All categories

3 years ago

Write this numbers down from the smallest to the largest 0.3, 0.01, 0.38, 0.127, 0.1

Mathematics

1 answer:

LiRa [457]3 years ago

8 0

Step-by-step explanation:

0.001 < 0.1 < 0.127 < 0.3 < 0.38

You might be interested in

What is 2 2/15 - 1 2/3=<br> I ended up with 2 3/21 <br> Really need help with this!

QveST [7]

Answer:

$2 \frac{2}{15} - 1 \frac{2}{3}$

$= \frac{32}{15} - \frac{5}{3}$

$= \frac{32 \times 1 - 5 \times 5}{15}$

$= \frac{32 - 25}{15}$

$= \frac{7}{15}$

5 0

2 years ago

Jarred wants to buy a go-cart for $1,200. His part-time job pays him $160 a week. He has already saved $400. Which inequality re

Oduvanchick [21]

The answer is B I believe!

8 0

2 years ago

Read 2 more answers

Which term describes the slope of the line below?

SashulF [63]

Answer:

B. Zero

Step-by-step explanation:

The slope of the line is in the the first two quadrants and is horizontal, meaning it has zero slope.

7 0

1 year ago

Read 2 more answers

Please help me with this math question will mark brainliest

LenaWriter [7]

Answer:

-9

Step-by-step explanation:

Using the sum/difference property of logarithms, we can rewrite the expression given as:

log b^3 + log c^3 - log √(a^3) --> log √(a^3) can also be written as log a^1.5

Next, we can use the power property of logarithms, and rewrite it again as:

3log b + 3log c - 1.5log a

Now, we can substitute the values of log a, log b, and log c:

3(11) + 3(-9) - 1.5(10)

33 - 27 - 15

-9

Simplifying, we get -9 as the answer.

3 0

2 years ago

Assume the heights in a female population are normally distributed with mean 65.7 inches and standard deviation 3.2 inches. Then

siniylev [52]

Answer:

0.620

Step-by-step explanation:

We know that 1 feet = 12 inches, so, 5 feet is equivalent to 60 inches. Then, we are looking for the probability that a typical female from this population is between 60 inches and 67 inches. We know that

$\mu$ = 65.7 inches and

$\sigma$ = 3.2 inches

and the normal density function for this mean and standard deviation is

$\frac{1}{\sqrt{2\pi } 3.2}exp[-\frac{(x-65.7)^{2}}{2(3.2)^{2}} ]$

The probability we are looking for is given by

$\int\limits^{67}_{60} {\frac{1}{\sqrt{2\pi } 3.2}exp[-\frac{(x-65.7)^{2}}{2(3.2)^{2}} ] } \, dx =0.620$

You can use a computer to calculate this integral. You can use the following instruction in the R statistical programming language

pnorm(67, 65.7, 3.2)-pnorm(60, 65.7, 3.2)

7 0

2 years ago