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castortr0y [4]
3 years ago
15

Megan has 7 coins that make one pound. The coins are of only two different kinds. What are the 7 coins?

Mathematics
2 answers:
arsen [322]3 years ago
7 0
I'm probably thinking about half dollar coins and quarters... Not 99% sure, but it's my lucky guess...
FinnZ [79.3K]3 years ago
4 0
The answer is 7 pounds
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Evaluate the integral. 3 2 t3i t t − 2 j t sin(πt)k dt
sveta [45]

∫(t = 2 to 3) t^3 dt

= (1/4)t^4 {for t = 2 to 3}

= 65/4.

----

∫(t = 2 to 3) t √(t - 2) dt

= ∫(u = 0 to 1) (u + 2) √u du, letting u = t - 2

= ∫(u = 0 to 1) (u^(3/2) + 2u^(1/2)) du

= [(2/5) u^(5/2) + (4/3) u^(3/2)] {for u = 0 to 1}

= 26/15.

----

For the k-entry, use integration by parts with

u = t, dv = sin(πt) dt

du = 1 dt, v = (-1/π) cos(πt).


So, ∫(t = 2 to 3) t sin(πt) dt

= (-1/π) t cos(πt) {for t = 2 to 3} - ∫(t = 2 to 3) (-1/π) cos(πt) dt

= (-1/π) (3 * -1 - 2 * 1) + [(1/π^2) sin(πt) {for t = 2 to 3}]

= 5/π + 0

= 5/π.

Therefore,

∫(t = 2 to 3) <t^3, t√(t - 2), t sin(πt)> dt = <65/4, 26/15, 5/π>.

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
What is the answer to this
allochka39001 [22]
678.58 that is the surface area to this problem
5 0
3 years ago
Hans wants to earn more than $51 trimming trees. He charges $7 per hour and pays $5 in equipment fees. What are the possible num
Flura [38]

Answer:

h ≤ 8

Step-by-step explanation:

Hans charges $7 per hour and pays $5 in equipment fees.

$5 is fixed in this case. Let the possible number of hours be h.

According to question,

Atleast means less than equal to. So,

7h-5≤51

Add 5 to both sides.

7h+5-5 ≤ 51+5

7h ≤ 56

h ≤ 8

So, the number of hours be less than equal to 8.

7 0
2 years ago
Find the center and radius of the circle.<br> (x + 3)2 + (y + 4)2 =9
Tpy6a [65]

The general equation for any circle with its center at point (a, b) is

                 (x - a)² + (y - b)²  =  (radius)²

So the circle in the question is centered at (-3, -4), and its radius is  3 .

7 0
3 years ago
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