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3 years ago

Help .....................

Mathematics

1 answer:

Snezhnost [94]3 years ago

6 0

Answer:

~ 2 kilometers in length of the actual path ~

Step-by-step explanation:

Let us plan out our steps, and solve for each:

1. Given the information, let us create a proportionality as such:

1 = 10,000 ⇒ x - centimeters in length of actual path

20 x

2. Now let us cross multiply, and solve through simple algebra for x:

10,000 * 20 = x,

x = 200,000 centimeters of the width of the item in reality

3. The answer demands in km, so let us convert 200,000 cm ⇒ km:

200,000/100,000 = 2 kilometers in length of the actual path

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Step-by-step explanation:

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3 years ago

Solve for x. Tx-ux=3t

Veronika [31]

x(t-u) = 3t
dividing both sides by (t-u):
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6 0

3 years ago

PLEASE HELP ASAP 25 PTS + BRAINLIEST TO RIGHT/BEST ANSWER

pishuonlain [190]

Answer:

3-i

Step-by-step explanation:

1-5i + 2 +4i

We add the real parts

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6 0

3 years ago

A researcher wishes to estimate the number of households with two cars. How large a sample is needed in order to be 99% confiden

solmaris [256]

Answer:

Sample size n = 1382

so correct option is D) 1382

Step-by-step explanation:

given data

confidence level = 99 %

margin of error = 3%

probability = 25 %

to find out

How large a sample size needed

solution

we know here P = 25 %

so 1 - P = 1 - 0.25

1 - P = 0.75

and we know E margin of error is 0.03 so value of Z for 99%

α = 1 - 99% = 1 - 0.99

α = 0.01

and $\frac{\alpha}{2}$ = $\frac{0.01}{2}$

$\frac{\alpha}{2}$ = 0.005

so Z is here

$Z_(\frac{\alpha}{2})$ = 2.576

sample size will be

Sample size n = $(\frac{(Z_(\frac{\alpha}{2})}{E})^2 * P * (1-P)$

put here value

Sample size n = (\frac{2.576}{0.03})^2 * 0.25 * 0.75

Sample size n = 1382

so correct option is D) 1382

3 0

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 A 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