All categories

3 years ago

What is 4.35 rounded to the nearest hundred

1 answer:

5 0

Any time you are rounding anything below 4 becomes a zero anything above a 5 is like a ten
ex.
Q. round to the nearest hundredth 3.24
A. 3.20

Q. round to the nearest hundrerg4 3.25
3.30

which means your answer is
4.40

You might be interested in

Solve for x. 5x/−6 = −40/3 Enter your answer in the box

eduard

Answer:

$x = 16$

That's your answer

8 0

2 years ago

Read 2 more answers

Good night lol what ever..........

frozen [14]

Answer:

night.....

Step-by-step explanation:

..............

4 0

2 years ago

A solid silver paperweight in the form of a square pyramid is shown below. If silver costs $0.12 per cubic centimeter, how much

Rufina [12.5K]

Answer:

It would cost $232.32, this answer is rounded.

Step-by-step explanation:

The formula for the volume of a square pyramid is:

$V=a^2\frac{h}{3}$

So its just:

$8.8^2\frac{12.5}{3} = 322.67 cm^3$

Then you just multiply the volume by the cost per cubic centimetre:

$322.67cm^3*$0.12=$38.704$

After just multiply the cost of one by how many you want, which is 6:

$38/704*6=232.32$

4 0

3 years ago

1.94x - 10/8x + 4.32 = [2(1.15x+8.5)]

kipiarov [429]

$1,94x-\frac{10}{8}x+4,32=[2(1,15x+8,5)]\\\\
1,94x-1,25x+4,32=2,3x+17\\\\
0,69x+4,32=2,3x+17\ \ \ |Subtract\ 2,3x\\\\
-1,61x+4,32=17\ \ \ |Subtract\ 4,32\\\\
-1,61x=12,68\ \ \ |Divide\ by\ -1,61\\\\\boxed{x\approx7,9}
$

8 0

3 years ago

A random sample of 28 statistics tutorials was selected from the past 5 years and the percentage of students absent from each on

SVEN [57.7K]

Answer:

Step-by-step explanation:

Hello!

X: number of absences per tutorial per student over the past 5 years(percentage)

X≈N(μ;σ²)

You have to construct a 90% to estimate the population mean of the percentage of absences per tutorial of the students over the past 5 years.

The formula for the CI is:

X[bar] ± $Z_{1-\alpha /2}$ * $\frac{S}{\sqrt{n} }$

⇒ The population standard deviation is unknown and since the distribution is approximate, I'll use the estimation of the standard deviation in place of the population parameter.

Number of Absences 13.9 16.4 12.3 13.2 8.4 4.4 10.3 8.8 4.8 10.9 15.9 9.7 4.5 11.5 5.7 10.8 9.7 8.2 10.3 12.2 10.6 16.2 15.2 1.7 11.7 11.9 10.0 12.4

X[bar]= 10.41

S= 3.71

$Z_{1-\alpha /2}= Z_{0.95}= 1.645$

[10.41±1.645* $(\frac{3.71}{\sqrt{28} } )$ ]

[9.26; 11.56]

Using a confidence level of 90% you'd expect that the interval [9.26; 11.56]% contains the value of the population mean of the percentage of absences per tutorial of the students over the past 5 years.

I hope this helps!

7 0

3 years ago