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3 years ago

7

A moving walkway at an airport moves at a pace of 1.751.75 feet per second. If you stand on the walkway as it moves, how long wi

ll it take to transport you 280280 feet?

1 answer:

NeX [460]3 years ago

8 0

160 seconds. should be the answer if your numbers gave are correct.

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17:3 and 15:60 Equivalent ratios

inessss [21]

Answer:

Step-by-step explanation:

17:3 is already in lowest terms. 17:3 = 34:6, 51:9, etc.

15:60 = 1:4

5 0

2 years ago

I need help Plzz & Thank You .

Afina-wow [57]

${x50506}^{?}$

6 0

3 years ago

Which of the following (-1,8) and (5,-2) are the distance of the line segment

olganol [36]

The formula of a distance between two points:

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

We have the points (-1, 8) and (5, -2). Substitute:

$d=\sqrt{(5-(-1))^2+(-2-8)^2}=\sqrt{6^2+(-10)^2}=\sqrt{36+100}=\sqrt{136}\\\\=\sqrt{4\cdot34}=\sqrt4\cdot\sqrt{34}=\boxed{2\sqrt{34}}$

8 0

3 years ago

If the probability of an outcome is 2/9, what are the odds against that outcome?

IrinaVladis [17]

Answer:

7/9

Step-by-step explanation:

If the chances of an outcome are 2/9, the odds against it can be found by subtracting that from 1.

1 - 2/9

= 7/9

So, the odds against that outcome are 7/9

7 0

2 years ago

If you know please help me

MrRa [10]

Average rate of change of the function $=\frac{75}{2}$

Solution:

Given function: $f(x)=5(2)^{x}$ from x = 1 to x = 5

Substitute x = 1 and x = 5 in f(x).

$f(1)=5(2)^{1}=10$

$f(5)=5(2)^{5}=160$

Let us find the average rate of change of the function.

Average rate of change

$$=\frac{f(b)-f(a)}{b-a}$

Here a = 1 and b = 5.

$$=\frac{f(5)-f(1)}{5-1}$

Substitute f(5) and f(1).

$$=\frac{160-10}{4}$

$$=\frac{150}{4}$

$$=\frac{75}{2}$

Average rate of change of the function $=\frac{75}{2}$

4 0

3 years ago