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3 years ago

6

In three more years Miguel's grandfather will be six times as old as Miguel was last year. When Miguel's grandfather present age

is added to his grandfathers age is added to his grandfathers present age the total is 68. How old is his grandfather now ?

1 answer:

lora16 [44]3 years ago

7 0

"In three more years,Miguel's grandfather will be six times as old as Miguel was last year. When Miguel's present age is added to his grandfather's present age, the total is 68. How old is each one now?"

Miguel is 11 years old and his grandfather is 57 years old.
Last year Miguel was 10. In three more years his grandfather would be 60. 60/6=10. 11+57=68

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What is the approximate value for the modal daily sales?

Aleksandr [31]

Answer:

Step-by-step explanation:

Hello!

<em>The table shows the daily sales (in $1000) of shopping mall for some randomly selected days </em>

<em>Sales 1.1-1.5 1.6-2.0 2.1-2.5 2.6-3.0 3.1-3.5 3.6-4.0 4.1-4.5 </em>

<em>Days 18 27 31 40 56 55 23 </em>

<em>Use it to answer questions 13 and 14. </em>

<em>13. What is the approximate value for the modal daily sales? </em>

To determine the Mode of a data set arranged in a frequency table you have to identify the modal interval first, this is, the class interval in which the Mode is included. Remember, the Mode is the value with most observed frequency, so logically, the modal interval will be the one that has more absolute frequency. (in this example it will be the sales values that were observed for most days)

The modal interval is [3.1-3.5]

Now using the following formula you can calculate the Mode:

$Md= Li + c[\frac{(f_{max}-f_{prev})}{(f_{max}-f_{prev})(f_{max}-f_{post})} ]$

Li= Lower limit of the modal interval.

c= amplitude of modal interval.

fmax: absolute frequency of modal interval.

fprev: absolute frequency of the previous interval to the modal interval.

fpost: absolute frequency of the posterior interval to the modal interval.

$Md= 3,100 + 400[\frac{(56-40)}{(56-40)+(56-55)} ]= 3,476.47$

<em>A. $3,129.41 B. $2,629.41 C. $3,079.41 D. $3,123.53 </em>

Of all options the closest one to the estimated mode is A.

<em>14. The approximate median daily sales is … </em>

To calculate the median you have to identify its position first:

For even samples: PosMe= n/2= 250/2= 125

Now, by looking at the cumulative absolute frequencies of the intervals you identify which one contains the observation 125.

F(1)= 18

F(2)= 18+27= 45

F(3)= 45 + 31= 76

F(4)= 76 + 40= 116

F(5)= 116 + 56= 172 ⇒ The 125th observation is in the fifth interval [3.1-3.5]

$Me= Li + c[\frac{PosMe-F_{i-1}}{f_i} ]$

Li: Lower limit of the median interval.

c: Amplitude of the interval

PosMe: position of the median

F(i-1)= accumulated absolute frequency until the previous interval

fi= simple absolute frequency of the median interval.

$Me= 3,100+400[\frac{125-116}{56} ]= 3164.29$

<em>A. $3,130.36 B. $2,680.36 C. $3,180.36 D. $2,664</em>

Of all options the closest one to the estimated mode is C.

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3 years ago

an outlier may be defined as a data point that is more than 1.5 times the interquartile range below the lower quartile or is mor

Ira Lisetskai [31]

Supposing a normal distribution, we find that:

The diameter of the smallest tree that is an outlier is of 16.36 inches.

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

We suppose that tree diameters are normally distributed with <u>mean 8.8 inches and standard deviation 2.8 inches.</u>

<u />

In a normal distribution with mean $\mu$ and standard deviation $\sigma$ , the z-score 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, which is the percentile of X.<u> </u>

<u />

In this problem:

Mean of 8.8 inches, thus $\mu = 8.8$ .
Standard deviation of 2.8 inches, thus $\sigma = 2.8$ .

<u />

The interquartile range(IQR) is the difference between the 75th and the 25th percentile.

<u />

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X when Z has a p-value of 0.25, so X when Z = -0.675.

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

$-0.675 = \frac{X - 8.8}{2.8}$

$X - 8.8 = -0.675(2.8)$

$X = 6.91$

75th percentile:

X when Z has a p-value of 0.75, so X when Z = 0.675.

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

$0.675 = \frac{X - 8.8}{2.8}$

$X - 8.8 = 0.675(2.8)$

$X = 10.69$

The IQR is:

$IQR = 10.69 - 6.91 = 3.78$

What is the diameter, in inches, of the smallest tree that is an outlier?

The diameter is <u>1.5IQR above the 75th percentile</u>, thus:

$10.69 + 1.5(3.78) = 16.36$

The diameter of the smallest tree that is an outlier is of 16.36 inches.

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A similar problem is given at brainly.com/question/15683591

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2 years ago

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$\huge \boxed{\mathbb{QUESTION} \downarrow}$

What is the solution to the inequality -3x-42>3 ?

$\large \boxed{\mathfrak{Answer \: with \: Explanation} \downarrow}$

$- 3 x - 42 \gt 3$

Add 42 to both sides.

$-3x>3+42$

Add 3 and 42 to get 45.

$-3x>45$

Divide both sides by -3. As -3 is <0, the inequality direction has changed.

$x$

Divide 45 by -3 to get -15.

$\huge \boxed{ \boxed{ \bf \: x$

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3 years ago

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Olegator [25]

Answer:

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Elis [28]

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