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Vesnalui [34]
3 years ago
7

600,000 gallons of water pass through a given point along a river every minute. Which equation represents the amount of water, y

, that passes through the point in x minutes?
Mathematics
2 answers:
11111nata11111 [884]3 years ago
8 0

Answer:

y=600,000  x=1

Step-by-step explanation:

y is equal to 600,000 which means that every minute (x) 600,000 gallons of water would pass that point in the river

Delvig [45]3 years ago
8 0

Answer: y=600,000 x=1

Step-by-step explanation: okay so if 600,000 is water then that would mean that u have y=600,000 and if x is minutes it does not say so it would come down to y=600,000 =x.

Hope this helps.

GOOD LUCK!!

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6 and 15 have 3 as a common factor. Thus 6:15 is 2:5. The other ratios are also the same as given in the question. Since the numbers in the second column increase in 15s, the numbers in the first row should increase in sixes (12:30 is the same as 6:15 and 2:5). Thus the answers are:

  1. 6:15
  2. 12:30
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A particular brand of dishwasher soap is sold in three sizes: 25 oz, 40 oz, and 65 oz. Twenty percent of all purchasers select a
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Answer:

Step-by-step explanation:

Given that:

There are 3 sizes of dishwasher soaps.

Twenty percent of all purchasers select a 25-oz box, 50% select a 40-oz box, and the remaining 30% choose a 65-oz box. Let X1 and X2 denote the package sizes selected by two independently selected purchasers.

Therefore, 3 \times 3 = 9 random samples of size 2 are taken from the population.

The information is well represented in the table below.

x_1     x_2     p(x_1 \ x_2) = p(x_1)p(x_2)    \bar x = \dfrac{x_1+x_2}{2}     s^2= (x_1-\bar x)^2 +(x_2-\bar x)^2

25    25    0.20×0.20=0.04        25                              0

25    40    0.20×0.50=0.10         32.5                         112.5

25    65    0.20×0.30= 0.06        45                            800

40    25    0.50×0.20=0.10         32.5                          112.5

40    40    0.50×0.50=0.25         40                             0

40    65    0.50×0.30=0.15          52.5                         312.5

65    25    0.30×0.20=0.06        45                             800

65    40    0.30×0.50=0.15          52.5                         312.5

65    65    0.30×0.30=0.09         65                              0

The sample distribution of \bar x is shown below as ;

\bar x           25        32.5              40       45                  52.5          65

p(\bar x)       0.04   0.10+0.10      0.25   0.06+0.06    0.15+0.15    0.09    

                       = 0.20                      =0.12              =0.30

The value of E(\bar x) = \sum \bar x  \ P(\bar x)

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the sampling distribution of {s^2} is be shown below;

{s^2}          0                          112.5            312.5             800

           0.04+0.25       0.10+0.10     0.15+0.15        0.06+0.06

          +0.09

p(s)^2   = 0.38                =0.20           = 0.30               0.12

The value of E(s)^2  =\sum s^2 .p(s)^2

= (0×0.38)+(112.5×0.20)+(312.5×0.30)+(800×0.12)

=0+112+93.75+96

=301.75

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