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3 years ago

HELP!! geometry homework

Mathematics

2 answers:

GaryK [48]3 years ago

8 0

Answer:

2794 in³

Step-by-step explanation:

Given

base (a) = 17 in

vertical height (h) = 29 in

Now

Volume of the square pyramid

= 1/3 a²h

= 1/3 * 17² * 29

= 2793.7 in³

= 2794 in³

hope it helps :)

ahrayia [7]3 years ago

6 0

Answer:

2794

Step-by-step explanation:

V=s^2 times height over three

17^2 times 29/3

PEMDAS, exponents first

17^2=289

PEMDAS, next up multiplication

289 times 29/3

8381/3 or 2793.7

round to nearest whole number

2794

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20 points I beg you help

svp [43]

Answer : -6

The common difference of a sequence is the value that changes every time the sequence continues.

In your sequence it is seen that 6 is subtracted from each term in order to get to the next in the sequence.

7 0

3 years ago

Read 2 more answers

Solve the simultaneous equation<br> 2x + 4y=1<br> 3x - 5y=7

nexus9112 [7]

Answer:

x = 3/2 or 1.5.

y = -1/2 or -0.5.

Step-by-step explanation:

Solve for one equation, one variable.

2x + 4y = 1

subtrract 4y from both sides.

2x = 1 - 4y

divide by 2 to get x by itself.

x = 1/2 - 2y

Now that we have this, we can put it into the other equation.

3x - 5y = 7

Substitute variable.

3(1/2 - 2y) - 5y = 7

Distribute the 3.

1.5 - 6y - 5y = 7

Combine terms.

-11y + 1.5 = 7

Subtract by 1.5.

-11y = 5.5

Divide by -11.

y = -1/2, or -0.5.

Now we can find x. Substitute this into the initial equation.

2x + 4y = 1

Substitute variable.

2x + 4(-1/2) = 1

Multiply.

2x + (-2) = 1

Simplify.

2x - 2 = 1

Add 2 to both sides.

2x = 3

Divide by 2.

x = 3/2, or 1.5.

x = 3/2 or 1.5.

y = -1/2 or -0.5.

6 0

3 years ago

A manufacturing company uses an acceptance scheme on items from a production line before they are shipped. The plan is a two-st

pishuonlain [190]

Answer:

a) The probability that a box containing 3 defectives will be shipped is $51.74\%$

b) The probability that a box containing only 1 defective will be sent back for screening is $17.39\%$

Step-by-step explanation:

a) The first step is to count the number of total possible random sets of taking a sample size of 4 items over 23 items of the box, so $\left[\begin{array}{ccc}23\\4\end{array}\right] =23C4=8855$

The second step is to count the number of total possible random sets of taking a sample size of 4 items over 20 items of the box (discounting the 3 defectives) as the possible ways to succeed, so $\left[\begin{array}{ccc}20\\4\end{array}\right] =20C4=4845$

Finally we need to compute $\frac{\# ways\ to\ succeed}{\# random\ sets\ of \ 4} =\frac{4845}{8855}=0.5471=P(S)$ , therefore the probability that a box containing 3 defectives will be shipped is $P(S)=54.71\%$

a) The first step is to count the number of total possible random sets of taking a sample size of 4 items over 23 items of the box, so $\left[\begin{array}{ccc}23\\4\end{array}\right] =23C4=8855$

The second step is to count the number of total possible random sets of taking a sample size of 4 items over 22 items of the box (discounting the defective 1) as the possible ways to succeed, so $\left[\begin{array}{ccc}22\\4\end{array}\right] =22C4=7315$

Then we need to compute $\frac{\# ways\ to\ succeed}{\# random\ sets\ of \ 4} =\frac{7315}{8855}=0.8260=P(S)$ , therefore the probability that a box containing 1 defective will be shipped is $P(S)=82.60\%$