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3 years ago

These box plots show daily low temperatures for a sample of days in two

Mathematics

1 answer:

vladimir2022 [97]3 years ago

3 0

It's D. the median for town A, 30, is less than the median for town B, 40.

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Solve for x <br>4^x/9^x=½<br>

kirill115 [55]

Answer:

x = 0.85476

Step-by-step explanation:

4ˣ/9ˣ=½

9ˣ / 4ˣ = 2

(9/4)ˣ = 2

2.25ˣ = 2 .... aⁿ = x logₐx = n

log₂.₂₅2 = x

x = 0.85476

3 0

3 years ago

Select the correct answer.

sladkih [1.3K]

Answer:

Step-by-step explanation:

Because this ia so simple I don't even need to explain to waste my last few brain cells.

8 0

3 years ago

Read 2 more answers

N/1.6 = 5 what does n equal

FinnZ [79.3K]

N= 8
b/c if u do it backwards 5*1.6=8- you get 8
and to check 8/1.6=5

so n= 5
hope this helps

6 0

3 years ago

Which one is lower 31.54 for 2 hours or 43.96 for 3.5 hours

pychu [463]

The answer would be 43.96 for 3.5 hours because for each hour it's 12.56

7 0

4 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

$dA=49\cdot 0.1+55 \cdot 0.1$

$dA=10.4$

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

4 0

3 years ago