c. C = 0.41(w - 5) + 10.8
(w - 5) will represent the number of pounds over 5. For example, for a weight of 6 pounds, 6-5 = 1 is the number of pounds over 5.
The cost is $0.41 for each pound over 5, so that cost can be represented by ...
... 0.41(w - 5)
This charge is in addition to the base charge of $10.80, so the total cost will be ...
... C = 0.41(w - 5) + 10.80
Answer:
Step-by-step explanation:
Make the denominators equal after simplifying the denominators*. The denominators are currently 2 and 4. So you can just make your life easier by making the denominator of the first number a 4. Multiply both sides by 2.
1. Simplify
12*2+1 = 25/2
25/2*2/2 = 50/4
2. Simplify second number
16*4+3 = 67/4
3. Add the improper fractions
50/4 + 67/4 = 117/4
Terms:
*denominators - the bottom number of a fraction
Hope this helped,
Kavitha Banarjee
number of people
6
3
30
A
B
40
C
28
What number goes in cell A?
Answer:
Cell A = 15
Step-by-step explanation:
If 6 spring rolls will serve 3 people, let x be the number of people 30 spring rolls will serve.
Thus,
6 spring rolls = 3 people
30 spring rolls = x
Cross multiply
6*x = 3*30
6x = 90
Divide both sides by 6
6x/6 = 90/6
x = 15
30 spring rolls will serve 15 people.
The question is asking for a proportion based on the initial data. The correct number for cell A is 15, calculated by dividing the total number of spring rolls by the ratio of spring rolls per person.
This question deals with proportions. Proportions look at the relationship between two pairs of numbers. If 6 spring rolls can serve 3 people, that means each person will get 2 spring rolls. This forms a ratio of 2:1, meaning each person gets 2 spring rolls.
Applying this ratio to 30 spring rolls, we would do the following calculation: 30 spring rolls / 2 spring rolls per person = 15 people. Therefore, cell A in the table should be filled with the number 15.
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691 648 967 959 826 573 598 790 954
711 515 649 960 949 802 507
a. Construct the frequency distribution using classes of 500 up to 600, 600 up to 700, etc.
Texts Frequency
500 up to 600
600 up to 700
700 up to 800
800 up to 900
900 up to 1000
Total
b. Construct the relative frequency distribution, the cumulative frequency distribution and the cumulative relative frequency distribution. (Round "Relative Frequency" and "Cumulative Relative Frequency" to 2 decimal places.)
Texts Relative Frequency Cumulative Frequency Cumulative Relative Frequency
500 up to 600
600 up to 700
700 up to 800
800 up to 900
900 up to 1000
c-1. How many of the 13-year-olds sent at least 600 but less than 700 text messages?
c-1. Number of 13-year-olds
Number of 13-year-olds
c-2. How many sent less than 900 text messages?
Number of 13-year-olds
d-1. What percent of the 13-year-olds sent at least 800 but less than 900 text messages? (Round your answer to the nearest whole percent.)
Percent of 13-year-olds %
d-5. What percent of the 13-year-olds sent less than 600 text messages? (Round your answer to the nearest whole percent.)
Percent of 13-year-olds %
Answer:
7 ; 19 ; 8% ; 28%
Step-by-step explanation:
Given the data:
791 542 671 672 555 582 616 961 639
691 648 967 959 826 573 598 790 954
711 515 649 960 949 802 507
How many of the 13-year-olds sent at least 600 but less than 700 text messages? = 7
c-2. How many sent less than 900 text messages? = (7 + 7 + 3 + 2) = 19
d-1. What percent of the 13-year-olds sent at least 800 but less than 900 text messages? =0.08 × 100 = 8% (from relative frequency)
d-5. What percent of the 13-year-olds sent less than 600 text messages? 0.28 × 100 = 28% (from relative frequency)
By sorting text messages into classes, we can get the frequency distribution. From there, we can determine the relative and cumulative frequencies. Finally, we can examine how many students sent texts within certain ranges and express these as percentages.
To answer this question, let's first classify the amount of text messages sent by each of the 25 13-year-olds into groups or classes of 100. Then we count the frequencies, or how many text messages fall into each class. This helps us construct the frequency distribution.
The classes are: 500-600, 600-700, 700-800, 800-900, and 900-1000.
Next, we calculate the relative frequency by dividing the frequency of each class by the total number of students. We round each relative frequency to 2 decimal places.
To calculate cumulative frequency, we keep an ongoing total of frequencies as we move up the classes. The cumulative relative frequency is computed similarly but applied to the relative frequencies.
In the last part, we determine how many 13-year-olds sent at least a certain number of texts but less than another number, and convert these to percentages.
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4 tons is maximum weight that could be supported by a beam that is 6 inches wide, 2 inches high, and 24 feet long
Gravitational force of attraction on an object, caused by the presence of a massive second object, such as the Earth or Moon.
Given,
Maximum weight that a certain type of rectangular beam can support varies inversely as it’s length and jointly as its width and the square of its height.
beam 5 inches wide, 2 inches high and 10 feet long can support a maximum weight of 8 tons.
W = k × w × h²/ L
8 = k × 5 × 2² / 10
8 = k ×20/10
8 = 2 × k
k = 4
The maximum weight that could be supported by a beam that is 6 inches wide, 2 inches high, and 24 feet long
w =4 ×6 ×2² / 24
= 24 ×4/ 24
W=4 tons
Hence 4 tons is maximum weight that could be supported by a beam that is 6 inches wide, 2 inches high, and 24 feet long
To learn more on Weight click:
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Way: W = k * w * h^2/ L
Find the measure of Z2.
Z2 = [?]°
Answer:
40 = <2
Step-by-step explanation:
<1 and <2 are corresponding angles and corresponding angles are equal when the lines are parallel
<1 = <2
40 = <2