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2 years ago

Please HELP!! find the area of the rhombus.

Mathematics

1 answer:

balandron [24]2 years ago

3 0

Answer:

32.5

Step-by-step explanation:

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G(n-7) if g(x) =x^2-5/3x

Leno4ka [110]

Instead of x, we write n - 7 in the function equation

$g(x)=x^2-\dfrac{5}{3}x\\\\g(n-7)=(n-7)^2-\dfrac{5}{3}(n-7)\\\\\text{use} (a-b)^2=a^2-2ab+b^2\\\text{and distributive property}\\\\=n^2-2(n)(7)+7^2+\left(-\dfrac{5}{3}\right)(n)+\left(-\dfrac{5}{3}\right)(-7)\\\\=n^2-14n+49-\dfrac{5}{3}n+\dfrac{35}{3}\\\\\text{combine like terms}\\\\=n^2+\left(-14n-1\dfrac{2}{3}n\right)+\left(49+11\dfrac{2}{3}\right)\\\\=n^2-15\dfrac{2}{3}n+60\dfrac{2}{3}$

8 0

2 years ago

Suppose that 50% of all young adults prefer McDonald's to Burger King when asked to state a preference. A group of 12 young adul

ddd [48]

Answer:

a) 0.194 = 19.4% probability that more than 7 preferred McDonald's

b) 0.787 = 78.7% probability that between 3 and 7 (inclusive) preferred McDonald's

c) 0.787 = 78.7% probability that between 3 and 7 (inclusive) preferred Burger King

Step-by-step explanation:

For each young adult, there are only two possible outcomes. Either they prefer McDonalds, or they prefer burger king. The probability of an adult prefering McDonalds is independent from other adults. So 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.

50% of all young adults prefer McDonald's to Burger King when asked to state a preference.

This means that $p = 0.5$

12 young adults were randomly selected

This means that $n = 12$

(a) What is the probability that more than 7 preferred McDonald's?

$P(X > 7) = P(X = 8) + P(X = 9) + P(X = 10) + P(X = 11) + P(X = 12)$

In which

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

$P(X = 8) = C_{12,8}.(0.5)^{8}.(0.5)^{4} = 0.121$

$P(X = 9) = C_{12,9}.(0.5)^{9}.(0.5)^{3} = 0.054$

$P(X = 10) = C_{12,10}.(0.5)^{10}.(0.5)^{2} = 0.016$

$P(X = 11) = C_{12,11}.(0.5)^{11}.(0.5)^{1} = 0.003$

$P(X = 12) = C_{12,12}.(0.5)^{12}.(0.5)^{0} = 0.000$

$P(X > 7) = P(X = 8) + P(X = 9) + P(X = 10) + P(X = 11) + P(X = 12) = 0.121 + 0.054 + 0.016 + 0.003 + 0.000 = 0.194$

0.194 = 19.4% probability that more than 7 preferred McDonald's

(b) What is the probability that between 3 and 7 (inclusive) preferred McDonald's?

$P(3 \leq X \leq 7) = P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7)$

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

$P(X = 3) = C_{12,3}.(0.5)^{3}.(0.5)^{9} = 0.054$

$P(X = 4) = C_{12,4}.(0.5)^{4}.(0.5)^{8} = 0.121$

$P(X = 5) = C_{12,5}.(0.5)^{5}.(0.5)^{7} = 0.193$

$P(X = 6) = C_{12,6}.(0.5)^{6}.(0.5)^{6} = 0.226$

$P(X = 7) = C_{12,7}.(0.5)^{7}.(0.5)^{5} = 0.193$

$P(3 \leq X \leq 7) = P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7) = 0.054 + 0.121 + 0.193 + 0.226 + 0.193 = 0.787$

0.787 = 78.7% probability that between 3 and 7 (inclusive) preferred McDonald's

(c) What is the probability that between 3 and 7 (inclusive) preferred Burger King?

Since $p = 1-p = 0.5$ , this is the same as b) above.

0.787 = 78.7% probability that between 3 and 7 (inclusive) preferred Burger King

7 0

2 years ago

30 divided by<br> 4,800

DedPeter [7]

30/4800 = 3/480

= 1/160 in fraction form

= 0.00625 in decimal form

8 0

3 years ago

Read 2 more answers

Please help!!! It's due rn!

nexus9112 [7]

Step-by-step explanation:

In triangles BAD and BCD ,

BD=BD (common)

angle BDA= angle BCD {90°each(given)}

AD=DC (given)

.•. traingle BAD is congruent to triangle BCD (SAS criterion)

Hence , angle A = angle C (CPCT)

8 0

1 year ago

15 POINTS PLEASE HELP IVE ONLY GOT 10 MINS

frosja888 [35]

C is what I think it is

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2 years ago

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