All categories

3 years ago

What is 2.85 tons rounded to the nearest hundredth?

2 answers:

6 0

2.85 tons is already rounded to the nearest hundredth since there aren't any numbers after the 5 in the hundredth value.

UkoKoshka [18]3 years ago

5 0

Since the number '5' is located in the hundredths place (the number to the right of the decimal), there is no number next to it.

So, 2.85 stays the same

You might be interested in

3×-18=42 Solve for x!

Keith_Richards [23]

3x - 18 = 42
3x = 42 + 18
3x = 60
x = 60/3
x = 20

7 0

4 years ago

PLEASE IM IN CLASS RIGHT NOWWWW

agasfer [191]

Answer:

102

Step-by-step explanation:

8 0

3 years ago

Chau will run at most 28 miles this week. So far, he has run 18 miles. What are the possible numbers of additional miles he will

AysviL [449]

Answer:

Step-by-step explanation:

18 + x ≤ 28

x ≤ 28 - 18

x ≤ 10

1 ≤ x ≤ 10

3 0

3 years ago

I need help with this..

Misha Larkins [42]

Answer:

a0) 2

x<0

a1) x

0≤x<3

a2) 3

x≥3

Step-by-step explanation:

As shown in the given graph

function of y is a straight line at y=2 line till x=0

hence a0:

y= 2 for x<0

Then function becomes linear line from x=0 till x=3

hence a1:

y= x for 0≤x<3

Now after that graph of function y again shift to straight line from x=3 onward with y-axis value of 3

hence a2:

y= 3 for x≥3 !

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