The diameter of a quarter is 0.955 inches. If you laid 1,000,000 quarters side to

What is the area of the figure?

Lynna [10]

Answer:

112 in^2

Step-by-step explanation:

10 x 10 = 100

4 x 6 / 2 = 12

112 inches squared

5 0

3 years ago

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Find the sum or different 11/18-1/6

ira [324]

Answer:

4/9

Step-by-step explanation:

11/18-1/6

The first step is to get a common denominator, which is 18. We can multiply 1/6 by 3/3 to get the denominator to 18

11/18 - 1/6 *3/3

11/18 - 3/18

8/18

Now we can simplify the fraction by dividing the top and bottom by 2

4/9

7 0

3 years ago

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Rhombus EFGH is shown. What is the measure of ∠HGJ?

MaRussiya [10]

Answer:

∠HGJ = 35°

Step-by-step explanation:

Although the question is somewhat incomplete, I have tried to reconstruct the question as much as possible to aid understanding (attached is the pictorial representation of the rhombus).

Firstly, it is important to note that rhombuses are parallelograms which have all sides equal and whose diagonals are perpendicular to each other. In addition to this, their diagonals bisect one other.

Mathematically, |EF| = |FG| = |GH| = |HE|

Another property which rhombuses have is that their opposite angles are equal.

Mathematically, ∠E = ∠G, ∠H = ∠F

The adjacent angles in a rhombus are supplementary. This means that the sum of angles closest to each other is 180°.

Mathematically, ∠E + ∠F = 180°, ∠F + ∠G = 180°, ∠G + ∠H = 180°,

∠H + ∠E = 180°

The question gave us ∠GHE and asked us to find ∠HGJ (denoted as θ in the picture attached). Note that ∠GHE means ∠H. That is, ∠GHE ⇒ ∠H

∠GHE or ∠H = 110°

From the properties of rhombuses (earlier stated), we know that adjacent angles in a rhombus are supplementary

Mathematically,

∠GHE + ∠FGH ⇒ 110° + ∠FGH = 180° ⇒ ∠FGH = 180° - 110°

∠FGH or ∠G = 70°

To calculate for ∠HGJ, we divide ∠FGH by 2, we have:

∠HGJ = ∠FGH ÷ 2 = 70° ÷ 2

∠HGJ = 35°

3 0

3 years ago

A survey on British Social Attitudes asked respondents if they had ever boycotted goods for ethical reasons (Statesman, January

Blababa [14]

Answer:

a) 27.89% probability that two have ever boycotted goods for ethical reasons

b) 41.81% probability that at least two respondents have boycotted goods for ethical reasons

c) 41.16% probability that between 3 and 6 have boycotted goods for ethical reasons

d) The expected number is 2.3 and the standard deviation is 1.33.

Step-by-step explanation:

We use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

23% of the respondents have boycotted goods for ethical reasons.

This means that $p = 0.23$

a) In a sample of six British citizens, what is the probability that two have ever boycotted goods for ethical reasons?

This is P(X = 2) when n = 6. So

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 2) = C_{6,2}.(0.23)^{2}.(0.77)^{4} = 0.2789$

27.89% probability that two have ever boycotted goods for ethical reasons

b) In a sample of six British citizens, what is the probability that at least two respondents have boycotted goods for ethical reasons?

Either less than two have, or at least two. The sum of the probabilities of these events is decimal 1. So

$P(X < 2) + P(X \geq 2) = 1$

$P(X \geq 2) = 1 - P(X < 2)$

In which

$P(X < 2) = P(X = 0) + P(X = 1)$

$P(X = 0) = C_{6,0}.(0.23)^{0}.(0.77)^{6} = 0.2084$

$P(X = 1) = C_{6,1}.(0.23)^{1}.(0.77)^{5} = 0.3735$

$P(X < 2) = P(X = 0) + P(X = 1) = 0.2084 + 0.3735 = 0.5819$

$P(X \geq 2) = 1 - P(X < 2) = 1 - 0.5819 = 0.4181$

41.81% probability that at least two respondents have boycotted goods for ethical reasons

c) In a sample of ten British citizens, what is the probability that between 3 and 6 have boycotted goods for ethical reasons?

Now n = 10.

$P(3 \leq X \leq 6) = P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6)$

$P(X = 3) = C_{10,3}.(0.23)^{3}.(0.77)^{7} = 0.2343$

$P(X = 4) = C_{10,4}.(0.23)^{4}.(0.77)^{6} = 0.1225$

$P(X = 5) = C_{10,5}.(0.23)^{5}.(0.77)^{5} = 0.0439$

$P(X = 6) = C_{10,6}.(0.23)^{6}.(0.77)^{4} = 0.0109$

$P(3 \leq X \leq 6) = P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) = 0.2343 + 0.1225 + 0.0439 + 0.0109 = 0.4116$

41.16% probability that between 3 and 6 have boycotted goods for ethical reasons

d) In a sample of ten British citizens, what is the expected number of people that have boycotted goods for ethical reasons? Also find the standard deviation.

$E(X) = np = 10*0.23 = 2.3$