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crimeas [40]
4 years ago
10

What is the following quotient? Sqrt120/sqrt30

Mathematics
2 answers:
SVETLANKA909090 [29]4 years ago
3 0

Since both numbers are under the square root sign they can be combined under 1 sqrt sign.

\dfrac{\sqrt{120}}{\sqrt{30}}=\sqrt{\dfrac{120}{30}}

Now all you need do is perform the division under the root sign

sqrt(4) is your answer but not the final one.

2 <<<<< Answer

Leona [35]4 years ago
3 0

Answer:  the required quotient is 2.

Step-by-step explanation:  We are given to find the following quotient :

Q=\dfrac{\sqrt{120}}{\sqrt{30}}~~~~~~~~~~~~~~~~~~~~~~~~~~(i)

We will be using the following properties of radicals and exponents :

(i)~\sqrt{x}=x^\frac{1}{2},\\\\(ii)~\sqrt{a\times b}=\sqrt a\times \sqrt b.

From equation (i), we have

Q\\\\\\=\dfrac{\sqrt{120}}{\sqrt{30}}\\\\\\=\dfrac{\sqrt{4\times30}}{\sqrt{30}}\\\\\\=\dfrac{\sqrt4\times\sqrt{30}}{\sqrt{30}}\\\\=\sqrt{2^2}\\\\=(2^2)^\frac{1}{2}\\\\=2^{2\times\frac{1}{2}}\\\\=2.

Thus, the required quotient is 2.

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According to a Yale program on climate change communication survey, 71% of Americans think global warming is happening.† (a) For
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a) 0.2741 = 27.41% probability that at least 13 believe global warming is occurring

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Binomial probability distribution

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

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Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that \mu = E(X), \sigma = \sqrt{V(X)}.

In this problem, we have that:

p = 0.71

(a) For a sample of 16 Americans, what is the probability that at least 13 believe global warming is occurring?

Here n = 16, we want P(X \geq 13). So

P(X \geq 13) = P(X = 13) + P(X = 14) + P(X = 15) + P(X = 16)

In which

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

P(X = 13) = C_{16,13}.(0.71)^{13}.(0.29)^{3} = 0.1591

P(X = 14) = C_{16,14}.(0.71)^{14}.(0.29)^{2} = 0.0835

P(X = 15) = C_{16,15}.(0.71)^{15}.(0.29)^{1} = 0.0273

P(X = 16) = C_{16,16}.(0.71)^{16}.(0.29)^{0} = 0.0042

P(X \geq 13) = P(X = 13) + P(X = 14) + P(X = 15) + P(X = 16) = 0.1591 + 0.0835 + 0.0273 + 0.0042 = 0.2741

0.2741 = 27.41% probability that at least 13 believe global warming is occurring

(b) For a sample of 160 Americans, what is the probability that at least 110 believe global warming is occurring?

Now n = 160. So

\mu = E(X) = np = 160*0.71 = 113.6

\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{160*0.71*0.29} = 5.74

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Z = \frac{X - \mu}{\sigma}

Z = \frac{109.5 - 113.6}{5.74}

Z = -0.71

Z = -0.71 has a pvalue of 0.2389

1 - 0.2389 = 0.7611

0.7611 = 76.11% probability that at least 110 believe global warming is occurring

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