All categories

3 years ago

Which is larger .2 or .04

Mathematics

2 answers:

sertanlavr [38]3 years ago

3 0

The decimal 0.2 is larger than 0.04.

BARSIC [14]3 years ago

3 0

.2 is larger because it is in the tens place decimal while the four is in the hundreds place.

You might be interested in

Point j (-8 -12) is reflected over the x axis where is j

stiv31 [10]

Answer:

J(0,2)

Step-by-step explanation:

When you reflect a point across the x-axis, the x-coordinate remains the same, but the y-coordinate is transformed into its opposite (its sign is changed). If you forget the rules for reflections when graphing, simply fold your paper along the x-axis (the line of reflection) to see where the new figure will be located.

4 0

3 years ago

The critical points of a rational inequality are -1, 3, and 4. Which set of points can be tested to find a complete solution

mars1129 [50]

Answer:

It would be D. I got it correct on Edgenuity.

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

Help answer and EXPLAIN STEP BY STEP

pychu [463]

15. $(-k^{3} -5k^{2} + 1) - (6k^{3}-k^{2} -2)$

Step 1: Distribute the subtraction sign to all the numbers in the second set of parentheses...

$(-k^{3} -5k^{2} + 1) - (6k^{3}-k^{2} -2)$

$-k^{3} -5k^{2} +1 - 6k^{3} +k^{2} +2$

Step 2: Combine like terms ( $k^{2}$ 's go with $k^{2}$ 's)

$-k^{3} -5k^{2} + 1 - 6k^{3}+k^{2} +2$

$-5k^{2} +k^{2}$

$-4k^{2}$

$-k^{3} -4k^{2} + 1 - 6k^{3}+2$

Step 3: Combine like terms ( $k^{3}$ 's go with $k^{3}$ 's)

$-k^{3} -4k^{2} + 1 - 6k^{3}+2$

$-k^{3} + (- 6k^{3})$

$-7k^{3}$

$-7k^{3} -4k^{2} + 1 +2$

Step 4: Combine like terms (normal numbers go with normal numbers)

$-7k^{3} -4k^{2} + 1 +2$

1 + 2

$-7k^{3} -4k^{2} + 3$

You were correct!

16. $49m^2 - 84mn + 36n^2$

I don't really know how to explain this one but I got: (7m-6n)(7m-6n)

7 0

4 years ago

Which fractions are equivalent to 2/8?

Hitman42 [59]

Answer:

1/4 and 3/12

Step-by-step explanation:

3 0

4 years ago

A manager at a local manufacturing company has been monitoring the output of one of the machines used to manufacture chromium sh

denpristay [2]

Answer:

0.682 = 68.2% probability that the average length of these 16 shells will be between 116 and 120 centimeters when the machine is operating "properly".

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

Normally distributed with a mean of 118 centimeters and a standard deviation of 8 centimeters.

This means that $\mu = 118, \sigma = 8$

Sample of 16 shells

This means that $n = 16, s = \frac{8}{\sqrt{16}} = 2$

What is the probability that the average length of these 16 shells will be between 116 and 120 centimeters when the machine is operating "properly?"

This is the pvalue of Z when X = 120 subtracted by the pvalue of Z when X = 116.

X = 120

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{120 - 118}{2}$

$Z = 1$

$Z = 1$ has a pvalue of 0.841

X = 116

$Z = \frac{X - \mu}{s}$

$Z = \frac{116 - 118}{2}$

$Z = -1$

$Z = -1$ has a pvalue of 0.159

0.841 - 0.159 = 0.682

0.682 = 68.2% probability that the average length of these 16 shells will be between 116 and 120 centimeters when the machine is operating "properly".

3 0

3 years ago