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Aneli [31]
1 year ago
7

The quotient property of radicals requires the indices of the radicals to be the same. Does this mean that it is not possible to

write es002-1.jpg as a single radical? Explain.
Mathematics
2 answers:
Troyanec [42]1 year ago
6 0

The radicands are powers of the same base, so they can be written using rational exponents.

Simplify the quotient of the exponential expressions by getting a common denominator and subtracting exponents.

The simplified expression is the 4th root of y

GaryK [48]1 year ago
4 0
Although the expression is not included, it is possible to assure that the answer is negative.

It is negative because you can express a quotient of radicals as a single radical using the least common index fo the radicals.

Look. This is the quotient property of radicals:
<span>
\frac{ \sqrt[n]{x} }{ \sqrt[n]{y} } = \sqrt[n]{ \frac{x}{y} }

Now, if you have the quotient of two radicals with different indices you drive the radicals to one common index, i.e.  least common index, as per this:

\frac{ \sqrt[n]{x} }{ \sqrt[m]{x} } =  \frac{ \sqrt[m.n]{x.x^m} }{ \sqrt[m.n]{y.y^n} } = \sqrt[m.n]{ \frac{x^{m+1}}{y^{n+1}} }

As, you see it is proved that two any quotient of radicals can be rewriten using the quotient property.
</span>
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5 0
2 years ago
According to a survey, 10% of Americans are afraid to fly. Suppose 1100 Americans are sampled. a) What is the probability that 1
JulijaS [17]

Answer:

a) 14.46%

b) 0.00%

c) 1.54%

Step-by-step explanation:

According to the survey, 10% of Americans are afraid to fly.

This means p=0.10 and q=1-0.10=0.90.

If 1100 Americans are sampled, then the sample size is n=1100 .

The mean of the distribution is \mu=np.

This means \mu=1100\times 0.10=110

The standard deviation is \sigma=\sqrt{npq}

We substitute the values to get:

\sigma=\sqrt{1100\times 0.1\times 0.9}=9.95

a) We want to find the probability that 121 or more Americans in the survey are afraid to fly.

We first apply the continuity correction factor to get: P(X\ge121)=P(X\:>\:121-0.5)\\P(X\ge121)=P(X\:>\:120.5)

We now convert to Z-scores to get:

P(X\:>\:120.5)=P(z\:>\:\frac{120.5-110}{9.95})=1.06\\

From the standard normal distribution table P(z>1.06)=0.1446

As a percentage, the probability is 14.46%

b) We want to find the probability that 165 or more Americans in the survey are afraid to fly.

We apply the CCF to get:

P(X\ge165)=P(X\:>\:165-0.5)\\P(X\ge165)=P(X\:>\:164.5)

We convert to z-scores:

P(X\:>\:164.5)=P(z\:>\:\frac{164.5-110}{9.95})=5.48

From the normal distribution, P(z>164.5)=0

c)  First, 8% of 1100 is 88.

We want to find the probability that 88 or less Americans in the survey are afraid to fly.

We apply the CCF to get:

P(X\le88)=P(X\:

We convert to z-scores:

P(X\:

From the normal distribution, P(z>\:-2.16)=0.0154

As a percentage, we get 1.54%

3 0
2 years ago
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