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2 years ago

A rectangular prism has length of 5 feet and a width of 9 feet. If the surface area of the prism is 174 square feet,

Mathematics

1 answer:

Masteriza [31]2 years ago

8 0

kung 5ft and a width of 9ft and the area of prism is 174 the answer is 5ft

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WHATS THE AREA OF THE TRAPIZOID?!

Helen [10]

Answer: 45.. i think.

Step-by-step explanation: I couldn't really get the measurements from this photo so here is what I got. I used the formula A=a+b over 2 times h. I used 7=a, b=11,h=5. I added 7 and 11 to get 18, then divide that by 2, which equals 9. Then 9 times 5 = 45. I hope this somewhat helps.

3 0

3 years ago

Solve <br><br> LaTeX: \frac{-x}{3}<5

I am Lyosha [343]

Answer:

x > -15

Step-by-step explanation:

-x / 3 < 5

(-x / 3) * 3 < 5 * 3

-x < 15

x > -15 (Remember to flip the sign when multiplying or dividing an inequality by a negative number.)

3 0

3 years ago

Read 2 more answers

Linear Relationships Study Guide 1.) Select all the equations for which (-6, -1) is a solution. Show your work to prove that eac

TiliK225 [7]

we have point (-6, - 1)

Now we will put these points in each equation,

y = 4x +23

put x = -6 and y = -1

-1 = 4 (-6) +23

-1 = -24 + 23

-1 = -1

LHS = RHS, so this equation has (-6 , -1) as solution.

y = 6x

put x = -6 and y = -1

-1 = 6 (-6)

-1 not= -36

LHS is not equal RHS, so (-6 , -1) is not a solution for that equation,

y = 3x - 5

put x = -6 and y = -1

-1 = 3 (-6) - 5

-1 = -18 - 5

-1 not= -23

LHS is not equal RHS, so (-6 , -1) is not a solution for that equation,

y= 1/6 x

put x = -6 and y = -1

-1 = -6/6

-1 = -1

LHS = RHS, so (-6 , -1) is a solution for that equation,

6 0

1 year ago

Calculate the sample mean and sample variance for the following frequency distribution of heart rates for a sample of American a

spin [16.1K]

Answer:

$Mean = 68.9$

$s^2 =18.1$ --- Variance

Step-by-step explanation:

Given

$\begin{array}{cccccc}{Class} & {51-58} & {59-66} & {67-74} & {75-82} & {83-90} \ \\ {Frequency} & {6} & {3} & {11} & {13} & {4} \ \end{array}$

Solving (a): Calculate the mean.

The given data is a grouped data. So, first we calculate the class midpoint (x)

For 51 - 58.

$x = \frac{1}{2}(51+58) = \frac{1}{2}(109) = 54.5$

For 59 - 66

$x = \frac{1}{2}(59+66) = \frac{1}{2}(125) = 62.5$

For 67 - 74

$x = \frac{1}{2}(67+74) = \frac{1}{2}(141) = 70.5$

For 75 - 82

$x = \frac{1}{2}(75+82) = \frac{1}{2}(157) = 78.5$

For 83 - 90

$x = \frac{1}{2}(83+90) = \frac{1}{2}(173) = 86.5$

So, the table becomes:

$\begin{array}{cccccc}{x} & {54.5} & {62.5} & {70.5} & {78.5} & {86.5} \ \\ {Frequency} & {6} & {3} & {11} & {13} & {4} \ \end{array}$

The mean is then calculated as:

$Mean = \frac{\sum fx}{\sum f}$

$Mean = \frac{54.5*4+62.5*3+70.5*11+78.5*13+86.5*4}{6+3+11+13+4}$

$Mean = \frac{2547.5}{37}$

$Mean = 68.9$ -- approximated

Solving (b): The sample variance:

This is calculated as:

$s^2 =\frac{\sum (x - \overline x)^2}{\sum f - 1}$

So, we have:

$s^2 =\frac{(54.5-68.9)^2+(62.5-68.9)^2+(70.5-68.9)^2+(78.5-68.9)^2+(86.5-68.9)^2}{37 - 1}$

$s^2 =\frac{652.8}{36}$

$s^2 =18.1$ -- approximated

5 0

3 years ago

Please help me with this

Naya [18.7K]

I think N? Because it’s took 100 kilo and The other it took less

6 0

2 years ago

Read 2 more answers