All categories

3 years ago

What is the following product

Mathematics

1 answer:

Genrish500 [490]3 years ago

4 0

For this case we must find the product of the following expression:

$\sqrt [3] {5} * \sqrt {2}$

By definition of properties of powers and roots we have:

$\sqrt [n] {a ^ m} = a ^ {\frac {m} {n}}$

We rewrite the expression using the lowest common index of 6, then:

$5 ^ {\frac {1} {3}} * 2 ^ {\frac {1} {2}} =$

We rewrite the terms in an equivalent way:

$5 ^ {\frac {2} {6}} * 2 ^ {\frac {3} {6}} =$

We rewrite the expression using the property mentioned:

$\sqrt [6] {5 ^ 2} * \sqrt [6] {2 ^ 3} =$

We combine using the product rule for radicals:

$\sqrt [n] {a} * \sqrt [n] {b} = \sqrt [n] {ab}$

So:

$\sqrt [6] {5 ^ 2 * 2 ^ 3} =\\\sqrt [6] {25 * 8} =\\\sqrt[6]{200}$

ANswer:

Option b

You might be interested in

I have a friend in need. Her name is Sienna and she has cancer along with co.vid. Please pray for her well being because she rea

velikii [3]

Answer:

She'll be in my prayers

Step-by-step explanation:

When two or three are gathered in His name

3 0

3 years ago

Jane wants to estimate the proportion of students on her campus who eat cauliflower. After surveying 24 students, she finds 2 w

irina1246 [14]

Answer:

A 95% confidence interval for the proportion of students who eat cauliflower on Jane's campus is [0.012, 0.270].

Step-by-step explanation:

We are given that Jane wants to estimate the proportion of students on her campus who eat cauliflower. After surveying 24 students, she finds 2 who eat cauliflower.

Firstly, the pivotal quantity for finding the confidence interval for the population proportion is given by;

P.Q. = $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ~ N(0,1)

where, $\hat p$ = sample proportion of students who eat cauliflower

n = sample of students

p = population proportion of students who eat cauliflower

Here for constructing a 95% confidence interval we have used a One-sample z-test for proportions.

So, 95% confidence interval for the population proportion, p is ;

P(-1.96 < N(0,1) < 1.96) = 0.95 {As the critical value of z at 2.5% level

of significance are -1.96 & 1.96}

P(-1.96 < $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < 1.96) = 0.95

P( $-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < ${\hat p-p}$ < $1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.95

P( $\hat p-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < p < $\hat p+1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.95

Now, in Agresti and Coull's method; the sample size and the sample proportion is calculated as;

$n = n + Z^{2}__(\frac{_\alpha}{2})$

n = $24 + 1.96^{2}$ = 27.842

$\hat p = \frac{x+\frac{Z^{2}__(\frac{\alpha}{2}_) }{2} }{n}$ = $\hat p = \frac{2+\frac{1.96^{2} }{2} }{27.842}$ = 0.141

95% confidence interval for p = [ $\hat p-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ , $\hat p+1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ]

= [ $0.141 -1.96 \times {\sqrt{\frac{0.141(1-0.141)}{27.842} } }$ , $0.141 +1.96 \times {\sqrt{\frac{0.141(1-0.141)}{27.842} } }$ ]

= [0.012, 0.270]

Therefore, a 95% confidence interval for the proportion of students who eat cauliflower on Jane's campus [0.012, 0.270].

The interpretation of the above confidence interval is that we are 95% confident that the proportion of students who eat cauliflower on Jane's campus is between 0.012 and 0.270.

7 0

3 years ago

Plz help i will report if your not trying to answer

skelet666 [1.2K]

1) 1 x 10^4
2) 1 x 10^0

5 0

3 years ago

Larry and Carol are both members of a population, and a simple random sample is being conducted. If the chance of Larry being se

blagie [28]

It is D. 1/800 as she is just as likely to be picked as Larry

3 0

4 years ago