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sdas [7]
4 years ago
9

What is the product?

Mathematics
2 answers:
anygoal [31]4 years ago
8 0

Answer:

15 r squared minus 14 r minus 8

Step-by-step explanation:

(5r+2)(3r-4)

distribute and you will get

15r^2-20r+6r-8

combine like terms

15r^2-14r-8

if you have any questions just comment. ^^ ill be glad it help!

frutty [35]4 years ago
6 0

Answer:

Option D.

Step-by-step explanation:

The given expression is (5r + 2)(3r minus 4)

Convert this expression into an algebraic expression.

(5r + 2)(3r - 4)

By distribution law,

= 5r(3r - 4) + 2(3r - 4)

= 15r² - 20r + 6r - 8

= 15r² - 14r - 8

Now convert the algebraic expression into the language form.

15r squared minus 14r minus 8

Therefore, option D is the answer.

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Simplify the expression 7×+3(2×-3)-×
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Answer:

12x-9

Step-by-step explanation:

7x+3(2x-3)-x

7x-6x-9-x

13x-9-x

12x-9

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2. Sam is climbing 2/3 meters every 1/3 of an hour. How many meters does
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2/9

Step-by-step explanation:

if u multiply 2/3 × 1/3 u get 2/9

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3 years ago
A scooter is 3 1/2 feet long. Find the length of a scale model of the scooter if the scale is 1 inch = 3/4 feet.
Dmitrij [34]

Answer:

4 2/3

Step-by-step explanation:

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3 years ago
Fernando bought 8 pints of milk. how many fluid ounces of milk Fernando buy?
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Read 2 more answers
A basketball player makes a free throw 82.6% of the time. The player attempts 5 free throws. Use a histogram of the binomial dis
DiKsa [7]

Answer:

The most likely outcome is exactly 4 free throws

Step-by-step explanation:

Given

n = 5 --- attempts

p = 82.6\% ---- probability of a successful free throw

p = 0.826

Required

A histogram to show the most likely outcome

From the question, we understand that the distribution is binomial.

This is represented as:

P(X = x) = ^nC_x * p^x * (1 - p)^{n-x}

For x = 0 to 5, where x represents the number of free throws; we have:

P(X = x) = ^nC_x * p^x * (1 - p)^{n-x}

P(X = 0) = ^5C_0 * 0.826^0 * (1 - 0.826)^{5-0}

P(X = 0) = ^5C_0 * 0.826^0 * (0.174)^{5}

P(X = 0) = 1 * 1 * 0.000159 \approx 0.0002

P(X = 1) = ^5C_1 * 0.826^1 * (1 - 0.826)^{5-1}

P(X = 1) = ^5C_1 * 0.826^1 * (0.174)^4

P(X = 1) = 5 * 0.826 * 0.000917 \approx 0.0038

P(X = 2) = ^5C_2 * 0.826^2 * (1 - 0.826)^{5-2}

P(X = 2) = ^5C_2 * 0.826^2 * (0.174)^{3}

P(X = 2) = 10 * 0.682 * 0.005268 \approx 0.0359

P(X = 3) = ^5C_3 * 0.826^3 * (1 - 0.826)^{5-3}

P(X = 3) = ^5C_3 * 0.826^3 * (0.174)^2

P(X = 3) = 10 * 0.5636 * 0.030276 \approx 0.1706

P(X = 4) = ^5C_4 * 0.826^4 * (1 - 0.826)^{5-4}

P(X = 4) = 5 * 0.826^4 * (0.174)^1

P(X = 4) = 5 * 0.4655 * 0.174 \approx 0.4050

P(X = 5) = ^5C_5 * 0.826^5 * (1 - 0.826)^{5-5}\\

P(X = 5) = ^5C_5 * 0.826^5 * (0.174)^0

P(X = 5) = 1 * 0.3845 * 1 \approx 0.3845

From the above computations, we have:

P(X = 0)  \approx 0.0002

P(X = 1) \approx 0.0038

P(X = 2)  \approx 0.0359

P(X = 3)  \approx 0.1706

P(X = 4)  \approx 0.4050

P(X = 5) \approx 0.3845

See attachment for histogram

<em>From the histogram, we can see that the most likely outcome is at: x = 4</em>

<em>Because it has the longest vertical bar (0.4050 or 40.5%)</em>

3 0
3 years ago
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