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3 years ago

Possible side lengths for a rectangular fence are 8 feet, 10 feet, and 12 feet. Marsha

Mathematics

1 answer:

andriy [413]3 years ago

8 0

8ft by 12ft and 10ft by10ft

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A line with a slope of 2 passes through (3,9). Which choice is an equation of this line?

Veronika [31]

Answer:

The answer to your question is number 3.

Step-by-step explanation:

Data

slope = m = 2

Point = (3, 9)

Process

1.- Write the equation of the line in the form point-slope

y - y1 = m(x - x1)

2.- Identify the values of x1 and y1

x1 = 3 y1 = 9

3.- Substitution

y - 9 = 2(x - 3)

4.- Simplify

y - 9 = 2x - 6

y = 2x - 6 + 9

5.- Result

y = 2x + 3 slope y-intercept form

6 0

3 years ago

What is the volume of this pyramid ,rounded to the nearest tenth ?

prohojiy [21]

It is a) 6.5 because the formula is lwh/3 and the length and width are both 2 and the height is rt24 because 25-1 is 24 (Pythagorean Theorem) and so when you solve, you get about 6.53 and that rounds to 6.5

8 0

4 years ago

(02.07 MC)

hjlf

The answer is 1.64 feet. 1.64 x 12 = 19.68 inches

19.68 x 2.54 cm= 49.9872 cm

7 0

3 years ago

Hii please help i’ll give brainly if you give a correct answer <br> - please show your work.

eduard

Step-by-step explanation:

1. In the picture down below.

2. Because they are = 20

100 ÷ 5 = 20

10 ÷ 0.5 = 10 x 2 = 20

1 ÷ 0.05 = 1 x 20 = 20

3 0

3 years ago

For a certain population of sea turtles, 18 percent are longer than 6.5 feet. A random sample of 90 sea turtles will be selected

nika2105 [10]

Answer:

The standard deviation of the sampling distribution of the sample proportion of sea turtles longer than 6.5 feet for samples of size 90 is of 0.0405

Step-by-step explanation:

Central Limit Theorem:

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean $\mu = p$ and standard deviation $s = \sqrt{\frac{p(1-p)}{n}}$

18 percent are longer than 6.5 feet.

This means that $p = 0.18$

A random sample of 90 sea turtles

This means that $n = 90$

What is the standard deviation of the sampling distribution of the sample proportion of sea turtles longer than 6.5 feet for samples of size 90

$s = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.18*0.82}{90}} = 0.0405$

The standard deviation of the sampling distribution of the sample proportion of sea turtles longer than 6.5 feet for samples of size 90 is of 0.0405

8 0

3 years ago