Ask question

Ask question

All categories

3 years ago

5

Compare.Use <,>,or =. 1,000 ft _____ 300 yd

2 answers:

Airida [17]3 years ago

8 0

1,000 ft > 300 yd
1,000 ft = 3 yards

frosja888 [35]3 years ago

5 0

1000ft > 300 yd

1ft=3yd
1000ft=333.33yd

You might be interested in

PLEASE HELPPP MEE.....

mariarad [96]

Answer:

C ⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

6 0

3 years ago

Answer quickly, please. I need this. I will give brainliest to whoever answers first. You also get 25 points for this

sukhopar [10]

** DISCLAMIER** am not completely sure. Please do not use my answer unless you are very desperate.

since O,R correspond with A,N I think you half F on each side and add it to 11.2 so
10 divided by 2 = 5
7 is already half.
5 + 7 is 12
12 + 11.2 = 23.2

IF THIS IS WRONG TRY 17
reason for 17:
10 + 7 = 17
If you look at both lines they look the same length as A, N.

3 0

3 years ago

Write an expression to match the statement below Add 5 to 48 divided by 12

Sergeeva-Olga [200]

Answer:

49

Step-by-step explanation:

48+ 5 = 53 divided by = 49

8 0

3 years ago

Suppose that an airline uses a seat width of 16.5 in. Assume men have hip breadths that are normally distributed with a mean of

Alexxx [7]

Answer:

a) 0.018

b) 0

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 14.4 in

Standard Deviation, σ = 1 in

We are given that the distribution of breadths is a bell shaped distribution that is a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

a) P(breadth will be greater than 16.5 in)

P(x > 16.5)

$P( x > 16.5) = P( z > \displaystyle\frac{16.5 - 14.4}{1}) = P(z > 2.1)$

$= 1 - P(z \leq 2.1)$

Calculation the value from standard normal z table, we have,

$P(x > 16.5) = 1 - 0.982 = 0.018 = 1.8\%$

0.018 is the probability that if an individual man is randomly selected, his hip breadth will be greater than 16.5 in.

b) P( with 123 randomly selected men, these men have a mean hip breadth greater than 16.5 in)

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\frac{\sigma}{\sqrt{n}}}$

P(x > 16.5)

$P( x > 16.5) = P( z > \displaystyle\frac{16.5-14.4}{\frac{1}{\sqrt{123}}}) = P(z > 23.29)$

$= 1 - P(z \leq 23.29)$

Calculation the value from standard normal z table, we have,

$P(x > 16.5) = 1 - 1 = 0$

There is 0 probability that 123 randomly selected men have a mean hip breadth greater than 16.5 in

4 0

3 years ago

What is the gradient of the graph shown?<br> Give your answer in its simplest form.

Minchanka [31]

Answer:

m=3/2

Step-by-step explanation:

Two points are shown

point 1: (0;-3). 4;3. -2-6

point 2: (2;0)

$m = \frac{y2 - y1}{x2 - x1}$

$m = \frac{0 - ( - 3)}{2 - 0}$

$m = \frac{3}{2}$

6 0

3 years ago