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3 years ago

What is the difference between exponent form and expanded form (repeated multiplication)?

1 answer:

5 0

Expanded FormExpanded form refers to a base and an exponent written as repeated multiplication. ExponentExponents are used to describe the number of times that a term is multiplied by itself. ExpressionAn expression is a mathematical phrase containing variables, operations and/or numbers.

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Is the function y = 6 + 9 linear or nonlinear?

Luba_88 [7]

Answer:

Linear

Step-by-step explanation:

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3 years ago

Find the slope of the line

Serhud [2]

Answer:

Slope is undefined.

Step-by-step explanation:

Slope of 0 is a horizontal line.

Slope of undefined is a vertical line.

Any other slope is something in between.

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3 years ago

Read 2 more answers

There are two glass 1 of them with salt solution and one with sugar solution

Brilliant_brown [7]

Answer:

how long question i did not found any answer

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2 years ago

PLLLZZ EXPLAIN HOW YOU GOT IT! I will give you branilest!

Nezavi [6.7K]

Answer:

Getting through ratio method

100-20=80 percent have to pay

18:80

x:100

cross multiply

80 x x=18 x 100

80x=1800

x=1800/80

x=22.5

So the original price was $22.50

Step-by-step explanation:

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2 years ago

The computers of nine engineers at a certain company are to be replaced. Four of the engineers have selected laptops and the oth

Gala2k [10]

Answer:

(a) There are 70 different ways set up 4 computers out of 8.

(b) The probability that exactly three of the selected computers are desktops is 0.305.

Step-by-step explanation:

Of the 9 new computers 4 are laptops and 5 are desktop.

Let X = a laptop is selected and Y = a desktop is selected.

The probability of selecting a laptop is = $P(Laptop) = p_{X} = \frac{4}{9}$

The probability of selecting a desktop is = $P(Desktop) = p_{Y} = \frac{5}{9}$

Then both X and Y follows Binomial distribution.

$X\sim Bin(9, \frac{4}{9})\\ Y\sim Bin(9, \frac{5}{9})$

The probability function of a binomial distribution is:

$P(U=k)={n\choose k}\times(p)^{k}\times (1-p)^{n-k}$

(a)

Combination is used to determine the number of ways to select k objects from n distinct objects without replacement.

It is denotes as: ${n\choose k}=\frac{n!}{k!(n-k)!}$

In this case 4 computers are to selected of 8 to be set up. Since there cannot be replacement, i.e. we cannot set up one computer twice or thrice, use combinations to determine the number of ways to set up 4 computers of 8.

The number of ways to set up 4 computers of 8 is:

${8\choose 4}=\frac{8!}{4!(8-4)!}\\=\frac{8!}{4!\times 4!} \\=70$

Thus, there are 70 different ways set up 4 computers out of 8.

(b)

It is provided that 4 computers are randomly selected.

Compute the probability that exactly 3 of the 4 computers selected are desktops as follows:

$P(Y=3)={4\choose 3}\times(\frac{5}{9})^{3}\times (1-\frac{5}{9})^{4-3}\\=4\times\frac{125}{729}\times\frac{4}{9}\\ =0.304832\\\approx0.305$

Thus, the probability that exactly three of the selected computers are desktops is 0.305.

(c)

Compute the probability that of the 4 computers selected at least 3 are desktops as follows:

$P(Y\geq 3)=1-P(Y$

Thus, the probability that at least three of the selected computers are desktops is 0.401.

6 0

3 years ago