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Murrr4er [49]
3 years ago
10

Find the volume 15m 16m 16m 16m the picture is below please help

Mathematics
1 answer:
kolezko [41]3 years ago
3 0

Answer:

5,376 m³

Step-by-step explanation:

cube - 16x16x16=4096

pyramid = 16x16x15/3=1280

4096+1280=5376

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Please help: (picture)
Slav-nsk [51]

We need to solve the speed formula for d. To do so, let's start by moving the number of the left hand side:


\frac{S}{356} = \sqrt{d}


Square both sides to get rid of the square root:


\frac{S^2}{126736} = d


Now plug the known value of the speed to find the distance:


d = \frac{140^2}{126736} = \frac{19600}{126736} \approx 0.15465


So the closest answer is the last one: d=0.155km

3 0
3 years ago
In an experiment, college students were given either four quarters or a $1 bill and they could either keep the money or spend it
gavmur [86]

Answer:

a) P(A|B) = \frac{15/83}{44/83} =\frac{15}{44}=0.341

b) P(B|A) = \frac{29/83}{44/83} =\frac{29}{44}=0.659

c)  A. A student given a $1 bill is more likely to have kept the money.

Because the probability 0.659 is atmoslt two times greater than 0.341

Step-by-step explanation:

Assuming the following table:

                                                     Purchased Gum      Kept the Money   Total

Students Given 4 Quarters              25                              14                      39

Students Given $1 Bill                       15                               29                    44

Total                                                   40                              43                     83

a. find the probability of randomly selecting a student who spent the money, given that the student was given a $1 bill.

For this case let's define the following events

B= "student was given $1 Bill"

A="The student spent the money"

For this case we want this conditional probability:

P(A|B) =\frac{P(A and B)}{P(B)}

We have that P(A)= \frac{40}{83} , P(B)= \frac{44}{83}, P(A and B)= \frac{15}{83}

And if we replace we got:

P(A|B) = \frac{15/83}{44/83} =\frac{15}{44}=0.341

b. find the probability of randomly selecting a student who kept the money, given that the student was given a $1 bill.

For this case let's define the following events

B= "student was given $1 Bill"

A="The student kept the money"

For this case we want this conditional probability:

P(A|B) =\frac{P(A and B)}{P(B)}

We have that P(A)= \frac{43}{83} , P(B)= \frac{44}{83}, P(A and B)= \frac{29}{83}

And if we replace we got:

P(B|A) = \frac{29/83}{44/83} =\frac{29}{44}=0.659

c. what do the preceding results suggest?

For this case the best solution is:

A. A student given a $1 bill is more likely to have kept the money.

Because the probability 0.659 is atmoslt two times greater than 0.341

3 0
3 years ago
I NEED THIS ANSWERED I DONT UNDERSTAND
Alex_Xolod [135]

Answer: 3.84

Step-by-step explanation:

If you time 6 by 0.8 you get 4.8 then you times 4.8 by 0.8 you get 3.84

Hope this helps :)

7 0
3 years ago
Read 2 more answers
I need help with this question. Write the equation in standard form. Identify A,B, and C.
frutty [35]
\frac{5}{2}x-9=8y\\
\frac{5}{2}x-8y=9\\\\
A=\frac{5}{2}\\
B=-8\\
C=9
6 0
3 years ago
Consider the dilation of pentagon ABCDE by a scale
weeeeeb [17]

Answer:

A'B'= 6

CD= 10

Step-by-step explanation: I did it on edgenuity! : )

8 0
3 years ago
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