The arena would like to estimate its annual running costs for the next twelve months. Use the following first 4 months' figures to estimate the running costs for 12 months. Month Costs January £14,889.51 February £22,936 March £9,856.88 April £6,777.77 The estimated running costs for 12 months is

The 95% confidence interval for the population mean ( μ) is approximately (8.46,11.54), calculated from a sample of 26 elements with a mean of 10 and a standard deviation of 4.

To calculate the 95% confidence interval for the population mean (μ), we can use the formula:

Confidence interval=Sample mean±(Critical value× Sample size/Standard deviation​ )

Given:

Sample mean (xˉ ) = 10

Standard deviation (σ) = 4

Sample size (n) = 26

Confidence level = 95%

Step 1: Find the critical value from the Z-table for a 95% confidence level.

Since it's a two-tailed test, we'll find the Z-value corresponding to a cumulative probability of 0.975.

Z α/2​ =1.96

Step 2: Plug the values into the formula:

Confidence interval=10±(1.96× 26​/4​ )

Step 3: Calculate the margin of error:

Margin of error≈1.96×45.099

Margin of error≈1.96× 5.0994

Margin of error≈1.96×0.785

Margin of error≈1.5376

Step 4: Calculate the confidence interval:

Lower limit=10−1.5376

Lower limit≈8.4624

Upper limit=10+1.5376

Upper limit≈11.5376

So, the 95% confidence interval for

μ is approximately (8.4624,11.5376).

Total weight of six identical toys = 1.8 Kg

Therefore, the weight of one such toy = 1.8 kg/ 6 = 0.3 kg

Further, 1 kg = 1000 gms

Answer to Ques B-

Weight of one toy in grams = 0.3 * 1000 grams

= 300 grams

Answer to Ques C-

Weight of 4 toys in kilograms = 0.3 * 4 kilograms

= 1.2 kilograms

Hope this helps..!!

Thank you :)

Answer:

(c) Factor by Grouping

To calculate Jennifer's monthly income, divide her annual income by 12. Then, subtract the deductions and taxes to find her net monthly income.

To calculate Jennifer's monthly income, we need to divide her annual income by 12, since there are 12 months in a year. So, Jennifer's monthly income would be $74,900 / 12 = $6,241.67.

To calculate Jennifer's net monthly income, we need to subtract the deductions and taxes from her monthly income. First, let's calculate the deductions:

Now, we can subtract these deductions from Jennifer's monthly income:

Net monthly income = $6,241.67 - $374.50 - $936.25 - $386.67 - $90.39 = $4,453.86

Therefore, Jennifer's net monthly income is $4,453.86.

Answer:

Step-by-step explanation:

Given: m∠H = 30°, m∠J = 50°, m∠P = 50°, m∠N = 100°

To prove: △HKJ ~ △LNP

Proof:

Step 1. m∠H = 30°, m∠J = 50°, m∠P = 50°, m∠N = 100° (Given)

Step 2. m∠H + m∠J + m∠K = 180° (Triangle angle sum theorem)

Step 3. 30° + 50° + m∠K = 180°(substitution property)

Step 4. 80° + m∠K = 180°(addition Property)

Step 5. m∠K = 100°(subtraction property of equality)

Step 6. m∠J = m∠P; m∠K = m∠N(substitution)

Step 7. ∠J ≅ ∠P; ∠K ≅ ∠N (If angles are equal then they are congruent)

Step 8. △HKJ ~ △LNP( AA similarity theorem)

Hence proved.

Thus, the missing step in 2 is (Triangle angle sum theorem)

Answer:

D) triangle angle sum theorem

Step-by-step explanation:

Just finished the test!!

Answer:

V = 54π cm³ ⇒ answer (C)

Step-by-step explanation:

∵ V = (πd²h)/8

∵ d = 6 cm

∵ h = 12 cm

∴ V = (π × 6² × 12) ÷ 8 = 54π cm³

Final answer:

To find the volume of a parabolic cone with a diameter of 6 cm and height of 12 cm using the formula V = πd²h/8, calculate the radius (3 cm), square it, multiply by the height, divide by 8, and then multiply by π, resulting in 54π cm³(Option C).

Explanation:

The student's question pertains to calculating the volume of a parabolic cone using a specific formula, which is V = πd²h/8, where d is the diameter and h is the height of the cone.

Given the cone's diameter (d) of 6 cm and height (h) of 12 cm, we can plug these values into the formula to find the volume:

Therefore, the volume of a parabolic cone with a diameter of 6 cm and a height of 12 cm is 54π cm³.

The given function is represented below:

F(x)=500e^(0.05)x with a value x=2

or, F(x) = 500 * e^(0.05*2)

( putting the value of x as 2 )

or, F(x) = 500 * e^(0.1)

or, F(x) = 500 * e^0.1

or, F(x) = 500 * e^(0.1)

or, F(x) = 500 * 1.10517092

or, F(x) = 552.585459

The value of the function at x = 2, rounded off to 3 decimal places is given by 552.585

Hope this helps..!!

Thank you :)

Answer:

The equation of the resulting parabola is:

y=(x+7)^2

Step-by-step explanation:

We, know that the transformation of the type:

f(x+a) is a translation of the parent function f(x) to the left or right depending on the sign of the constant 'a'.

if:

a<0 then the translation is to the right.

and if a>0 then the translation is to the left.

It is given that the graph of the function,

let f(x)=y=x^2 is translated 7 units to the left.

This means that the equation of the resulting function will be:

y=(x+7)^2

Original Price of the shirt is $38.

Double discount :

Given that it was reduced by 20%, means

[tex] \\ Cost \; of \; Shirt\; after\; 1^{st}\; discount (20\%)\\ \\= \; (100\; - \; 20) \%\; of \; Original Price=\; 80\%\; of\; \$38\\ \\ = \frac{80}{100} \times 38\; =\; \frac{3040}{100}\; =\; \$30.40\\ \\ Cost \; of \; Shirt \; after\; 2^{nd}\; discount(10\%)\\ \\ (100 - 10)\%\; of\; $30.40=90\%\; of\; \$30.40\\ [/tex]

[tex] =\frac{90}{100} \times 30.40 = \frac{2736}{100}= \$27.36 \\ \\ After \; double \; \; discount,(20\%\; of\; Original\; Price, again\; 10\%\; reduced\; )\\\\ Double \; Discounted \; Price = \$ 27.36\\ \\ But \; After\; single \; discount\; of\; 30\%\; of\; Original \; Price, \; we\; get\\ \\ (100-30)\%\; of \; \$38=70\% \times \$38=\; \frac{70}{100} \times 38 =\frac{2660}{100}=\$26.6 [/tex]

Answer:

Let X the random variable that represent the mean finish time for a yearly amateur auto race a population, and for this case we know the distribution for X is given by:

[tex]X \sim N(185.19,0.341)[/tex]  

Where [tex]\mu=185.19[/tex] and [tex]\sigma=0.341[/tex]

The z score is given by this formula:

[tex]z=\frac{x-\mu}{\sigma}[/tex]

And for a time of 184.14 we have the following z score:

[tex] z = \frac{184.14-185.19}{0.341}= -3.08[/tex]

Let Y the random variable that represent the mean finish time for the previous year auto race a population, and for this case we know the distribution for X is given by:

[tex]Y \sim N(110.4,0.137)[/tex]  

Where [tex]\mu=110.4[/tex] and [tex]\sigma=0.137[/tex]

The z score is given by this formula:

[tex]z=\frac{x-\mu}{\sigma}[/tex]

And for a time of 110.05 we have the following z score:

[tex] z = \frac{110.05-110.4}{0.137}=-2.557[/tex]

As we can see we have a higher z score for the case of the previous year so then we have a more convincing victory on this case since represent a higher quantile in the normal standard distribution.

Step-by-step explanation:

Previous concepts

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".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Solution to the problem

Let X the random variable that represent the mean finish time for a yearly amateur auto race a population, and for this case we know the distribution for X is given by:

[tex]X \sim N(185.19,0.341)[/tex]  

Where [tex]\mu=185.19[/tex] and [tex]\sigma=0.341[/tex]

The z score is given by this formula:

[tex]z=\frac{x-\mu}{\sigma}[/tex]

And for a time of 184.14 we have the following z score:

[tex] z = \frac{184.14-185.19}{0.341}= -3.08[/tex]

Let Y the random variable that represent the mean finish time for the previous year auto race a population, and for this case we know the distribution for X is given by:

[tex]Y \sim N(110.4,0.137)[/tex]  

Where [tex]\mu=110.4[/tex] and [tex]\sigma=0.137[/tex]

The z score is given by this formula:

[tex]z=\frac{x-\mu}{\sigma}[/tex]

And for a time of 110.05 we have the following z score:

[tex] z = \frac{110.05-110.4}{0.137}=-2.557[/tex]

As we can see we have a higher z score for the case of the previous year so then we have a more convincing victory on this case since represent a higher quantile in the normal standard distribution.

Answer:

19.63

Step-by-step explanation:

Answer:

6

Step-by-step explanation:

Replace x with -10

[tex]b(-10) = |(-10) +4| = |-6| = 6[/tex]

Answer: b) [tex]f(x)=4x+2[/tex]

Step-by-step explanation:

The given table :-

Input           4               5                7                8

Output        18              22              30              34

Let's check all the options

a) [tex]f(x)=5x-2[/tex]

At x= 4, [tex]f(4)=5(4)-2=18[/tex]

At x= 5 , [tex]f(5)=5(5)-2=25-2=23\neq22[/tex]  , thus it not the correct function.

b) [tex]f(x)=4x+2[/tex]

At x= 4 ,  [tex]f(4)=4(4)+2=18[/tex]

At x= 5 ,  [tex]f(5)=4(5)+2=22[/tex]

At x= 7 ,  [tex]f(7)=4(7)+2=30[/tex]

At x= 8 ,  [tex]f(8)=4(8)+2=34[/tex] , thus it is the function rule that  produces the output.

c) [tex]f(x)=3x+6[/tex]

At x= 4 ,  [tex]f(4)=3(4)+6=18[/tex]

At x= 5 ,  [tex]f(5)=3(5)+6=21\neq22[/tex]  , thus it not the correct function.

d)  [tex]f(x)=2x+10[/tex]

At x= 4 ,  [tex]f(4)=2(4)+10=18[/tex]

At x= 5 ,  [tex]f(5)=2(5)+10=20\neq22[/tex]  , thus it not the correct function.

Hence, the correct function rule that produces the output = b) [tex]f(x)=4x+2[/tex]

Answer:

-14a-18b+22c+26d

The water level in a paper cup shaped like a cone with radius 3 cm and height 10 cm, filled at a rate of 2 cm³/s, rises at approximately 0.025 cm per second when the water is 5 cm deep.

This question involves related rates, a concept in Calculus. We know that the volume, V, of a cone with a radius r and height h is given by the formula V = (1/3)πr²h. Given that the shape of the cup is conical, the radius and the height of the water in the cup are proportional, so we can express r as r=3h/10.

Thus, we can rewrite the volume formula in terms of h: V = 1/3 * π * (3h/10)² * h = πh³/100. Differentiating both sides with respect to time t, we get dV/dt = πh² dh/dt. We want to find dh/dt (the rate at which the water level rises) when h=5 cm and given that dV/dt (the rate at which water is poured into the cup) is 2 cm³/s.

Plugging these values into the differentiated formula, we get: 2 = π(5)² * dh/dt. Solving for dh/dt, we find that dh/dt = 2/(25π) or about 0.025 cm/s.  So, the water level is rising at a rate of approximately 0.025 cm per second when the water is 5 cm deep.

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Final answer:

The integral of y/((y^2)-1) is (1/2) ln|y^2 - 1| + C.

Explanation:

To evaluate the integral of y/((y^2)-1), we can perform a substitution by letting u = y^2 - 1. Then, du = 2y dy, and the integral becomes:

∫ y/((y^2)-1) dy = (1/2) ∫ 1/u du = (1/2) ln|u| + C

Substituting back u = y^2 - 1, the final answer is (1/2) ln|y^2 - 1| + C.