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Ralph is three times as old is Sarah. In six years Ralph will be only twice as old as Sarah will be then. Find Ralphs age now. W

andrew11 [14]

Let's begin by calling Sarah's age now as X. As Ralph is 3 times as old as Sarah, X times 3 = 3X. Hence, Ralph's age is 3X. In six years, Ralph will be twice as old as Sarah. To calculate six years from now, add 6 to X for Sarah, and 6 to 3X for Ralph. As Ralph is twice as old as Sarah and we want to find the difference between the ages to calculate X, multiply X+6 by 2. You'll get 2X+12. Therefore, 2X+12=3X+6. Deduct 6 from 3X+6 as we want to isolate the variable. Because you did that to one side, you have to deduct 6 from 2X+12. Hence, now you have 2X+6=3X. X=6. Ralph's age is 3X, so 6 times 3 is 18. Ralph is 18 years old.

5 0

2 years ago

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One angle is five degrees less than three times the degree measure of another angle. Together, the angles measures have a sum of

Radda [10]

Answer:

a

Step-by-step explanation:

5 0

2 years ago

What is the approximate value for the modal daily sales?

Aleksandr [31]

Answer:

Step-by-step explanation:

Hello!

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

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 

Days 18 27 31 40 56 55 23 

Use it to answer questions 13 and 14. 

13. What is the approximate value for the modal daily sales? 

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.

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

A. $3,129.41 B. $2,629.41 C. $3,079.41 D. $3,123.53 

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

14. The approximate median daily sales is … 

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.

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

A. $3,130.36 B. $2,680.36 C. $3,180.36 D. $2,664

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

5 0

2 years ago

Maya is packing her backpack for hike in one pocket she puts in 1/4 pound bag of trail mix a water bottle weighing 2 1/5 pounds

xz_007 [3.2K]

This problem is simply asking us to add the weights which are presented as fractions. You can easily get the answer if you input them directly in a calculator. But I think the challenge here is to manually add fractions.

Let's add the two 1/4 lb weights because it is easier for they have a common denominator. Just simply add the numerators.

1/4 + 1/4 = 2/4

Then, add with this the fraction part of 2 1/5.

2/4 + 1/5 = ?
The least common denominator is 4*5 = 20. Then divide 20 with each denominator and multiply to their respective numerator.

2/4 + 1/5 = [(20/4 * 2) + (20/5 *1)]/20 = 14/20

14/20 is simplified to 7/10. Then add the whole number 2. Therefore, the sum of all weights is 2 7/10 pounds.

5 0

3 years ago

Suppose you are doing a study on the wealth of the population in your town. You take a random survey of 100 people. Which would

Angelina_Jolie [31]

Answer:

D) The total number of people in the population.

Step-by-step explanation:

The answer is D) The total number of people in the population. because it is a quantitative data measurement. All other options, Age, Gender and Education are qaualitative measures, which are aligned with the Objective of the study, Wealth - A qualitative measure too!

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3 years ago

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