If you are given a 2% raise and the inflatoin is 3%, you are

Mathematics

1 answer:

lozanna [386]3 years ago

6 0

Making less money in today’s dollars (option c)

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What is the interquartile range of the data shown in the box plot? Please answer fast

cupoosta [38]

(First one)Low - 18
Q1 - 20
Median - 25
Q3 - 30
(Last one) High - 34

5 0

3 years ago

What is the common ratio of the geometric sequence: -1/12.-1/2,-3,18,...

Minchanka [31]

$\frac{1}{6}$ : Common ratio

Check:
= $-\frac{1}{12}$ ÷ $\frac{1}{6}$
= $-\frac{1}{12}$ × 6
= $-\frac{1}{2}$

= $-\frac{1}{2}$ ÷ $\frac{1}{6}$
= $- \frac{1}{2}$ × 6
=-3

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3 years ago

The length and width of a rectangle are measured as 55 cm and 49 cm, respectively, with an error in measurement of at most 0.1 c

valentina_108 [34]

Answer:

The maximum error in the calculated area of the rectangle is $10.4 \:cm^2$

Step-by-step explanation:

The area of a rectangle with length $L$ and width $W$ is $A= L\cdot W$ so the differential of <em>A</em> is

$dA=\frac{\partial A}{\partial L} \Delta L+\frac{\partial A}{\partial W} \Delta W$

$\frac{\partial A}{\partial L} = W\\\frac{\partial A}{\partial W}=L$ so

$dA=W\Delta L+L \Delta W$

We know that each error is at most 0.1 cm, we have $|\Delta L|\leq 0.1$ , $|\Delta W|\leq 0.1$ . To find the maximum error in the calculated area of the rectangle we take $\Delta L = 0.1$ , $\Delta W = 0.1$ and $L=55$ , $W=49$ . This gives