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emmainna [20.7K]
2 years ago
6

Which expression is equivalent to

Mathematics
1 answer:
Scilla [17]2 years ago
6 0

The equivalent expression of \frac{(5ab)^3}{30ab^{-7}} is \frac{25a^{2}b^{10}}{6}

<h3>How to determine the equivalent expression?</h3>

The expression is given as:

\frac{(5ab)^3}{30ab^{-7}}

Expand the numerator

\frac{(5ab)^3}{30ab^{-7}} =\frac{125a^3b^3}{30ab^{-7}}

Divide 125 and 30 by 5

\frac{(5ab)^3}{30ab^{-7}} =\frac{25a^3b^3}{6ab^{-7}}

Apply the law of indices

\frac{(5ab)^3}{30ab^{-7}} =\frac{25a^{3-1}b^{3+7}}{6}

Evaluate the exponent

\frac{(5ab)^3}{30ab^{-7}} =\frac{25a^{2}b^{10}}{6}

Hence, the equivalent expression of \frac{(5ab)^3}{30ab^{-7}} is \frac{25a^{2}b^{10}}{6}

Read more about equivalent expression at:

brainly.com/question/2972832

#SPJ1

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Can somebody give an example of how to identify a polynomial
Nikitich [7]
It has to obey all the polynomial rules, like 3x^4 or 6x, it has to have the base then the power.
4 0
3 years ago
If AM = x + 8 and MB= 6x – 2, find MB.
IrinaK [193]

Answer:

10

Step-by-step explanation:

Firstly, AM=x+8

or, A=x+8-M......eqn (i)

MB=6x-2

or, B=6x-2-M......eqn (ii)

Now,

from eqn (i) and (ii)

x+8-M=6x-2-M

or, 8+2=6x-x-M+M

or, 10=5x

or, 10÷5=x

therefore, x=2

Again,

MB=6x-2

=6×2-2

=12-2

=10

is it correct

please mark me as BRAINLIEST

8 0
3 years ago
Thirty percent of all customers who enter a department store will make a purchase. Suppose that 6 customers enter the store and
torisob [31]

Answer:

Step-by-step explanation:

Let's use binomial distribution, because we are interested in find the total number of successes in a sequence of n independent experiments.

In probability, a binomial distribution is used in a Bernoulli process. That is, a sequence of experiments of n independent  trials, each asking a dichotomous question, that means it only has two answers. One answer is a success (probability: p) and the other is a failure (probability: 1-p). In this case, every customer who enters in the store is the experiment. If they make a purchase is a success event, taking into account that every purchase decision has to be independent.

P(X=x) = C(n,x)p^{x} (1-p)^{n-x}

x= total number of success

n= total number of experiments

p=probability of success on an indivual trial

C(n,x) = \frac{n!}{x!(n-x)!}

a) x= 5 ------> customers that make a purchase

n= 6 ------> total customers

p=0.3

p(X=5)= C(6,5)(0.3)^{5} (1-0.3)^{6-5} \\p(X=5)=0.102

b) At least 3 customers make a purchase. x ≥ 3

P(X\geq 3)=1-P (X=0)-P(X=1)-P(X=2)\\P(X\geq 3)=1-(C(6,0)(0.3^{0})(0.7)^{6}) - (C(6,1)(0.3^{1})(0.7)^{5}) - (C(6,2)(0.3^{2})(0.7)^{4})\\ P(X\geq 3) = 1-0.11765-0.30253-0.32414\\P(X\geq3)= 0.2557

c) At most 2 customers make a purchase. x ≤ 2

P(X\leq2) = P(X=0) + P(X=1) + P(X=2)\\P(X\leq2) = (C(6,0)(0.3^{0})(0.7)^{6}) + (C(6,1)(0.3^{1})(0.7)^{5}) + (C(6,2)(0.3^{2})(0.7)^{4})\\P(X\leq2) = 0.11765-0.30253-0.32414\\P(X\leq2)=0.7443

d)At least 1 customer makes a purchase. x ≥ 1

P(X\geq 1)=1-P(X=0)\\P(X\geq 1)= 1-C(0,6)(0.3)^{0}(0.7)^{6} \\P(X\geq 1)= 1-0.11765 = 0.8824

e)They must be kind to customers, always willing to help and give information about products clearly and honestly.

Have enough inventory and offer promotions for the purchase of more than one product

Have an agile payment system that avoids rows and delays in purchases.

6 0
4 years ago
Can someone please help me
Anettt [7]

Answer:

6, 12, 21, 10, then 4

Step-by-step explanation:

This is more difficult to answer than you would think, you can't actually see the picture when typing the answer

5 0
3 years ago
Read 2 more answers
How many ways can 5 people be arranged in a line?
Rasek [7]

Answer: 120 ways

Step-by-step explanation: In this problem, we're asked how many ways can 5 people be arranged in a line.

Let's start by drawing 5 blanks to represent the 5 different positions in the line.

Now, we know that 5 different people can fill the spot in the first position. However, once the first position is filled, only 4 people can fill the second spot and once the second spot is filled, only 3 people can fill the third spot and so on. So we have <u>5</u> <u>4</u> <u>3</u> <u>2</u> <u>1</u>.

Now, based on the counting principle, there are 5 x 4 x 3 x 2 x 1 ways for all 5 spots to be filled.

5 x 4 is 20, 20 x 3 is 60, 60 x 2 is 120, and 120 x 1 is 120.

So there are 120 ways for all 5 spots to be filled which means that there are 120 ways that 5 people can be arranged in a line.

I have also shown my work on the whiteboard in the image attached.

7 0
4 years ago
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