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lesya692 [45]
3 years ago
7

the surface area of a rectangular prism is 190 square inches the length is 10 inches and the width is three inches find the heig

ht
Mathematics
1 answer:
Nastasia [14]3 years ago
7 0

Answer:

h=5in

Step-by-step explanation:

The following areas are equal:

1. Front and Back.

2. Left and right.

3. Top and Bottom.

Therefore, we calculate 3 areas and

multiply each by 2 to get the 6 areas:

As=2(3*h) + 2(10*h) + 2(10*3)=1901n^2.

6h + 20h + 60 = 190,

26h = 190 - 60 = 130,

h = 5in.

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QveST [7]

Answer:

4.8

Step-by-step explanation:

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3 years ago
The sum of a number and 8 is -26
ArbitrLikvidat [17]

<h2><u>PLEASE MARK BRAINLIEST!</u></h2>

Answer:

What equals ?

Step-by-step explanation:

? + 8 = -26

To find the ?, all you have to do is subtract 8 from -26.

-26 - 8     = ?

-26 + (-8) = ?

-34           = ?

<h3>Your answer is -34.</h3>

I hope this helps!

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3 0
4 years ago
Please help me
puteri [66]
I will rounded up too 45 right answer
3 0
3 years ago
A 2,768–foot–long straight fence has posts that are set 8 feet on center; that is, the distance between the centers of two adjac
salantis [7]

Answer:

No.P=347

Step-by-step explanation:

The fence is 2768 feet long, how many posts should be placed along the fence if the adjacent distance between them is 8 feet?

How to start and end with a post the number of them (No.P) will always be the distance of the fence (f) between the adjacent distance of the posts (p) plus one     No.P = \frac{f}{p}+1= \frac{2768}{8}+1 = 346+1=347

As an example we have graph 1

5 0
3 years ago
Please answer all parts as I know the answers but need the work to go with them. Thus, I believe the above answers are correct.
vampirchik [111]

As the sample size n is less than 30 normal distribution is used.

The values of x are converted into z by using the formula z= x-u/ s and then the z values are found out from the table.

The limits are found by using the formula x±σz or x±sz where s= σ

As the sample size is 10 which is less than 30 the normal distribution is used.

The probability of x< 2.59 is 0.3446

The probability 2.60<X <2.63 is 0.9484

So lower and upper limits are 2.607 and 2.612

Part A

As the sample size is 10 which is less than 30 the normal distribution is used.

Part B

For given value of x= 2.59 z is obtained =0.4

x= 2.59

z= x-u/ s

z= 2.59-2.61/0.05

z= -0.02/0.05

z=- 0.4

P (X<2.59) = P(-0.4 <Z<0) = 0.5 -0.1554= 0.3446

The probability of x< 2.59 is 0.3446

Part C

For two given values of x= 2.60 and 2.63 z is obtained as =0.2 and 0.4

x1= 2.60

z1= x-u/ s

z= 2.60-2.61/0.05

z= -0.01/0.05

z=- 0.2

x2= 2.63

z2= x-u/ s

z= 2.63-2.61/0.05

z= 0.02/0.05

z= 0.4

P (2.60<X<2.63) = P(-0.2 <Z<0.4)

= P(-0.2 <Z<0)+ P(0 <Z<0.4)

=0.793 + 0.1554= 0.9484

The probability 2.60<X <2.63 is 0.9484

Part D"

p= 0.57

From the table z= 0.045

z= x-u/ s

zs= x-u

zs+u = x

x1= 0.045*0.05 +2.61= 2.61225

x2= 2.61- 0.00225= 2.60775

So lower and upper limits are 2.607 and 2.612

For further understanding of probability calculation click

brainly.com/question/25638875

#SPJ1

8 0
2 years ago
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