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3 years ago

17cm per minute = meters per hour

1 answer:

7 0

Answer: 10.2 meters per hour; or, write as: 10 ⅕ meters per hour.
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Explanation:
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Note: "h" = time in "hour(s)" ;
"cm" = length in "centimeter(s)" ;
"min" = time in "minute(s)";
"m" = length in "meter(s)"
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Note these EXACT conversions:
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100 cm = 1 m ;
60 min = 1 h ;
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To solve:

Given: 17 cm / min ; convert to: ________ m / h
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(17 cm / min)* (1 m /100cm) * (60 min / 1 h) = _______ m / h
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The "cm" units cancel to "1"; the "min" units cancel to "1" and we are left with units of "m/h" {"meters per hour"}.
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We are left with (17 *1 m * 60) / (100 * 1 h) = [(17 * 60) m] / [100 h]
= [(17 *60) / 100 ] meters per hour.
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To simply: (17 * 60) / 100 ;
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Method 1): To simplify: (17 * 60) / 100 ;

→Rewrite as: (17*60) / 100 = (17*20*3)/(20*5) ;
→Cancel out the "20's "; and rewrite as:

→ (17*60) / 100 = (17*3)/5 = 51/5
= 10.2 meters per hour; or, write as: 10 ⅕ meters per hour.
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Method 2) To simplify: (17 * 60) / 100 ; Use calculator (or by hand):
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→ (17 * 60) / 100 = 1,020 /100
= 10.2 meters per hour; or, write as: 10 ⅕ meters per hour.
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Method 3) To simplify: (17 * 60) / 100 ;
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→ (17 * 60) / 100 = ? ;

→ Divide BOTH the "100" AND the "60" by "10";

→ 60÷10 = 6 ; 100÷10 =10; and rewrite—replace the "60" with a "6"; and replace the "100" with a "10" ;
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→ (17*60)/100 = (17*6)/10 = 102/10

= 10.2 meters per hours; or, write as: 10 ⅕ meters per hour.
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1 year ago

The radius of a semicircle is 1 kilometer. what is the semicircle's perimeter? use 3.14 for pie.

Aleksandr-060686 [28]

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3.14 * 1

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3 years ago

Chris has 115$ to spend at a shopping mall . If he decides to spend $42 on clothing and $37 on food how much money will he have

Katena32 [7]

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3 years ago

A telemarketer is successful at getting people to donate money for her organization in 55% of all calls she makes. She must get

Tems11 [23]

An interesting twist to a binomial distribution problem.

Given:
p=55%=0.55 for probability of success in solicitation
x=4=number of successful solicitations
n=number of calls to be made
P(x,n,p)>=89.9%=0.899 (from context, it is >= and not =, which is almost impossible)

From context of question, all calls are assumed independent, with constant probability of success, so binomial distribution is applicable.

The number of successes, x, is then given by
$P(x)=C(n,x)p^x(1-p)^{n-x}$ where
p=probability of success
n=number of trials
x=number of successes $C(n,x)=\frac{n!}{x!(n-x)!}$

Here we need n such that
P(x,n,p)>=0.899
given
x>=4, p=0.55, which means we need to find

Method 1: if a cumulative binomial distribution table is available, we can look up n=9,10,11 and find
P(x>=4,9,0.55)=0.834
P(x>=4,10,0.55)=0.898
P(x>=4,11,0.55)=0.939
So she must make (at least) 11 calls to make sure the probability of meeting her quota is 89.9% or more.

Method 2: using technology.
Similar to method 1, we can look up the probabilities directly, for n=9,10,11
P(x>=4,9,0.55)=0.834178
P(x>=4,10,0.55)=0.8980051
P(x>=4,11,0.55)=0.9390368

Method 3: using simple calculator
Here we need to calculate the probabilities for each value of n=10,11 and sum the probabilities of FAILURE S=P(0,n,0.55)+P(1,n,0.55)+P(2,n,0.55)+P(3,n,0.55)
so that the probability of success is 1-S.
For n=10,
P(0,10,0.55)=0.000341
P(1,10,0.55)=0.004162
P(2,10,0.55)=0.022890
P(3,10,0.55)=0.074603
So that
S=0.000341+0.004162+0.022890+0.074603
=0.101995
and Probability of getting 4 successes (or more)
=1-S
=0.898005, missing target by 0.1%

So she will have to make 11 phone calls, bring up the probability to 93.9%. The work is similar to that of n=10.

8 0

3 years ago