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3 years ago

A playground is shaped like a rectangle whose length is 25m and breadth is 20

Mathematics

1 answer:

FrozenT [24]3 years ago

8 0

Answer:

500m square

Step-by-step explanation:

first we know that to find the area of something we have to multiply so I did 25×20 and i got 500

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Tony needs one gallon of orange juice to make punch for a party. Gallons of orange juice cost $3.99 each. Quarts of orange juice

Travka [436]

Answer

C because 2 quarts make's one gallon

Step-by-step explanation:

5 0

3 years ago

Directions: Calculate the percent increase or decrease between the starting and ending

valentinak56 [21]

Answer:233.333%

Step-by-step explanation:

1. 10-3=7 subtract end by start

2.7/3=2.333 divide difference by absolute value of start

32.333x100=233.333% multiply by 100 if negative its decreasing if positive it increasing

hoped this helped

8 0

3 years ago

Read 2 more answers

Factor the expression completely 8x^2-18 Please help

nirvana33 [79]

Answer:

2 (2x - 3) x (2x + 3)

4 0

3 years ago

Please help me and If your answer is wrong or not appropreat I will report it

arsen [322]

Answer:

972 x^16 y^24

Step-by-step explanation:

Simplify the following:

(-2 x^3 y^7)^2 (3 x^2 y^2)^5

Multiply each exponent in -2 x^3 y^7 by 2:

(-2)^2 x^(2×3) y^(2×7) (3 x^2 y^2)^5

2×7 = 14:

(-2)^2 x^(2×3) y^14 (3 x^2 y^2)^5

2×3 = 6:

(-2)^2 x^6 y^14 (3 x^2 y^2)^5

(-2)^2 = 4:

4 x^6 y^14 (3 x^2 y^2)^5

Multiply each exponent in 3 x^2 y^2 by 5:

4 x^6 y^14×3^5 x^(5×2) y^(5×2)

5×2 = 10:

4×3^5 x^6 y^14 x^(5×2) y^10

5×2 = 10:

4×3^5 x^6 y^14 x^10 y^10

3^5 = 3×3^4 = 3 (3^2)^2:

4×3 (3^2)^2 x^6 y^14 x^10 y^10

3^2 = 9:

4×3×9^2 x^6 y^14 x^10 y^10

9^2 = 81:

4×3×81 x^6 y^14 x^10 y^10

3×81 = 243:

4×243 x^6 y^14 x^10 y^10

4 x^6 y^14×243 x^10 y^10 = 4 x^(6 + 10) y^(14 + 10)×243:

4×243 x^(6 + 10) y^(14 + 10)

14 + 10 = 24:

4×243 x^(6 + 10) y^24

6 + 10 = 16:

4×243 x^16 y^24

4×243 = 972:

Answer: 972 x^16 y^24

8 0

2 years ago

Read 2 more answers

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.

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

$\sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{10*0.23*0.77} = 1.33$

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

5 0

3 years ago