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3 years ago

What is the simplified value of the exponential expression 16^1/4 1/2 1/4 2 4

Mathematics

2 answers:

Kruka [31]3 years ago

7 0

Answer:

Step-by-step explanation:

16 ^ (1/4)

Rewriting 16 as 2^4

2^4 ^ 1/4

We know that a^b^c = a^(b*c)

2 ^ (4*1/4)

2 ^1

sergejj [24]3 years ago

5 0

The right answer is 2.

please see the attached picture for full solution

Hope it helps

Good luck on your assignment

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Find the quotient: 3)21 A. 6 B. 7 C. 8 D. 9

finlep [7]

The answer is B:7!!!!!!!

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3 years ago

A cone is placed inside a cylinder as shown. The radius of the cone is half the radius of the cylinder. The height of the cone i

Dovator [93]

we can use

volume of cone formula

$V=\frac{1}{3} \pi r'^2 h'$

where

r' is radius of cone

h' is height of cone

we are given

The radius of the cone is half the radius of the cylinder

so,

$r'=\frac{r}{2}$

The height of the cone is equal to the radius of the cylinder

so,

$h'=r$

now, we can plug values into formula

we get

$V=\frac{1}{3} \pi (\frac{r}{2})^2*r$

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so, option-C.........Answer

7 0

3 years ago

there are students in a school. This number is 15% more than it was last year. Calculate the number of students last year

ANTONII [103]

Answer: See explanation

Step-by-step explanation:

Your question isn't complete but let's help out by giving some values to the question.

Let's say there are 230 students in a school. This number is 15% more than it was last year. Calculate the number of students last year.

Let the number of students last year be x.

Since there's a 15% increase, this implies that (100% + 15%) = 115% of x equals to 230. This will be:

115% of x = 230

115% × x = 230

115/100 × x = 230

1.15x = 230

x = 230/1.15

x = 200

There were 200 students last year.

Just assign the missing values and use the above method and you'll get your answer.

8 0

3 years ago

Time spent using e-mail per session is normally distributed, with mu equals 11 minutes and sigma equals 3 minutes. Assume that

liq [111]

Answer:

a) 0.259

b) 0.297

c) 0.497

Step-by-step explanation:

To solve this problem, it is important to know the normal probability distribution and the central limit theorem.

Normal probability distribution

Problems of normally distributed samples are 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.

Central limit theorem

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$ , a large sample size can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$