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ale4655 [162]
2 years ago
7

Caitlyn pays $2.20 in postage to mail a CD to a friend. She uses 43-cent stamps and 6-cent stamps. How many of each stamp did Ca

itlyn use?
Mathematics
1 answer:
aksik [14]2 years ago
8 0

Caitlyn used 4 43-cent stamps and 8 6-cent stamps

<h3>How many of each stamp did Caitlyn use?</h3>

The given parameters are:

Amount paid = $2.20

Stamp used: 43-cent stamps and 6-cent stamps

Represent the amount of stamps in dollars

$0.43 and $0.06

Let the count of these stamps be x and y

So, we have:

0.43x + 0.06y = 2.20

Multiply through by 100

43x + 6y = 220

Let x = 4.

So, we have:

43* 4 + 6y = 220

This gives

6y = 48

Divide by 6

y = 8

Hence, Caitlyn used 4 43-cent stamps and 8 6-cent stamps

Read more about linear equation at:

brainly.com/question/14323743

#SPJ1

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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)

=  25(0.04) + 32.5(0.20) + 40(0.25) + 45(0.12)  +  52.5(0.30)  +  65(0.09)

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From the same table above ;

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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