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krek1111 [17]
3 years ago
9

The radius of a cylindrical water tank is 1.5 ft, and its height is 16 ft. What is the volume of the tank?

Mathematics
2 answers:
sesenic [268]3 years ago
7 0
The first answer is right yes
Anni [7]3 years ago
5 0

Answer:

volume \: of \: the \: pipe = \pi {r}^{2} h \\  = 3.14 \times  {1.5}^{2}  \times 16 \\ =  113.04 {ft}^{3}

113 ft³ to the nearest whole number

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astraxan [27]

Answer: yes.

Step-by-step explanation:

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Find the coordinates of the midpoint of HX given that H(-1,3) and X(7,-1).
Elis [28]
A. find the average of both x's for the x value, and the average of both y's for the y value. (-1+7)=3  (3+(-1))=1
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What is the solution for p in the equation?
Nookie1986 [14]

Answer:

p=-6

Step-by-step explanation:

the first thing to do is to add all like terms for example \frac{1}{3}p+\frac{1}{2}p would turn into \frac{2}{6}p+\frac{3}{6}p to make it so we can add them together which after we add both of those p's together we get \frac{5}{6}p which now we have \frac{5}{6}p+7=\frac{7}{6}p+5+4 which now we add 5 and 4 which we get 9 which now the equation looks like \frac{5}{6}p+7=\frac{7}{6}p+9 which now we subtract 7 from 9 and we are left with 2 so now the equation is \frac{5}{6}p=\frac{7}{6}p+2 which now we subtract \frac{7}{6}p from \frac{5}{6}p and we are left with -\frac{2}{6}p=2 which now we divide both sides by -\frac{2}{6} which makes the equation look like p=2÷-\frac{2}{6} which equals -6 so all we are left with is p=-6

6 0
3 years ago
1) On a standardized aptitude test, scores are normally distributed with a mean of 100 and a standard deviation of 10. Find the
Musya8 [376]

Answer:

A) 34.13%

B)  15.87%

C) 95.44%

D) 97.72%

E) 49.87%

F) 0.13%

Step-by-step explanation:

To find the percent of scores that are between 90 and 100, we need to standardize 90 and 100 using the following equation:

z=\frac{x-m}{s}

Where m is the mean and s is the standard deviation. Then, 90 and 100 are equal to:

z=\frac{90-100}{10}=-1\\ z=\frac{100-100}{10}=0

So, the percent of scores that are between 90 and 100 can be calculated using the normal standard table as:

P( 90 < x < 100) = P(-1 < z < 0) = P(z < 0) - P(z < -1)

                                                =  0.5 - 0.1587 = 0.3413

It means that the PERCENT of scores that are between 90 and 100 is 34.13%

At the same way, we can calculated the percentages of B, C, D, E and F as:

B) Over 110

P( x > 110 ) = P( z>\frac{110-100}{10})=P(z>1) = 0.1587

C) Between 80 and 120

P( 80

D) less than 80

P( x < 80 ) = P( z

E) Between 70 and 100

P( 70

F) More than 130

P( x > 130 ) = P( z>\frac{130-100}{10})=P(z>3) = 0.0013

8 0
3 years ago
Can someone please help me with this? Right answer gets brainliest!!
Kipish [7]

Answer:

Part 1: y= 49

Part 2: y=12

Step-by-step explanation:

Part 1: The y is going up by 7, so if you add 7 to 42, you get 49

Part 2: The y is going up by 4, so if you add 4 to 8, you get 12

8 0
3 years ago
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