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VLD [36.1K]
3 years ago
7

How do you do this? Someone give me the answer

Mathematics
1 answer:
Gemiola [76]3 years ago
4 0
It would last 13.5 weeks because you would make a proportion:
12/16 = x/18 and then cross multiply. You would get 216=16x, then you have to divide both sides by 16 to get x = 13.5
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Which is more 8.83 or 8.9
Artemon [7]

Answer:

8.9

Step-by-step explanation:

If you think of 8.9 as 8.90, then it's obvious that 8.90 (or 8.9) is greater than 8.83.

5 0
2 years ago
Read 2 more answers
Last week 24,000 fans attended a football match. this week three times as many bought tickets, but one-sixth of them cancelled t
lukranit [14]

Answer:

C. 60,000

Step-by-step explanation:

<u>Given</u>

  • 24,000 fans attended a football match
  • three times as many bought tickets
  • one-sixth of them cancelled their tickets

<u>Three times as many bought tickets</u>

24,000 * 3 = 72,000

<u>one-sixth of them cancelled their tickets</u>

72,000 * 1/6 = 12,000

<u>Subtract</u>

72,000 - 12,000 = 60,000

<u>Answer</u>

60,000 people are attending this week

6 0
4 years ago
Will spent $3.75 for 3 pounds or granola. what is the unit rate in dollars per pound
MissTica
The unit rate per pound is
$1.25
( i divided 3.75 into 3 )
6 0
3 years ago
Read 2 more answers
Find the indicated probability or percentage for the sampling error. The distribution of weekly salaries at a large company is r
Flauer [41]

Answer:

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 80 employees will be at most $75 is 0.9297.

Step-by-step explanation:

According to the Central Limit Theorem if we have a non-normal population with mean <em>μ</em> and standard deviation <em>σ</em> and appropriately huge random samples (<em>n</em> > 30) are selected from the population with replacement, then the distribution of the sample means will be approximately normally distributed.

Then, the mean of the distribution of sample means is given by,

\mu_{\bar x}=\mu

And the standard deviation of the distribution of sample means is given by,

\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}

The information provided is:

\mu=\$1000\\\sigma=\$370\\n=80

As <em>n</em> = 80 > 30, the central limit theorem can be used to approximate the sampling distribution of sample mean weekly salaries.

Let \bar X represent the sample mean weekly salaries.

The distribution of \bar X is: \bar X\sim N(\$1000,\ \$41.37)

Now we need to compute the probability of the sampling error made in estimating the mean weekly salary to be at most $75.

The sampling error is the the difference between the estimated value of the parameter and the actual value of the parameter, i.e. in this case the sampling error is, |\bar X-\mu|= 75.

Compute the probability as follows:

P(-75

                                     =P(-1.81

Thus, 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 80 employees will be at most $75 is 0.9297.

3 0
3 years ago
Please please help me on this one!
gizmo_the_mogwai [7]

Answer:

3422 x232

Step-by-step explanation:

3 0
3 years ago
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