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svetlana [45]
2 years ago
10

Which of the following represents fifth root of x to the fourth power in exponential form? (6 points) a 4x5 b 5x4 c x to the fiv

e fourths power d x to the four fifths power
Mathematics
2 answers:
nirvana33 [79]2 years ago
4 0

Answer:

x^{\frac{4}{5} }

Step-by-step explanation:

Using the rule of radicals/ exponents

\sqrt[n]{x^{m} } ⇔ x^{\frac{m}{n} } , thus

\sqrt[5]{x^{4} } = x^{\frac{4}{5} }

sleet_krkn [62]2 years ago
3 0

Answer:

B) 5x4

one its not A nor its C

its either B or D but i would go with B because D doesn't make sense all because D is saying x4^5 but its 5x^4

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a shopping channel shipped 467,023 packaged last year what is a reasonable estimate of the number of packages shipped explain
WITCHER [35]
467,000 because if you round higher to 468,000 it would be to far from the original number and 467,000 is closer.

Hope that made sence!
8 0
3 years ago
B a C Which of these angles are adjacent angles?​
BigorU [14]

Answer: a and c

Step-by-step explanation:

I know it’s not b but isn’t adjacent angel two angels?

Correct me if I’m wrong

3 0
2 years ago
Exercise 11.21) Many smartphones, especially those of the LTE-enabled persuasion, have earned a bad rap for exceptionally bad ba
stiv31 [10]

Answer:

Step-by-step explanation:

Hello!

The researcher suspects that the battery life between charges for the Motorola Droid Razr Max differs if its primary use is talking or if its primary use is for internet applications.

Since the means for talk time usage (20hs) is greater than the mean for internet usage (7hs) the main question is if the variance in hours of usage is also greater when the primary use is talk time.

Be:

X₁: Battery duration between charges when the primary usage of the phone is talking. (hs)

n₁= 12

X[bar]₁= 20.50 hs

S₁²= 199.76hs² (S₁= 14.13hs)

X₂: Battery duration between charges when the primary usage of the phone is internet applications.

n₂= 10

X[bar]₂= 8.50

S₂²= 33.29hs² (S₂= 5.77hs)

Assuming that both variables have a normal distribution X₁~N(μ₁;σ₁²) and X₂~N(μ₂; σ₂²)

The parameters of interest are σ₁² and σ₂²

a) They want to test if the population variance of the duration time of the battery when the primary usage is for talking is greater than the population variance of the duration time of the battery when the primary usage is for internet applications. Symbolically: σ₁² > σ₂² or since the test to do is a variance ratio: σ₁²/σ₂² > 1

The hypotheses are:

H₀: σ₁²/σ₂² ≤ 1

H₁: σ₁²/σ₂² > 1

There is no level of significance listed so I've chosen α: 0.05

b) I've already calculated the sample standard deviations using a software, just in case I'll show you how to calculate them by hand:

S²= \frac{1}{n-1}*[∑X²-(∑X)²/n]

For the first sample:

n₁= 12; ∑X₁= 246; ∑X₁²= 7240.36

S₁²= \frac{1}{11}*[7240.36-(246)²/12]= 199.76hs²

S₁=√S₁²=√199.76= 14.1336 ≅ 14.13hs

For the second sample:

n₂= 10; ∑X₂= 85; ∑X₂²= 1022.12

S₂²= \frac{1}{9}*[1022.12-(85)²/10]= 33.2911hs²

S₂=√S₂²=√33.2911= 5.7698 ≅ 5.77hs

c)

For this hypothesis test, the statistic to use is a Snedecors F:

F= \frac{S_1^2}{S_2^2} *\frac{Sigma_1^2}{Sigma_2^2} ~~F_{n_1-1;n_2-1}

This test is one-tailed right, wich means that you'll reject the null hypothesis to big values of F:

F_{n_1-1;n_2-1; 1-\alpha }= F_{11;9;0.95}= 3.10

The rejection region is then F ≥ 3.10

F_{H_0}= \frac{199.76}{33.29} * 1= 6.0006

p-value: 0.006

Considering that the p-value is less than the level of significance, the decision is to reject the null hypothesis.

Then at a 5% level, there is significant evidence to conclude that the population variance of the duration time of the batteries of Motorola Droid Razr Max smartphones used primary for talk is greater than the population variance of the duration time of the batteries of Motorola Droid Razr Max smartphones used primary for internet applications.

I hope this helps!

8 0
2 years ago
During the early 20th century, in what way was the radio better than the phonograph in encouraging the concept of crosspollinati
ololo11 [35]

C. Phonograph were inferior in sound quality.

Step-by-step explanation:

Researches have shown that plants respond to the sounds of their pollinator.

Some plants begin to produce a greater amount of nectar and some begin to vibrate to attract the attention of the pollinators towards them.

Greater the intensity of the sound of the pollinator, the faster the reaction will be.

Phonographs have a sound of lower intensity when compared to the intensity of sound produced by the radio.

Thus the radio was better than the phonograph in encouraging the concept of cross-pollination.

6 0
2 years ago
Given the sequence rn defined recursively below, find r3. r1rn=2=−rn−1+n−2
kifflom [539]

Answer:

r_3=3

Step-by-step explanation:

Given:

r_1=2\\r_n=-r_{n-1}+n-2

We want to find the value of r_3 .

\\r_2=-r_{2-1}+2-2\\r_2=-r_1\\r_2=-2\\\\r_3=-r_{3-1}+3-2\\r_3=-r_{2}+1\\r_3=-(-2)+1\\r_3=2+1\\r_3=3

7 0
3 years ago
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