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aleksandrvk [35]
3 years ago
7

Adrian wants his lawn to be mown 3 men apply for the task the first man can mow the lawn in 6 hours the second man can mow the l

awn in 4 hours the third man can mow the lawn in 3 hours how long will it take for the 3 men to complete the task if all of them mowed the lawn together
SHOW WORK PLS
Mathematics
1 answer:
matrenka [14]3 years ago
3 0

First man mows the lawn in 6 hours so he mows 1/6 of the lawn per hour

Second man mows at 1/4 lawn per hour

Third man mows at 1/3 lawn per hour

The total of the three: 1/6 + 1/4 + 1/3 = 2/12 + 3/12 + 4/12 = 9/12 = 3/4 of a lawn per hour.

1 lawn / 3/4 lawn per hour =  1 and 1/3 hours all together.

Answer: 1 and 1/3 hours

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If log(base4)x=2.5 and log(base Y)125=-3/2, find the numerical value of x/y, in simplest form.
Naya [18.7K]
X = (2^2)^(2.5) 
<span>x = 2^(2 * 2.5) </span>
<span>x = 2^5 </span>
<span>x = 32 
</span>y^(-3/2) = 125 
<span>y^(-3) = 125^2 </span>
<span>y^(-3) = (5^3)^2 </span>
<span>y^(-3) = (5^2)^3 </span>
<span>y^(-3) = 25^3 </span>
<span>y = 25^(-1) </span>
<span>y = 1/25 </span>


<span>x/y => </span>
<span>32 / (1/25) => </span>
<span>32 * 25 => </span>
<span>800 is the simplest form of above
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3 years ago
What is 3/8 of 200?<br> pls show work<br> 35 POINTS!!!
yanalaym [24]

Answer:

The answer is 75

Step-by-step explanation:

3/8*200=75.

3 0
3 years ago
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7 0
4 years ago
8. GradPoint question
Naddika [18.5K]

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Step-by-step explanation:

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3 years ago
Historically, there is a 40% chance of having clear sunny skies in Seattle in July. Let's assume that each day is independent fr
Igoryamba

Answer: a)  0.0058

b) 0.0026

Step-by-step explanation:

Given : The probability of having clear sunny skies in Seattle in July : p= 0.40

The number of days spent in Seattle  in July: n= 18

a) Using, Binomial probability formula : P(x)=^nC_xp^x(1-p)^{n-x}

The probability of having clear sunny skies on at least 13 of those days:-

P(x\geq13)=P(13)+P(14)+P(15)+P(16)+P(17)+P(18)\\\\=^{18}C_{13}(0.4)^{13}(0.6)^5+^{18}C_{14}(0.4)^{14}(0.6)^4+^{18}C_{15}(0.4)^{15}(0.6)^3+^{18}C_{16}(0.4)^{16}(0.6)^2+^{18}C_{17}(0.4)^{17}(0.6)^1+^{18}C_{18}(0.4)^{18}(0.6)^0

=\dfrac{18!}{13!5!}(0.4)^{13}(0.6)^5+\dfrac{18!}{14!4!}(0.4)^{14}(0.6)^4+\dfrac{18!}{15!3!}(0.4)^{15}(0.6)^3+\dfrac{18!}{16!2!}(0.4)^{16}(0.6)^2+(18)(0.4)^{17}(0.6)^1+(1)(0.4)^{18}

=0.00447111249474+0.00106455059399+0.000189253438931+0.0000236566798664+0.00000185542587187+0.000000068719476736

=0.00575049735288\approx0.0058

b) On converting binomial to normal distribution, we have

\text{Mean=}\mu=np= 18\times0.40=7.2\\\\\text{Standard deviation}=\sigma=\sqrt{np(1-p)}\\\\=\sqrt{18(0.40)(1-0.40)}=2.07846096908\approx2.08

Let x be the number of days having clear sunny skies in Seattle in July.

Then, using z=\dfrac{x-\mu}{\sigma} we have

z=\dfrac{13-7.2}{2.08}=2.78846153846\approx2.79

P-value = P(x\geq13)=P(z\geq2.79)=1-P(z

=1-0.9973645=0.0026355\approx0.0026

6 0
3 years ago
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