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igor_vitrenko [27]
3 years ago
13

PLEASE HELP PLEASE I WILL GIVE 20 POINTS AND BRAINLIEST The club will base its decision about whether to increase the budget for

the indoor rock climbing facility on the analysis of its usage. The decision to increase the budget will depend on whether members are using the indoor facility at least two times a week. Use the best measure of center for both data sets to determine whether the club should increase the budget. Assume there are four weeks in a month. If you think the data is inconclusive, explain why.
Mathematics
1 answer:
iren2701 [21]3 years ago
5 0

Answer:

They should in crease budget.

Step-by-step explanation:

divide each number by 2 to see how often they go equally/use the data table you have available of the two climbers.

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Solve the proportion
mario62 [17]

Answer:

\boxed{\sf x = 5}

Step-by-step explanation:

\sf Solve  \: for  \: x  \: over  \: the  \: real \:  numbers:  \\ \sf \implies  \frac{2}{x - 3}   =  \frac{5}{x}  \\  \\  \sf Take  \: the \:  reciprocal  \: of  \: both \:  sides:  \\ \sf \implies  \frac{x - 3}{2}  =  \frac{x}{5}  \\  \\  \sf Expand  \: out \:  terms \:  of \:  the \:  left  \: hand \:  side:  \\  \\ \sf \implies \frac{x}{2}  -  \frac{3}{2}  =  \frac{x}{5}  \\  \\  \sf Subtract \:  \frac{x}{5}   -  \frac{3}{2}  \: from  \: both  \: sides: \\  \sf \implies \frac{x}{2}  -  \frac{3}{2} - ( \frac{x}{5}   -  \frac{3}{2} ) =  \frac{x}{5} - ( \frac{x}{5}  -  \frac{3}{2} ) \\  \\  \sf \implies \frac{x}{2}  -  \frac{3}{2} -  \frac{x}{5}    +   \frac{3}{2} =  \frac{x}{5} -  \frac{x}{5}  +  \frac{3}{2}  \\  \\  \sf \frac{x}{5}  -  \frac{x}{5}  = 0 :  \\  \sf \implies \frac{x}{2}  -  \frac{x}{5}  -  \frac{3}{2}  +  \frac{3}{2}  =  \frac{3}{2}  \\  \\  \sf  \frac{3}{2}   -   \frac{3}{2}   = 0:  \\  \sf \implies \frac{x}{2}  -  \frac{x}{5}  =  \frac{3}{2}   \\  \\ \sf \frac{x}{2}  -  \frac{x}{5} =  \frac{5x - 2x}{10}  =  \frac{3x}{10} :  \\   \sf \implies \frac{3x}{10}  =  \frac{3}{2}   \\  \\ \sf Multiply \:  both  \: sides \:  by \:  \frac{10}{3}  : \\   \sf \implies \frac{3x}{10}  \times  \frac{10}{3}  =  \frac{3}{2 }  \times  \frac{10}{3}   \\  \\ \sf \frac{3x}{10}  \times  \frac{10}{3}  =   \cancel{\frac{3}{10} } \times( x) \times  \cancel{ \frac{10}{3} } = x :  \\  \sf \implies x =  \frac{3}{2}  \times  \frac{10}{3} \\  \\   \sf  \frac{3}{2}  \times  \frac{10}{3}  = \cancel{ \frac{3}{2} }  \times \cancel{ \frac{3}{2} }  \times 5 :   \\ \sf \implies x = 5

8 0
3 years ago
James works as a waiter. He served meals with bills of $23.59, $40.65, $30.50, and $15.68. If his total for tips was $17.67, wha
SIZIF [17.4K]

Answer:

16%

Step-by-step explanation:

To find the percent, add the total for each meal to find the total dollars spent.

23.59 + 40.65 + 30.50 + 15.68 = 110.42

Divide the tip amount 17.67 by 110.42.

17.67/110.42 = 0.16

Multiply the decimal by 100 to convert to a percent.

0.16*100 = 16%

6 0
3 years ago
Nemecek Brothers make a single product on two separate production lines, A and B. Its labor force is equivalent to 1000 hours pe
frez [133]

Answer:

(a) The inequality for the number of items, x, produced by the labor, is given as follows;

250 ≤ x ≤ 600

(b) The inequality for the cost, C is $1,000 ≤ C ≤ $3,000

Step-by-step explanation:

The total time available for production = 1000 hours per week

The time it takes to produce an item on line A = 1 hour

The time it takes to produce an item on line B = 4 hour

Therefore, with both lines working simultaneously, the time it takes to produce 5 items = 4 hours

The number of items produced per the weekly labor = 1000/4 × 5 = 1,250 items

The minimum number of items that can be produced is when only line B is working which produces 1 item per 4 hours, with the weekly number of items = 1000/4 × 1 = 250 items

Therefore, the number of items, x, produced per week with the available labor is given as follows;

250 ≤ x ≤ 1250

Which is revised to 250 ≤ x ≤ 600 as shown in the following answer

(b) The cost of producing a single item on line A = $5

The cost of producing a single item on line B = $4

The total available amount for operating cost = $3,000

Therefore, given that we can have either one item each from lines A and B with a total possible item

When the minimum number of possible items is produced by line B, we have;

Cost = 250 × 4 = $1,000

When the maximum number of items possible, 1,250, is produced, whereby we have 250 items produced from line B and 1,000 items produced from line A, the total cost becomes;

Total cost = 250 × 4 + 1000 × 5 = 6,000

Whereby available weekly outlay = $3000, the maximum that can be produced from line A alone is therefore;

$3,000/$5 = 600 items = The maximum number of items that can be produced

The inequality for the cost, C, becomes;

$1,000 ≤ C ≤ $3,000

The time to produce the maximum 600 items on line A alone is given as follows;

1 hour/item × 600 items = 600 hours

The inequality for the number of items, x, produced by the labor, is therefore, given as follows;

250 ≤ x ≤ 600

8 0
3 years ago
120-3/4y=60 solve for y
navik [9.2K]
120 - 3/4y=60
lowest common multiple=4y
480y-3=240y
480y-240y=3
240y=3
y=3/240
y=1/80

Answer: y=1/80
5 0
3 years ago
What does x equal if the equation is x+2.62=6.37
KengaRu [80]

Answer:

x=3.75

The drawing will help

5 0
3 years ago
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