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3 years ago

What does the equal sign with the wavy line above it mean?

Mathematics

1 answer:

Yanka [14]3 years ago

6 0

It means "approximately" instead of "equal to"

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What is the answer for "a boat travels 4 miles in 4 hours (with a constant speed). how far can it travel in 2 hours (with the sa

ElenaW [278]

4 miles in 4 hrs.....4/4 =1...thats 1 mile per hr

for 2 hrs.....at 1 mile per hr = 2 miles <==

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3 years ago

Why do purely competitive markets tend to benefit consumers over producers? Companies control quality and supply. Companies work

Tom [10]

In general, purely competitive markets tend to benefit consumers over producers because "<span>Consumers control price through demand". Although perfect competition is practically impossible. </span>

3 0

3 years ago

Read 2 more answers

Which is the best first step when solving the following system of equations? 5 c + 4 d = 10. 3 c + 2 d = 16. Multiply the first

Virty [35]

Answer:

5c+4d=10

3c+2d=16

8c+6c=26

8 0

2 years ago

How do i solve this?

bixtya [17]

Answer:

5.x^8.y^3

Step-by-step explanation:

15.x^10.y^5.z^2 ÷ 3.x^2.y^2.z^2

= 15/3. x^(10-2). y^(5-2). z^(2-2)

= 5.x^8.y^3.z^0

= 5.x^8.y^3

7 0

3 years ago

A company has a policy of retiring company cars; this policy looks at number of miles driven, purpose of trips, style of car and

pav-90 [236]

Answer:

ans=13.59%

Step-by-step explanation:

The 68-95-99.7 rule states that, when X is an observation from a random bell-shaped (normally distributed) value with mean $\mu$ and standard deviation $\sigma$ , we have these following probabilities

$Pr(\mu - \sigma \leq X \leq \mu + \sigma) = 0.6827$

$Pr(\mu - 2\sigma \leq X \leq \mu + 2\sigma) = 0.9545$

$Pr(\mu - 3\sigma \leq X \leq \mu + 3\sigma) = 0.9973$

In our problem, we have that:

The distribution of the number of months in service for the fleet of cars is bell-shaped and has a mean of 53 months and a standard deviation of 11 months

So $\mu = 53, \sigma = 11$

So:

$Pr(53-11 \leq X \leq 53+11) = 0.6827$

$Pr(53 - 22 \leq X \leq 53 + 22) = 0.9545$

$Pr(53 - 33 \leq X \leq 53 + 33) = 0.9973$

-----------

$Pr(42 \leq X \leq 64) = 0.6827$

$Pr(31 \leq X \leq 75) = 0.9545$

$Pr(20 \leq X \leq 86) = 0.9973$

-----

What is the approximate percentage of cars that remain in service between 64 and 75 months?

Between 64 and 75 minutes is between one and two standard deviations above the mean.

We have $Pr(31 \leq X \leq 75) = 0.9545 = 0.9545$ subtracted by $Pr(42 \leq X \leq 64) = 0.6827$ is the percentage of cars that remain in service between one and two standard deviation, both above and below the mean.

To find just the percentage above the mean, we divide this value by 2

So:

$P = {0.9545 - 0.6827}{2} = 0.1359$

The approximate percentage of cars that remain in service between 64 and 75 months is 13.59%.

4 0

4 years ago