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3 years ago

0.346 rounded to the nearest hundredth

2 answers:

6 0

0.35 is rounded to the nearest hundredth 3would be in the tenth place 4is in the hundredth place 6is in the thousandths place hope this helps

Annette [7]3 years ago

4 0

0.346 rounded to the nearest hundredth is .30.

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Miles brought the Russian novel Anna Kerenina. His paperback edition contains 992 page. Miles takes 3 days to read the first 50

Nesterboy [21]

Answer:

59.52 days

Step-by-step explanation:

So this is a simple problem we can solve easily

If it takes 3 days to read 50 pages then obviously his rate of reading is 50 pages per day

If we divide 992 by 50, then we can find how many 3 day increments it will take him

(992/50 is 19.84)

Now is we multiply 19.84 by 3, we get the total amount of days which is 59.52

I hope I helped and have a wonderful day!

7 0

3 years ago

(b) Carlos removed 56 fish from his pond over a period of 7 days. He removed the same number

Pavlova-9 [17]

Well supposing that you want all fish gone -

The answer would be 8

56 (divided) 7 = 8
Day 1 - 48
Day 2 - 40
Day 3 - 32
Day 4 - 24
Day 5 - 16
Day 6 - 8
Day 7 - 0

Hope this was what you were looking for :)

6 0

3 years ago

If Ben invests $4500 at 4% interest per year, how much additional money must he invest at 5 1/2% annual interest to ensure that

pogonyaev

$Step \; 1: \; Assing \; variable \; for \; the \; unknown \; that \; we \; need \; to \; find.\\Let \; x \; be \; additional \; money \; invested$

$Step \; 2: \; Use \; the \; appropriate \; formula \; of \; simple \; interest \; set \; up \; an \; equation\\Interest \; earned \; each \; year = \; Amount \; invested \times annual \; rate \; of \; interest\\\\Amount \; invested \; in \; 4 \% \; rate \; of \; interest=\$4500\\Amount \; invested \; in \; 5\frac{1}{2}\% \; rate \; of \; interest= \$x\\Amount \; invested \; totally \; in \; 4\frac{1}{2}\% \; rate \; of \; interest = \$(4500+x)$

$Interest \; earned \; in \; 4\% \; interest \; rate\\ = \; 4500(4\%)=180\\Interest \; earned \; in \; 5\frac{1}{2}\% \; interest \; rate\\ = x(5\frac{1}{2}\%)=0.055x\\Interest \; earned \; totally \; in \; 4\frac{1}{2}\% \; interest \; rate = (4500+x)(4\frac{1}{2}\%)\\=0.045(4500+x)$

$\\\\So, \; the \; equation \; would \; be:\\180+0.055x=0.045(4500+x)$

$Step \; 1: \; Distribute \; 0.045 \; in \; the \; right \; side\\180+0.055x=202.5+0.045x\\\\Step \; 2: \; Subtract \; 0.045x \; and \; 180 \; on \; both \; sides \; to \; get \; x \; alone\\0.055x-0.045x=202.5-180\\\\Step \; 3: \; Combine \; Like \; Terms\\0.01x=22.5\\\\Step \; 4: Divide \; 0.01 \; on \; both \; sides\\\frac{0.01x}{0.01}=\frac{22.5}{0.01}\\\\Step \; 5: \; Simplifying \; fraction \; on \; both \; sides\\x=2250$

$\underline{Conclusion:}\\Ben \; should \; invest \; additional \; money \; of \; \$2250$

5 0

4 years ago

Since Spring started, Kareem has been surveying the growth of leaves on his neighborhood trees. He goes out every day and comput

Otrada [13]

Answer:

y=1x-b is thou answer

4 0

3 years ago

Read 2 more answers

A random sample of n = 64 observations is drawn from a population with a mean equal to 20 and standard deviation equal to 16. (G

dezoksy [38]

Answer:

a) The mean of a sampling distribution of $\\ \overline{x}$ is $\\ \mu_{\overline{x}} = \mu = 20$ . The standard deviation is $\\ \frac{\sigma}{\sqrt{n}} = \frac{16}{\sqrt{64}}=2$ .

b) The standard normal z-score corresponding to a value of $\\ \overline{x} = 16$ is $\\ Z = -2$ .

c) The standard normal z-score corresponding to a value of $\\ \overline{x} = 23$ is $\\ Z = 1.5$ .

d) The probability $\\ P(\overline{x}$ .

e) The probability $\\ P(\overline{x}>23) = 1 - P(Z$ .

f) $\\ P(16 < \overline{x} < 23) = P(-2 < Z < 1.5) = P(Z$ .

Step-by-step explanation:

We are dealing here with the concept of a sampling distribution, that is, the distribution of the sample means $\\ \overline{x}$ .

We know that for this kind of distribution we need, at least, that the sample size must be $\\ n \geq 30$ observations, to establish that:

$\\ \overline{x} \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

In words, the distribution of the sample means follows, approximately, a normal distribution with mean, $\mu$ , and standard deviation (called standard error), $\\ \frac{\sigma}{\sqrt{n}}$ .

The number of observations is n = 64.

We need also to remember that the random variable Z follows a standard normal distribution with $\\ \mu = 0$ and $\\ \sigma = 1$ .

$\\ Z \sim N(0, 1)$

The variable Z is

$\\ Z = \frac{\overline{x} - \mu}{\frac{\sigma}{\sqrt{n}}}$ [1]

With all this information, we can solve the questions.

Part a

The mean of a sampling distribution of $\\ \overline{x}$ is the population mean $\\ \mu = 20$ or $\\ \mu_{\overline{x}} = \mu = 20$ .

The standard deviation is the population standard deviation $\\ \sigma = 16$ divided by the root square of n, that is, the number of observations of the sample. Thus, $\\ \frac{\sigma}{\sqrt{n}} = \frac{16}{\sqrt{64}}=2$ .

Part b

We are dealing here with a random sample. The z-score for the sampling distribution of $\\ \overline{x}$ is given by [1]. Then

$\\ Z = \frac{\overline{x} - \mu}{\frac{\sigma}{\sqrt{n}}}$

$\\ Z = \frac{16 - 20}{\frac{16}{\sqrt{64}}}$

$\\ Z = \frac{-4}{\frac{16}{8}}$

$\\ Z = \frac{-4}{2}$

$\\ Z = -2$

Then, the standard normal z-score corresponding to a value of $\\ \overline{x} = 16$ is $\\ Z = -2$ .

Part c

We can follow the same procedure as before. Then

$\\ Z = \frac{\overline{x} - \mu}{\frac{\sigma}{\sqrt{n}}}$

$\\ Z = \frac{23 - 20}{\frac{16}{\sqrt{64}}}$

$\\ Z = \frac{3}{\frac{16}{8}}$

$\\ Z = \frac{3}{2}$

$\\ Z = 1.5$

As a result, the standard normal z-score corresponding to a value of $\\ \overline{x} = 23$ is $\\ Z = 1.5$ .

Part d

Since we know from [1] that the random variable follows a standard normal distribution, we can consult the cumulative standard normal table for the corresponding $\\ \overline{x}$ already calculated. This table is available in Statistics textbooks and on the Internet. We can also use statistical packages and even spreadsheets or calculators to find this probability.

The corresponding value is Z = -2, that is, it is two standard units below the mean (because of the negative value). Then, consulting the mentioned table, the corresponding cumulative probability for Z = -2 is $\\ P(Z$ .