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spin [16.1K]
2 years ago
11

Sam is eating a Big Hamburger. The first bite was 20% of the Hamburger, the second bite was 20% of what is left and so every nex

t bite is 20% of what is left. a How much did the Hamburger weigh originally, if after the second bite there was 160g of it left?
Mathematics
1 answer:
lakkis [162]2 years ago
6 0
The weight of the hamburger is 160g after two bites.

Each bite takes away 20% of the weight of the burger, so we'll divide the initial weight by a percentage to find the weight before the bite.

The starting weight is 160g. Because 20% of the weight was removed from a bite, we'll divide this weight by the percentage of the burger remaining after the bite:

100 - 20 = 80

80% of the burger is left after a bite. Convert this fraction into a decimal by dividing by 100:

80 \div 100 = 0.8

Divide the initial weight by this decimal:

160 \div 0.8 = 200

The weight of the burger before the second bite was 200.

We'll do the same procedure for the first bite. The initial weight is 200, and there is 80% of the burger remaining after the first bite. Because 80% is 0.8 in decimal form, divide 200 by 0.8:

200 \div 0.8 = 250

The starting weight of the burger before the bites was 250 grams.
You might be interested in
Find the coefficient of variation for each of the two sets of data, then compare the variation. Round results to one decimal pla
svp [43]

Here is  the correct computation of the question given.

Find the coefficient of variation for each of the two sets of data, then compare the variation. Round results to one decimal place. Listed below are the systolic blood pressures (in mm Hg) for a sample of men aged 20-29 and for a sample of men aged 60-69.

Men aged 20-29:      117      122     129      118     131      123

Men aged 60-69:      130     153      141      125    164     139

Group of answer choices

a)

Men aged 20-29: 4.8%

Men aged 60-69: 10.6%

There is substantially more variation in blood pressures of the men aged 60-69.

b)

Men aged 20-29: 4.4%

Men aged 60-69: 8.3%

There is substantially more variation in blood pressures of the men aged 60-69.

c)

Men aged 20-29: 4.6%

Men aged 60-69: 10.2 %

There is substantially more variation in blood pressures of the men aged 60-69.

d)

Men aged 20-29: 7.6%

Men aged 60-69: 4.7%

There is more variation in blood pressures of the men aged 20-29.

Answer:

(c)

Men aged 20-29: 4.6%

Men aged 60-69: 10.2 %

There is substantially more variation in blood pressures of the men aged 60-69.

Step-by-step explanation:

From the given question:

The coefficient of variation can be determined by the relation:

coefficient \ of  \ variation = \dfrac{standard \ deviation}{mean}*100

We will need to determine the coefficient of variation both men age 20 - 29 and men age 60 -69

To start with;

The coefficient of men age 20 -29

Let's first find the mean and standard deviation before we can do that ;

SO .

Mean = \dfrac{\sum \limits^{n}_{i-1}x_i}{n}

Mean = \frac{117+122+129+118+131+123}{6}

Mean = \dfrac{740}{6}

Mean = 123.33

Standard deviation  = \sqrt{\dfrac{\sum (x_i- \bar x)^2}{(n-1)} }

Standard deviation =\sqrt{\dfrac{(117-123.33)^2+(122-123.33)^2+...+(123-123.33)^2}{(6-1)} }

Standard deviation  = \sqrt{\dfrac{161.3334}{5}}

Standard deviation = \sqrt{32.2667}

Standard deviation = 5.68

The coefficient \ of  \ variation = \dfrac{standard \ deviation}{mean}*100

coefficient \ of  \ variation = \dfrac{5.68}{123.33}*100

Coefficient of variation = 4.6% for men age 20 -29

For men age 60-69 now;

Mean = \dfrac{\sum \limits^{n}_{i-1}x_i}{n}

Mean = \frac{   130 +    153    +  141  +    125 +   164  +   139}{6}

Mean = \dfrac{852}{6}

Mean = 142

Standard deviation  = \sqrt{\dfrac{\sum (x_i- \bar x)^2}{(n-1)} }

Standard deviation =\sqrt{\dfrac{(130-142)^2+(153-142)^2+...+(139-142)^2}{(6-1)} }

Standard deviation  = \sqrt{\dfrac{1048}{5}}

Standard deviation = \sqrt{209.6}

Standard deviation = 14.48

The coefficient \ of  \ variation = \dfrac{standard \ deviation}{mean}*100

coefficient \ of  \ variation = \dfrac{14.48}{142}*100

Coefficient of variation = 10.2% for men age 60 - 69

Thus; Option C is correct.

Men aged 20-29: 4.6%

Men aged 60-69: 10.2 %

There is substantially more variation in blood pressures of the men aged 60-69.

4 0
2 years ago
Jeanne babysits for $6 per hour. She also works as a reading tutor for $10 per hour. She is only allowed to work 20 hours per we
Maru [420]
A) System of inequalities

Income from babysitting: 6x
Income from tutoring: 10y

Goal for this week: at least $ 75 => 6x + 10y ≥ 75
Number of hours: maximum $20 => x + y ≤ 20

x and y have to be zero or positvie => x≥0 and y ≥ 0

Answer of part A.

6x + 10y ≥ 75
x+y ≤ 20
x ≥ 0
y ≥ 0

Part B. Graph

The solution is a region closed by 4 lines and you can find it in the file attached.

In that graph:

1) the line x + y = 20 is the upper limit
2) the line 6x + 10 y = 75 is the lower incclined line
3) y = 0 is the botton horizontal line it is the x-axis
4) x = 0 is the left vertical line, in is the y-axis

Part C.

Use, for example the point (7,10).

It is in the region of the graph limited by the above equations.

It implies:

Number of hours worked = 7 + 10 = 17 which is less than 20.

Income: 6(7) + 10(10) = 42 + 100 = 142, which is greater than 75

Of course, both are greater than 0.

In this way we proved that the point (7,10) is a true solution for the model.






Download pdf
8 0
2 years ago
Ivan and kate live 42 miles apart. ivan leaves his house at 8:00 a.m. and bikes towards kate's house at a constant speed of 12 m
valentinak56 [21]

Answer:

10:18am

Step-by-step explanation:

7 0
2 years ago
Vladimir says that the equation of the line that passes through points (negative 5, negative 3) and (10, 9) is y = four-fifths x
Vilka [71]

Answer:

C. both Vladimir and Robyn

Step-by-step explanation:

Just took the test. Hope this helps

5 0
2 years ago
Read 2 more answers
A ship's sonar finds that the angle of depression to a ship wrack on the bottom of the oceanis 12.5°. If a point on the ocean fl
Travka [436]

Answer:

The 13.2 meters from that point on the ocean floor to the wreck.

Step-by-step explanation:

As given

A ship's sonar finds that the angle of depression to a ship wrack on the bottom of the ocean is 12.5°.

If a point on the ocean floor is 60 meters.

Now by using the trigonometric identity.

tan\theta = \frac{Perpendicular}{Base}

As shown in the figure given below.

Perpendicular = CB

Base = AC = 60 meters

\theta = 12.5^{\circ}

Put in the identity

tan\ 12.5^{\circ} = \frac{CB}{AC}

tan\ 12.5^{\circ} = \frac{CB}{60}

tan\ 12.5^{\circ} = 0.22

tan\ 12.5^{\circ} = 0.22

0.22= \frac{CB}{60}

CB = 60 × 0.22

CB = 13.2 meters

Therefore the 13.2 meters from that point on the ocean floor to the wreck.







8 0
2 years ago
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