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Taya2010 [7]
3 years ago
5

Simplify. Your answer should contain only positive exponents −ab^2•(2a^0b^-4)^-4•2a^4b^-1

Mathematics
1 answer:
nordsb [41]3 years ago
6 0

Answer:

-a^5b^17/8

Step-by-step explanation:

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Please please help me out!!
Roman55 [17]

Answer:

And step-by-step explanation

7 0
4 years ago
Read 2 more answers
Use the identity a^3+b^3=(a+b)^3−3ab(a+b) to determine the sum of the cubes of two numbers if the sum of the two numbers is 4 an
Butoxors [25]

a^3+b^3=(a+b)^3-3ab(a+b)\\\\a+b=4,\ ab=1\\\\\text{Substitute:}\\\\a^3+b^3=(4)^3-3(1)(4)=64-12=52

5 0
3 years ago
Which of the sets of ordered pairs represents a function? (4 points) A = {(−4, 5), (1, −1), (2, −2), (2, 3)} B = {(2, 2), (3, −2
elena55 [62]
A function will not have any repeating x values....it can have repeating y values, just not the x ones

(-4,5),(-1,1),(2,-2),(2,3)
this is not a function because it has 2 sets of points that has x as 2...so it has repeating x values

(2,2),(3,-2),(9,3),(9,-3)
this is not a function because it has 2 sets of points that has an x value of 9...so it also has repeating x values

so both of these are not functions.....neither of them
6 0
3 years ago
What is the area of a 320 ft semicircle?
Sever21 [200]

Answer: they do not have enough info but I will try my best to answer  if im wrong its not my fault  so please add more info

Solution

2. I know that for a Semi-Circle Area = 320 . From this find out Area of the Semi-Circle.

Solution:

Here Area=320(Given)

Area=

πr2

2

r2=

Area⋅2

π

r2=

320⋅2

22

7

r=14.2701

Diameter(d)=2r

=2⋅14.2701

=28.5402

Circumference=πr

=

22

7

⋅14.2701

=44.849

Perimeter=πr+2r

=

22

7

⋅14.2701+2⋅14.2701

=44.849+28.5402

=89.6979

Step-by-step explanation:

Byee stay safe: Wear your mask, drink water and stan bts

also can I have brainliest

4 0
2 years ago
The alkalinity level of water specimens collected from the Han River in Seoul, Korea, has a mean of 50 milligrams per liter and
Sati [7]

Answer:

a) 94.06% probability that a water specimen collected from the river has an alkalinity level exceeding 45 milligrams per liter.

b) 94.06% probability that a water specimen collected from the river has an alkalinity level below 55 milligrams per liter.

c) 50.98% probability that a water specimen collected from the river has an alkalinity level between 48 and 52 milligrams per liter.

Step-by-step explanation:

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.

In this problem, we have that:

\mu = 50, \sigma = 3.2

a. exceeding 45 milligrams per liter.

This probability is 1 subtracted by the pvalue of Z when X = 45. So

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

Z = \frac{45 - 50}{3.2}

Z = -1.56

Z = -1.56 has a pvalue of 0.0594.

1 - 0.0594 = 0.9406

94.06% probability that a water specimen collected from the river has an alkalinity level exceeding 45 milligrams per liter.

b. below 55 milligrams per liter.

This probability is the pvalue of Z when X = 55.

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

Z = \frac{55 - 50}{3.2}

Z = 1.56

Z = 1.56 has a pvalue of 0.9604.

94.06% probability that a water specimen collected from the river has an alkalinity level below 55 milligrams per liter.

c. between 48 and 52 milligrams per liter.

This is the pvalue of Z when X = 52 subtracted by the pvalue of Z when X = 48. So

X = 52

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

Z = \frac{52 - 50}{3.2}

Z = 0.69

Z = 0.69 has a pvalue of 0.7549

X = 48

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

Z = \frac{48 - 50}{3.2}

Z = -0.69

Z = -0.69 has a pvalue of 0.2451

0.7549 - 0.2451 = 0.5098

50.98% probability that a water specimen collected from the river has an alkalinity level between 48 and 52 milligrams per liter.

4 0
3 years ago
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