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tensa zangetsu [6.8K]
3 years ago
12

Please Help!!!

Mathematics
1 answer:
alekssr [168]3 years ago
4 0
<span>A) About 25% of the fishermen caught between 25 and 30 fish.
 >Since the boxplot looks symmetric, the parts are each 25%. 

</span><span>B) The spread is symmetrical about the median.
 >The median is 20. 
20-15 = 5; 15-10 = 5
25-20 = 5; 30-25 = 5

</span><span>E) The range is 20 fish.
 > The largest data is 30, smallest data is 10
Range = largest - smallest = 30-10 = 20</span>
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Vugar has a maximum of 30 batteries for toys. Each toy helicopter uses the same number of batteries, and each toy car uses the s
Leno4ka [110]

Answer: 3H+4C<_ 30

The coefficients of the variables H and C represent the number of batteries each toy helicopter and each toy car uses.

Step-by-step explanation:

Each toy helicopter uses 3 batteries, and each toy car uses 4 batteries.

Now let's check whether Vugar has enough batteries for 5 toy helicopters and 4 toy cars. To do this, we substitute  H=5 and C= 4 in the given inequality:

Does Vugar have enough batteries to play with 5 toy helicopters and 4 toy cars?

No, because if you plug in the value for H and C:

3H + 4C<_ 30

3(5) + 4(4) <_30

15 + 16 <_ 30

31 <_ 30; false

Since the inequality is false, Vugar does not have enough batteries to play with 555 toy helicopters and 444 toy cars.

Each toy helicopter uses 333 batteries, and each toy car uses 444 batteries.

No, Vugar does not have enough batteries to play with 555 toy helicopters and 444 toy cars.

6 0
4 years ago
Read 2 more answers
considering the leading term of the polynomial function. what is the end behavior of the graph? describe the end behavior and pr
mario62 [17]

The leading term of polynomial function is the the term contain highest degree so here in the given question leading term is  -3x^5

and leading coefficient is the coefficient of the term with greatest exponent -3

RULES for End behaviour

we have following four cases

CASE1: Even degree and positive leading coefficient

x\rightarrow -\infty      f(x)\rightarrow \infty

x\rightarrow \infty        f(x)\rightarrow \infty

CASE2: Even degree and negative leading coefficient

x\rightarrow -\infty        f(x)\rightarrow -\infty

x\rightarrow \infty           f(x)\rightarrow -\infty

CASE3: Odd degree and positive leading coefficient

x\rightarrow -\infty          f(x)\rightarrow -\infty

x\rightarrow \infty              f(x)\rightarrow \infty

CASE4: Odd degree and negative leading coefficient

x\rightarrow -\infty              x\rightarrow \infty  

x\rightarrow \infty                  f(x)\rightarrow -\infty


Here in the given case we have odd degree and negative leading coefficient

x\rightarrow -\infty              x\rightarrow -\infty  

x\rightarrow \infty                  f(x)\rightarrow -\infty


3 0
3 years ago
4% of what days is 56 days
leva [86]
The answer would be 14 because 4x__=56 ? 4 x14 =56
6 0
3 years ago
Marla has $95.50 in her savings account. She deposits $5 every week. Her mother also deposits $20 into the account every time Ma
tiny-mole [99]
Coefficient: 20 or 5
Variable:  c or w
Constant:  95.50

Pt B:  20(5) + 5(20) + 95.50
          100 + 100 + 95.50
          295.50

Pt C:The coefficient of c would change to 10
8 0
3 years ago
A cylinder shaped can needs to be constructed to hold 600 cubic centimeters of soup. The material for the sides of the can costs
PSYCHO15rus [73]

Answer:

the dimensions that minimize the cost of the cylinder are R= 3.85 cm and L=12.88 cm

Step-by-step explanation:

since the volume of a cylinder is

V= π*R²*L → L =V/ (π*R²)

the cost function is

Cost = cost of side material * side area  + cost of top and bottom material * top and bottom area

C = a* 2*π*R*L + b* 2*π*R²

replacing the value of L

C = a* 2*π*R* V/ (π*R²) + b* 2*π*R²  = a* 2*V/R + b* 2*π*R²

then the optimal radius for minimum cost can be found when the derivative of the cost with respect to the radius equals 0 , then

dC/dR = -2*a*V/R² + 4*π*b*R = 0

4*π*b*R = 2*a*V/R²

R³ = a*V/(2*π*b)

R=  ∛( a*V/(2*π*b))

replacing values

R=  ∛( a*V/(2*π*b)) = ∛(0.03$/cm² * 600 cm³ /(2*π* 0.05$/cm²) )= 3.85 cm

then

L =V/ (π*R²) = 600 cm³/(π*(3.85 cm)²) = 12.88 cm

therefore the dimensions that minimize the cost of the cylinder are R= 3.85 cm and L=12.88 cm

5 0
4 years ago
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