[tex]df=n-1=50-1=49[/tex]

The Confidence level provided is 0.90 or 90%, the value for the significance is [tex]\alpha=1-0.9=0.1[/tex] and [tex]\alpha/2 =0.05[/tex], and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.05,49)".And we see that [tex]t_{\alpha/2}=2.01[/tex]

i. Point of estimate:

[tex] \hat \mu = \bar X =143.0[/tex]

ii. Margin of error:

[tex] ME = 2.01 *\frac{56.7}{\sqrt{50}}= 16.12[/tex]

iii. The 90% confidence interval

Replacing in the confidence interval formula we got

[tex]143.0-16.12=126.88[/tex]    

[tex]143.0+16.12=159.12[/tex]    

The 90% confidence interval is 126.88 to 159.12

To find the area of the surface generated by revolving the curve about the x-axis, you need to evaluate the definite integral of the square of the given function within the specified interval.

Definite integral representing the area:

Given curve: y = 81 - x^2, -5 ≤ x ≤ 5

Revolving this curve about the x-axis generates a surface. The definite integral representing the area is obtained by integrating the square of the function and applying the limits of integration.

The total area of the regions between the curves is 726 square units

Calculating the total area of the regions between the curves

From the question, we have the following parameters that can be used in our computation:

y = 81 - x²

Also, we have the interval to be

−5 ≤ x ≤ 5

So, the area of the regions between the curves is

Area = ∫y dx

This gives

Area = ∫(81 - x²) dx

Integrate

Area =  81x - x³/3

Recall that  −5 ≤ x ≤ 5

So, we have

Area =  |[81(-5) - (-5)³/3] - [81(5) - (5)³/3]|

Evaluate

Area =  726

Hence, the total area of the regions between the curves is 726 square units

Answer:

Step-by-step explanation:

This is a test of 2 independent groups. The population standard deviations are not known. it is a two-tailed test. Let c b the subscript for students using computer and n be the subscript for students not using computer.

Therefore, the population means would be μc and μn.

The random variable is xc - xn = difference in the sample mean scores of students who used computers and those who didn't use computers

We would set up the hypothesis.

The null hypothesis is

H0 : μc = μn H0 : μc - μn = 0

The alternative hypothesis is

Ha : μc ≠ μn Ha : μc - μn ≠ 0

Since sample standard deviation is known, we would determine the test statistic by using the t test. The formula is

(xc - xn)/√(sc²/nc + sn²/nn)

From the information given,

xc = 309

xn = 303

sc = 29

sn = 32

nc = 60

nn = 40

t = (309 - 303)/√(29²/60 + 32²/40)

t = 0.953

The formula for determining the degree of freedom is

df = [sc²/nc + sn²/nn]²/(1/nc - 1)(sc²/nc)² + (1/nn - 1)(sn²/nn)²

df = [29²/60 + 32²/40]²/(1/60 - 1)(29²/60)² + (1/40 - 1)(32²/40)² = 1569.48/20.13

df = 78

We would determine the probability value from the t test calculator. It becomes

p value = 0.344

Assuming a level of significance of 0.05, we would not reject the null hypothesis because the p value, 0.344 is > 0.05

Therefore, we cannot conclude that the population mean scores differ.

Answer:

110 utensils

Step-by-step explanation:

Set up equal fractions, and cross-multiply.

[tex]\frac{22 forks}{33 spoons} = \frac{44 forks}{x spoons}[/tex]

Solve for spoons.

22x = 44*33

22x = 1452

x = 1452/22 = 66 spoons

Now add forks and spoons to get total utensils:

44 forks + 66 spoons = 110 utensils

Answer:

There would be 73 utensils altogether.

Step-by-step explanation:

We have the ratio 22:33 and x:44

in order to determine x we need to divide 44 by 22 which leaves us with 1.33333 or 1 and 1/3

we then need to multiply 22 by our quotient of 1.333 in which leaves us with 29.

so x=29

We then have to add 44 and 29 which leaves us with 73;)

Answer:

14

Step-by-step explanation: