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3 years ago

Is it possible for a 5 digit number to be rounded to a 6 digit number

Mathematics

2 answers:

Alexandra [31]3 years ago

6 0

No because no matter what you round it to you will always be multiplying the 5 digit number by itself that many times

IrinaK [193]3 years ago

5 0

When we round a number the new rounded number is simpler but the value is kept close to what it was. The common method for rounding numbers is the following: we decide which is the last digit to keep and leave this digit the same the same if the next digit is less than 5 or increase it by 1 if the next digit is bigger than 5.
Because the next digit matter, it is not possible for a 5 digit number to be rounded to 6 digit number.

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Work out (5 x 10^3) x (9 x 10^7)

Sveta_85 [38]

Step-by-step explanation:

5×10³) × (9×10^7)

= 45 × 10^10

= 4.5 × 10¹¹

6 0

3 years ago

Read 2 more answers

A warehouse distributor of carpet keeps 6,000 yards of deluxe shag carpet in stock during a month. The average demand for carpet

Bond [772]

Answer:

The probability that a customer’s order will not be met during a month is 1.67.

Step-by-step explanation:

Given : A warehouse distributor of carpet keeps 6,000 yards of deluxe shag carpet in stock during a month. The average demand for carpet from the stores that purchase from the distributor is 4,500 yards per month, with a standard deviation of 900 yards.

To find : What is the probability that a customer’s order will not be met during a month?

Solution :

Average Mean $\mu=4500$

Sample Mean $x=6000$

Standard deviation $\sigma=900$

The formula is given by,

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

Substitute the value in the formula,

$Z=\frac{6000-4500}{900}$

$Z=\frac{1500}{900}$

$Z=1.67$

The probability that a customer’s order will not be met during a month is 1.67.

7 0

3 years ago

What is the measure of each exterior angle of a decagon?

NemiM [27]

The sum of all of the exterior angles of any regular polygon is 360 (I'm assuming by decagon you mean a regular dodecagon).

Because the total has to be 360, and a decagon has 10 sides, you divide 360 by 10 to get each exterior angle.

360 ÷ 10 = 36, so each exterior angle of a decagon is 36 degrees.

3 0

4 years ago

Complete the problems. Show your work.

timama [110]

1. =4.8
2. =40%
3. = 160

8 0

3 years ago

Ocean fishing for billfish is very popular in the Cozumel region of Mexico. In the Cozumel region about 41% of strikes (while tr

Klio2033 [76]

Answer:

a) 59.10% probability that 12 or fewer fish were caught.

b) 99.74% probability that 5 or more fish were caught.

c) 58.84% probability that between 5 and 12 fish were caught.

Step-by-step explanation:

I am going to use the normal approximation to the binomial to solve this question.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

$E(X) = np$

The standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore 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 pvalue, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that $\mu = E(X)$ , $\sigma = \sqrt{V(X)}$ .

In this problem, we have that:

$n = 29, p = 0.41$

$\mu = E(X) = np = 29*0.41 = 11.89$

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = 2.6486$

Find the following probabilities.

a) 12 or fewer fish were caught.

Using continuity correction, this is $P(X \leq 12 + 0.5) = P(X \leq 12.5)$ , which is the pvalue of Z when X = 12.5. So

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

$Z = \frac{12.5 - 11.89}{2.6486}$

$Z = 0.23$

$Z = 0.23$ has a pvalue of 0.5910

59.10% probability that 12 or fewer fish were caught.

b) 5 or more fish were caught.

Using continuity correction, this is $P(X \geq 5 - 0.5) = P(X \geq 4.5)$ , which is 1 subtracted by the pvalue of Z when X = 4.5. So

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

$Z = \frac{4.5 - 11.89}{2.6486}$

$Z = -2.79$

$Z = -2.79$ has a pvalue of 0.0026

1 - 0.0026 = 0.9974

99.74% probability that 5 or more fish were caught.

c) between 5 and 12 fish were caught.

Using continuity correction, this is $P(5 - 0.5 \leq X \leq 12 + 0.5) = P(4.5 \leq X \leq 12.5)$ , which is the pvalue of Z when X = 12.5 subtracted by the pvalue of Z when X = 4.5. So.

From a), when X = 12.5, Z has a pvalue of 0.5910

From b), when X = 4.5, Z has a pvalue of 0.0026.

0.5910 - 0.0026 = 0.5884

58.84% probability that between 5 and 12 fish were caught.

8 0

4 years ago