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3 years ago

What is 108 times 8.5

Mathematics

2 answers:

valina [46]3 years ago

5 0

108 times 8.5 is 918

kvasek [131]3 years ago

3 0

Answer:

918

Step-by-step explanation:

108 x 8.5

108 x 5 = 540

108 x 8 = 864

540

+ 864

-----------------

918.0

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The library has 27 chairs and 5 tables. If the same number of chairs is placed at each tables, how many chairs can be placed eac

Alik [6]

Answer:5

Step-by-step explanation:

27 chairs

5 tables

27/5

=5.4

6 0

3 years ago

A half-century ago, the mean height of women in a particular country in their 20s was 64.7 inches. Assume that the heights of

Ainat [17]

Answer:

99.5% of all samples of 21 of today's women in their 20's have mean heights of at least 65.86 inches.

Step-by-step explanation:

We are given that a half-century ago, the mean height of women in a particular country in their 20's was 64.7 inches. Assume that the heights of today's women in their 20's are approximately normally distributed with a standard deviation of 2.07 inches.

Also, a samples of 21 of today's women in their 20's have been taken.

Let $\bar X$ = sample mean heights

The z-score probability distribution for sample mean is given by;

Z = $\frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\mu$ = population mean height of women = 64.7 inches

$\sigma$ = standard deviation = 2.07 inches

The Z-score measures how many standard deviations the measure is away from the mean. After finding the Z-score, we look at the z-score table and find the p-value (area) 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.

Now, Probability that the sample of 21 of today's women in their 20's have mean heights of at least 65.86 inches is given by = P( $\bar X$ $\geq$ 65.86 inches)

P( $\bar X$ $\geq$ 65.86 inches) = P( $\frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n} } }$ $\geq$ $\frac{65.86-64.7}{\frac{2.07}{\sqrt{21} } }$ ) = P(Z $\geq$ -2.57) = P(Z $\leq$ 2.57)

= 0.99492 or 99.5%

The above probability is calculated by looking at the value of x = 2.57 in the z table which has an area of 0.99492.

Therefore, 99.5% of all samples of 21 of today's women in their 20's have mean heights of at least 65.86 inches.

3 0

3 years ago

Suppose that $9500 is placed in an account that pays 8% interest compounded each year.

Blababa [14]

$\bf ~~~~~~ \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\dotfill &\$9500\\ r=rate\to 8\%\to \frac{8}{100}\dotfill &0.08\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{each year, thus once} \end{array}\dotfill &1\\ t=years\dotfill &1 \end{cases}$

$\bf A=9500\left(1+\frac{0.08}{1}\right)^{1\cdot 1}\implies A=9500(1.08)\implies \boxed{A=10260} \\\\\\ \stackrel{\textit{after 2 years, t = 2}}{A=9500\left(1+\frac{0.08}{1}\right)^{1\cdot 2}}\implies A=9500(1.08)^2\implies \boxed{A=11080.8}$

6 0

3 years ago

Winona’s bowling scores for the past nine weeks are 195, 180, 195, 212, 208, 231, 179, 246, and 195. Find the mean, median, and

belka [17]

Answer: Mean = 204.6 , Median = 195 and mode =195

Step-by-step explanation:

The given data for Winona’s bowling scores for the past nine weeks : 195, 180, 195, 212, 208, 231, 179, 246, 195.

The formula to find Mean is given by :-

$\text{Mean}=\dfrac{\text{Sum of all observations}}{\text{Number of observations}}\\\\=\dfrac{1841}{9}=204.555555556\approx204.6$

For median , first we arrange data values in order as

179, 180, 195, 195, 195, 208, 212, 231, 246

Since the number of data values is 9 (odd).

∴ Median = Middle most data value = 195

Also, Mode = Most occurring data value = 195

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3 years ago

There are 18 pieces of chalk. The ratio if chalk that is yellow to chalk that is not yellow is 1:2. How many more pieces of whit

katen-ka-za [31]

The amount of chalk are equal

5 0

3 years ago