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just olya [345]
1 year ago
8

A figure is made up of a square and a rectangle. The area of the square is 64 cm². The breadth of the rectangle is the same leng

th of the square. If the total area of the figure is 232 cm², what is the perimeter of the whole figure? ​
Mathematics
1 answer:
antoniya [11.8K]1 year ago
6 0

Answer:

90 cm

Step-by-step explanation:

Let the side of the square be S

Area of the square = S² = 64

So, S = √64 = 8 cm

Let L be the length of the rectangle and B be the breadth

We know B = S = 8

Area of rectangle = BL = 8L

Total area = Area of square + area of rectangle

Total area   = 64 + 8L  = 232  (given)

Subtracting 64 from both sides yields

8L = 232-64 = 168 or L = 168/8 = 21

Perimeter of square = 4S = 4 x 8 = 32

Perimeter of rectangle = 2(B + L) = 2(8 + 21) = 2 x 29 = 58

Total perimeter = 32 + 58 = 90 cm

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Marina86 [1]

The answer is 52,000.


You have to add all of the values and divide them by the number of values. In this case you would do 30,000 + 50,000 + 30,000 + 80,000 + 70,000 which equals 260,000 then you would divide that number by 5 since there are 5 values.

5 0
3 years ago
The mayor is interested in finding a 98% confidence interval for the mean number of pounds of trash per person per week that is
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Solution :

Given :

Sample mean, $\overline X = 34.2$

Sample size, n = 129

Sample standard deviation, s = 8.2

a. Since the population standard deviation is unknown, therefore, we use the t-distribution.

b. Now for 95% confidence level,

   α = 0.05, α/2 = 0.025

  From the t tables, T.INV.2T(α, degree of freedom), we find the t value as

  t =T.INV.2T(0.05, 128) = 2.34

  Taking the positive value of t, we get  

  Confidence interval is ,

  $\overline X \pm t \times \frac{s}{\sqrt n}$

 $34.2 \pm 2.34 \times \frac{8.2}{\sqrt {129}}$

 (32.52, 35.8)

 95% confidence interval is  (32.52, 35.8)

So with $95 \%$ confidence of the population of the mean number of the pounds per person per week is between 32.52 pounds and 35.8 pounds.

c. About $95 \%$ of confidence intervals which contains the true population of mean number of the pounds of the trash that is generated per person per week and about $5 \%$ that doe not contain the true population of mean number of the pounds of trashes generated by per person per week.  

4 0
2 years ago
Samples of 20 parts from a metal punching process are selected every hour. Typically, 1% of the parts require rework. Let X deno
Roman55 [17]

Answer:

a) P(X>np+3\sqrt{np(1-p)}=0.017

b) P(x>1)=0.190

c) P(Y>1)=0.651

Step-by-step explanation:

This a binomial experiment where success is denoted by parts that need rework.

X ∼ B(n, p); n = 20; p = 0.01

The expected value of X is: E(X) = np =20×0.01= 0.2

The variance is: Var(X) = np(1 − p) = 0.2 × 0.99 = 0.198,

The standard deviation SD(X)= \sqrt{0.198} ≈ 0.445

a) P(X>np+3\sqrt{np(1-p)}=P(X>0.2+3×0.445)=P(X>1.535)=P(X≥2)

Probability function is given by:

\frac{n!}{x!(n-x)!} *p^x*(1-p)^{(n-x)}

P(X≥2)=1-P(X<2)=1-P(X=1)-P(X=0)= 1 - \frac{20!}{1!(20-1)!} *(0.01)^{1}*(1-0.01)^{(20-1)}-\frac{20!}{0!(20-0)!} *(0.01)^{0}*(1-0.01)^{(20-0)}

P(X≥2)=1-0.165-0.818=0.017

b) p=0.04

P(x>1)=P(x≥2)= 1 - P(x=1) - P(x=0)= 1 - \frac{20!}{0!(20-1)!} *(0.04)^{1}*(1-0.04)^{(20-1)} - \frac{20!}{0!(20-0)!} *(0.04)^{0}*(1-0.04)^{(20-0)}

P(x>1)= 1 - 0.368 - 0.442=0.190

c) In this case we consider p=0.19 (Probability that X exceeds 1)

In this experiment Y is the number of hours and n= 5 hours.

Then, we check the probability in each hour:

P(Y>1)=1- P(Y=0)

P(Y=0)=\frac{5!}{0!(5-0)!} *(0.19)^{0}*(1-0.19)^{(5-0)}=0.349

P(Y>1)=1-0.349=0.651

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Mr. Greg had 8 sandwiches. He gave 1/4 to Nehamiah, 1/4 to Elijah, and 1/4 to Elizabeth. How many sandwiches did Mr. Greg have l
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There are many ways to work this out. One way could be:

Add up:
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Find what's remaining:
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So Mr.Greg has sandwiches left.
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iogann1982 [59]

The function would be h = 270 - 2.5s, where h is the height (in feet) and s it the time (in seconds) that has passed.

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2.5s represents the feet that it has descended. We know the block is lowered at a rate of 2.5 feet per second. Multiplying this by s, the time, will give us how much it has descended.

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