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horrorfan [7]
3 years ago
5

Jhons garden has a big planter that mensures 18 2/3 in by 8 5/6 in. What Is the area of Jhons planter

Mathematics
1 answer:
klemol [59]3 years ago
8 0

Area of Jhons planter is 164.89 square inches

<em><u>Solution:</u></em>

Given that,

Jhons garden has a big planter that measures 18\frac{2}{3} in by 8 \frac{5}{6} inches

To find: area of Jhons planter

From given information,

\text{ length } = 18\frac{2}{3} = \frac{3 \times 18 + 2}{3} = \frac{56}{3} \text{ inches }

\text{ width } = 8\frac{5}{6} = \frac{6 \times 8 + 5}{6} = \frac{53}{6} \text{ inches }

The area of planter is given as:

area = length \times width

area = \frac{56}{3} \times \frac{53}{6} = \frac{2968}{18} = 164.89

Thus area of Jhons planter is 164.89 square inches

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Inessa05 [86]
First divide bot sides of the formula by 1/2(3.14):-

2A / 3.14 =  r^2

r = sqrt (2A/3.14)
7 0
3 years ago
You always need some time to get up after the alarm has rung. You get up from 10 to 20 minutes later, with any time in that inte
Mama L [17]

Answer:

a) P(x<5)=0.

b) E(X)=15.

c) P(8<x<13)=0.3.

d) P=0.216.

e) P=1.

Step-by-step explanation:

We have the function:

f(x)=\left \{ {{\frac{1}{10},\, \, \, 10\leq x\leq 20 } \atop {0, \, \, \, \, \, \,  otherwise }} \right.

a)  We calculate  the probability that you need less than 5 minutes to get up:

P(x

Therefore, the probability is P(x<5)=0.

b) It takes us between 10 and 20 minutes to get up. The expected value is to get up in 15 minutes.

E(X)=15.

c) We calculate  the probability that you will need between 8 and 13 minutes:

P(8\leq x\leq 13)=P(10\leqx\leq 13)\\\\P(8\leq x\leq 13)=\int_{10}^{13} f(x)\, dx\\\\P(8\leq x\leq 13)=\int_{10}^{13} \frac{1}{10} \, dx\\\\P(8\leq x\leq 13)=\frac{1}{10} \cdot [x]_{10}^{13}\\\\P(8\leq x\leq 13)=\frac{1}{10} \cdot (13-10)\\\\P(8\leq x\leq 13)=\frac{3}{10}\\\\P(8\leq x\leq 13)=0.3

Therefore, the probability is P(8<x<13)=0.3.

d)  We calculate the probability that you will be late to each of the 9:30am classes next week:

P(x>14)=\int_{14}^{20} f(x)\, dx\\\\P(x>14)=\int_{14}^{20} \frac{1}{10} \, dx\\\\P(x>14)=\frac{1}{10} [x]_{14}^{20}\\\\P(x>14)=\frac{6}{10}\\\\P(x>14)=0.6

You have 9:30am classes three times a week.  So, we get:

P=0.6^3=0.216

Therefore, the probability is P=0.216.

e)  We calculate the probability that you are late to at least one 9am class next week:

P(x>9.5)=\int_{10}^{20} f(x)\, dx\\\\P(x>9.5)=\int_{10}^{20} \frac{1}{10} \, dx\\\\P(x>9.5)=\frac{1}{10} [x]_{10}^{20}\\\\P(x>9.5)=1

Therefore, the probability is P=1.

3 0
3 years ago
7. Find the distance between the points (13, 8) and (-12, 6)
Alik [6]

9514 1404 393

Answer:

  √629 ≈ 25.08

Step-by-step explanation:

The distance formula is useful for this.

  d = √((x2 -x1)² +(y2 -y1)²)

  d = √((-12 -13)² +(6 -8)²) = √(625 +4) = √629 ≈ 25.08

The distance between the points is about 25.08 units.

3 0
2 years ago
From a random sample of size 18, a researcher states that (11.1, 15.7) inches is a 90% confidence interval for mu, the mean leng
alexandr402 [8]

Complete Question

From a random sample of size 18, a researcher states that (11.1, 15.7) inches is a 90% confidence interval for mu, the mean length of bass caught in a small lake. A normal distribution was assumed. Using the 90% confidence interval obtain:

a. A point estimate of \mu and its 90% margin of error.

b. A 95% confidence interval for \mu.

Answer:

a

\= x  = 13.4   .   E = 2.3

b

10.7 <  \mu < 16.1

Step-by-step explanation:

From the question we are told that

  The sample size is  n = 18

  The 90% confidence interval is  (11.1, 15.7)

Generally the point estimate of  \mu is mathematically  evaluated  as

       \= x  = \frac{11.1 + 15.7 }{2}

=>    \= x  = 13.4

Generally the margin of error is mathematically evaluated  as

     E = \frac{15.7 - 11.1}{2 }

=> E = 2.3

  From the question we are told the confidence level is  90% , hence the level of significance is    

      \alpha = (100 - 90 ) \%

=>   \alpha = 0.10

Generally from the normal distribution table the critical value  of  \frac{\alpha }{2} is  

   Z_{\frac{\alpha }{2} } =  1.645

Generally the equation for the lower limit of the confidence interval is  

      \= x  -  Z_{\frac{\alpha }{2} } * \frac{s}{\sqrt{18} } = 11.1

=> 13.4   -  0.3877 s  = 11.1

=>  s = 5.932

  From the question we are told the confidence level is  95% , hence the level of significance is    

      \alpha = (100 - 95 ) \%

=>   \alpha = 0.05

Generally from the normal distribution table the critical value  of  \frac{\alpha }{2} is  

   Z_{\frac{\alpha }{2} } =  1.96

Generally the margin of error is mathematically represented as  

      E = Z_{\frac{\alpha }{2} } *  \frac{\sigma }{\sqrt{n} }

=>    E =  1.96 *  \frac{5.932}{\sqrt{18} }

=>    E =  2.7      

Generally 95% confidence interval is mathematically represented as  

      \= x -E <  \mu <  \=x  +E

=>    13.4  -  2.7  <  \mu < 13.4  +   2.7

=>    10.7 <  \mu < 16.1

3 0
3 years ago
I need help on this question
Alex777 [14]
The awser for this problem im preety sure is going to be aswer c
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3 years ago
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