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Dafna11 [192]
2 years ago
13

Find the area of the circle. 12 ft

Mathematics
2 answers:
Ratling [72]2 years ago
3 0

Answer:

  • Area of the circle is 113.04 ft²

Step-by-step explanation:

Given:

  • Diameter of circle = 12 feet

To Find:

  • Area of the circle

Using formula:

\\ \:  \: \dashrightarrow \:  \:  \:{\underline{\boxed{  \bf{Area_{( circle)} = \pi r^2}}}} \\ \\

<em>Where,</em>

  • π = 3.14
  • Diameter = 12 feet
  • Radius = 12/2 = 6 feet

On substituting the required values, we get:

\\   \:  \: \dashrightarrow \:  \:  \:  \sf Area_{(Circle)} =    3.14 \times  {(6)}^{2}  \\  \\  \\ \:  \: \dashrightarrow \:  \:  \:  \sf Area_{(Circle)} =   3.14 \times 36 \\  \\  \\ \:  \: \dashrightarrow \:  \:  \:  \sf Area_{(Circle)} =  113.04 \:  {ft}^{2}  \\  \\  \\

Hence,

  • Area of the circle is<u> </u>113.04 ft²
alexgriva [62]2 years ago
3 0

                    \rule{50}{1}\large\blue\textsf{\textbf{\underline{Question:-}}}\rule{70}{1}

           <em>Find the area of the circle (diameter = 12 ft)</em>

<em>        </em>\rule{50}{1}\large\blue\textsf{\textbf{\underline{Answer and how to solve:-}}}\rule{50}{1}

   We're asked to find the area of the circle, which can be found using the

     following formula:-

\longmapsto\underline{\boxed{\bold{A=\pi r^2}}}

Where:-

  • A=area
  • π=pi (3.14...)
  • r=radius

Provided Information:-

  • diameter (d) = 12 ft

What to do:-

     \bigstar Notice that we have the diameter and not the radius.

       \bigstar Since the radius is exactly one-half of the diameter, we divide the

             diameter by 2 and get the radius:-

\longmapsto\sf{d=2r}

   \bigstar Rearranging the formula,

\longmapsto\large\text{$r=\displaystyle\frac{d}{2}$}

\bigstar Substituting the required value (d=12) in lieu of d:-

\longmapsto\large\text{r=$\displaystyle\frac{12}{2}$}

\bigstar Circle's Radius:-

\longmapsto\large\text{\boxed{\boxed{r=6}}}

\triangle\rule{300}{1}\triangle

Now that we have the radius, let's find the area.

Remember, we need the following formula:-

\longmapsto\underline{\boxed{\sf{A=\pi r^2}}}

\bigstar Substituting the required value (r=6) into the formula,

\longmapsto\underline{\boxed{\sf{A=\pi (6)^2}}}

\bigstar Squaring the radius,

\longmapsto\underline{\boxed{\sf{A=\pi \times36}}}

\bigstar Now, Substitute 3.14 for pi:-

\longmapsto\sf{A=3.14\times36}

\bigstar Multiplying the values,

\longmapsto\sf{A=113.04\:ft^2}

  • Henceforth, We conclude that the right option is:-

\longmapsto\Large\underline{\boxed{\text{Option\:D}}}

<h3>Good luck with your studies.</h3>

#TogetherWeGoFar

\rule{300}{1}

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For each, list three elements and then show it is a vector space.
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Answer:

(a) Three polynomials of degree 1 with real coefficients belong to the set P_1=\{a_0+a_1x\ | a_0, a_1 \in \mathbb{R} \}, then:

2+3x \in P_1

4.5+\sqrt2 x \in P_1

\log5+78x \in P_1

(b) Three polynomials of degree 1 with real coefficients that hold the relation a_0 - 2a_1 = 0 belong to the set P_2=\{a_0+a_1x\ | a_0-2 a_1 =0 \}. The relation between the coefficients is equivalent to a_1 = \frac{a_0}{2}, then:

4+2x \in P_2

13+6.5x \in P_2

10.5+5.25x \in P_2

Step-by-step explanation:

(a) Three polynomials of degree 1 with real coefficients belong to the set P_1=\{a_0+a_1x\ | a_0, a_1 \in \mathbb{R} \}, then:

  • 2+3x \in P_1
  • 4.5+\sqrt2 x \in P_1
  • \log5+78x \in P_1

A vector space is any set whose elements hold the following axioms for any \vec{u}, \vec{v} and \vec{w} and for any scalar a and b:

  1. (\vec{u} + \vec{v} )+\vec{w} = \vec{u} +( \vec{v} +\vec{w})
  2. There is the <em>zero element </em>such that: \vec{0} + \vec{u} = \vec{u} + \vec{0}
  3. For all element \vec{u}of the set, there is an element -\vec{u} such that: -\vec{u} + \vec{u} = \vec{u} + (-\vec{u}) = \vec{0}
  4. \vec{u} + \vec{v} = \vec{v} + \vec{u}
  5. a(b\vec{v}) = (ab)\vec{v}
  6. 1\vec{u} = \vec{u}
  7. a(\vec{u} + \vec{v} ) = a\vec{u} + a\vec{v}
  8. (a+b)\vec{v} = a\vec{v}+b\vec{v}

Let's proof each of them for the first set. For the proof, I will define the polynomials a_0+a_1x, b_0+b_1x and c_0+c_1x and the scalar h and g.

  1. (a_0+a_1x + b_0+b_1x)+c_0+c_1x = a_0+a_1x +( b_0+b_1x+c_0+c_1x)\\(a_0+b_0+c_0) + (a_1+b_1+c_1)x = (a_0+b_0+c_0) + (a_1+b_1+c_1)x and defining a_0+b_0+c_0 = \alpha_0 and a_1+b_1+c_1 = \alpha_1, we obtain \boxed{\alpha_0+\alpha_1x= \alpha_0+\alpha_1x} which is another polynomial that belongs to P_1
  2. A null polynomial is define as the one with all it coefficient being 0, therefore: \boxed{0 + a_0+a_1x = a_0+a_1x + 0 = a_0+a_1x}
  3. Defining the inverse element in the addition as -a_0-a_1x, then -a_0-a_1x + a_0 + a_1x = a_0+a_1x + (-a_0-a_1x)\\\boxed{(-a_0+a_0)+(-a_1+a_1)x = (a_0-a_0)+(a_1-a_1)x = 0}
  4. (a_0+a_1x) +( b_0+b_1x) =( b_0+b_1x) +( a_0+a_1x)\\(a_0+b_0)+(a_1+b_1)x = (b_0+a_0)+(b_1+a_1)x\\\boxed{(a_0+b_0)+(a_1+b_1)x = (a_0+b_0)+(a_1+b_1)x}
  5. a[b(a_0+a_1x)] = ab (a_0+a_1x)\\a[ba_0+ba_1x] = aba_0+aba_1x\\\boxed{aba_0+aba_1x = aba_0+aba_1x}
  6. \boxed{1 \cdot (a_0+a_1x) = a_0+a_1x}
  7. \boxed{a[(a_0+a_1x)+(b_0+b_1x)] = a(a_0+a_1x) + a(b_0+b_1x)}
  8. (a+b)(a_0+a_1x)=aa_0+aa_1x+ba_0+ab_1x\\\boxed{(a+b)(a_0+a_1x)= a(a_0+a_1x) + b (a_0+a_1x)}

With this, we proof the set P_1 is a vector space with the usual polynomial addition and scalar multiplication operations.

(b) Three polynomials of degree 1 with real coefficients that hold the relation a_0 - 2a_1 = 0 belong to the set P_2=\{a_0+a_1x\ | a_0-2 a_1 =0 \}. The relation between the coefficients is equivalent to a_1 = \frac{a_0}{2}, then:

  • 4+2x \in P_2
  • 13+6.5x \in P_2
  • 10.5+5.25x \in P_2

Let's proof each of axioms for this set. For the proof, I will define again the polynomials a_0+a_1x, b_0+b_1x and c_0+c_1x and the scalar h and g. Again the relation a_1 = \frac{a_0}{2} between the coefficients holds

  1. [(a_0+a_1x) +( b_0+b_1x)]+(c_0+c_1x) = (a_0+a_1x) +[( b_0+b_1x)+(c_0+c_1x)]\\(a_0+b_0+c_0) + (a_1+b_1+c_1)x = (a_0+b_0+c_0) + (a_1+b_1+c_1)x and considering the coefficient relation and defining a_0+b_0+c_0 = \alpha_0 and a_1+b_1+c_1 = \alpha_1, we have (a_0+b_0+c_0) + (a_1+b_1+c_1)x = (a_0+b_0+c_0) + (a_1+b_1+c_1)x\\(a_0+b_0+c_0) + \frac{1}{2} (a_0+b_0+c_0)x = (a_0+b_0+c_0) + \frac{1}{2} (a_0+b_0+c_0)x\\\boxed{\alpha_0 + \alpha1x = \alpha_0 + \alpha1x} which is another element of the set since it is a degree one polynomial whose coefficient follow the given relation.

The proof of the other axioms can be done using the same logic as in (a) and checking that the relation between the coefficients is always the same.

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A study by Consumer Reports showed that 64% of supermarket shoppers believe supermarket brands to be as good as national name br
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Answer:

Step-by-step explanation:

Given that:

A study is conducted and the study by Consumer Reports showed that 64% of supermarket shoppers believe supermarket brands to be as good as national name brands.

a)

Thus; we formulate the null and the alternative hypotheses as follows:

The proportion is ; \frac{64}{100} = 0.64

Null hypothesis:    {H_0}:p = 0.64  

The Null hypothesis states that there is no evidence that “the percentage of supermarket shoppers believes that the supermarket ketchup is good as the national brand ketchup differ form 64%”.

Alternative hypothesis: {H_a}:p \ne 0.64H

The Alternative hypothesis states that there is evidence that “the percentage of supermarket shoppers believes that the supermarket ketchup is good as the national brand ketchup differ form 64%”.

b)

The proportion of the population = 0.64

The sample size n = 100

The number of shoppers = 52  stating that the supermarket brand was as good as the national brand

The sample proportion \hat  p = \frac{52}{100}

\hat  p = 0.52

The z - value is calculated by the formula:

z = \dfrac{\hat p-p}{\frac{\sqrt{ p(1-p)}}{100 }}

z = \dfrac{0.52-0.64}{\frac{\sqrt{ 0.64(1-0.64)}}{100 }}

z = \dfrac{-0.12}{0.048 }}

z = -2.50

Since z = -2.50

the p-value = 2P(Z ≤ -2.50)

p-value = 2 × 0.0062

The p-value = 0.0124

c)

At  significance level,  ∝ = 0.05 ; The p-value = 0.0124

According to  the rejection rule, if p-value is less than 0.05 then we will reject null hypothesis at ∝ = 0.05

Hence, the p-value =0.0124 < ∝ (=0.05)

According to the reject rule; reject null hypothesis.

Conclusion: There is no evidence at all that the percentage of supermarket shoppers believes that the supermarket ketchup is good as the national brand ketchup differ form 64%.

d) we know that the significance level of ∝  = 0.05

The value{z__{0.05}}  is obtained as:

P\left( {\left| Z \right| \le z} \right) = 0.05

Now; to determine Z ; we locate the probability value of 0.05 form the table of standard normal distribution. Then we proceed to the left until the  first column is reached and  the value is 1.90. Also , we move upward until we reach the top and the value is 0.06. Now; the intersection of the row and column results the area  to the left of z

This implies that :P(Z \leq -1.96)=0.05

The critical value for left tail is -1.96 and the critical value for right tail is 1.96.

Conclusion:

The critical value is -1.96 and the value of test statistic is - 2.50.  Here, we can see that  the value of test statistic is lesser than the critical value. Hence, we can be concluded that there is evidence that reject the null hypothesis.

Therefore, Yes, the national brand ketchup manufacturer is pleased with the conclusion.

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Answer:

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Step-by-step explanation:

The interquartile range defines the difference between the third and the first quartile.

Let's put this data set in ascending order:

24, 36, 48, 52, 52, 60, 72

The median is 52. The median of both the lower (1st quartile) and upper half (3rd quartile) of the data is 36 and 60.

Now we can subtract the first quartile from the third quartile (60 - 36) to get the interquartile range of 24.

Hope this helped! :)

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