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lana [24]
3 years ago
10

What is 91,284 rounded to the nearest hundred thousand

Mathematics
1 answer:
otez555 [7]3 years ago
5 0
If this is asking for the nearest hundred thousand, and not ten thousand, than you would round up to 100,000. 91,284 is close to 100,000.
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52,300,000 in scientific rotation
tatiyna

Answer:

-5.230,0000

Step-by-step explanation:

Your number in decimal form is -52,300,000 . The negative sign remains unchanged. To get to "standard" scientific notation, we move the decimal point so there is only one non-zero digit in front of the decimal point. So, -52,300,000 becomes -5.230,0000

6 0
3 years ago
Read 2 more answers
21. what are the excluded values of the function? y= 5/6x-72
tatiyna

you have to calculate the SID number and then take it upon yourself and divided X it's actual number and then there's a chance and the answer would be c.72

4 0
3 years ago
A cube made of an unknown material has a height of 9 cm. The mass of this cube is 3.645 grams. What is the density of this cube?
ehidna [41]

Answer:

density =  \frac{mass}{volume}

mass = 3.645 g

volume = 9×9×9

= 729 cm³

density = 3.645/729

= 0.005 gcm^-3

Step-by-step explanation:

the volume is 729 cm³ cause the given height is 9cm and it is a cube. cube has equal sides so their lengths are the same.

volume = length × height × width

8 0
3 years ago
Which of the following shows the division problem??
elena-14-01-66 [18.8K]

we are given

\frac{3x^2-4x+9}{x-2}

we can see that

numerator is

3x^2-4x+9

and denominator is

x-2

We always put coefficient of numerator inside box

we can see that coefficient of numerators are 3 , -4 and 9

and in the left side , we put value of x after putting denominator =0

x-2=0

x=2

so, left side will be 2

so, option-D..................Answer

7 0
3 years ago
The distribution of weekly salaries at a large company is right skewed with a mean of $1000 and a standard deviation of $350. Wh
Shtirlitz [24]

Answer:

68.76% probability that the sampling error made in estimating the mean weekly salary for all employees of the company by the mean of a random sample of weekly salaries of 50 employees will be at most $50

Step-by-step explanation:

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

Normal probability distribution:

Problems of normally distributed samples are 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.

Central limit theorem:

The Central Limit Theorem estabilishes that, for a random variable X, with mean \mu and standard deviation \sigma, a large sample size can be approximated to a normal distribution with mean \mu and standard deviation, which is also called standard error s = \frac{\sigma}{\sqrt{n}}

In this problem, we have that:

\mu = 1000, \sigma = 350, n = 50, s = \frac{350}{\sqrt{50}} = 49.5

What is the probability that the sampling error made in estimating the mean weekly salary for all employees of the company by the mean of a random sample of weekly salaries of 50 employees will be at most $50

This is the pvalue of Z when X = 1000+50 = 1050 subtracted by the pvalue of Z when X = 1000-50 = 950. So

X = 1050

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

By the Central Limit Theorem

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

Z = \frac{1050-1000}{49.5}

Z = 1.01

Z = 1.01 has a pvalue of 0.8438

X = 950

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

Z = \frac{950-1000}{49.5}

Z = -1.01

Z = -1.01 has a pvalue of 0.1562

0.8438 - 0.1562 = 0.6876

68.76% probability that the sampling error made in estimating the mean weekly salary for all employees of the company by the mean of a random sample of weekly salaries of 50 employees will be at most $50

4 0
3 years ago
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