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MA_775_DIABLO [31]
2 years ago
15

In a bag, there are a total of 50 dimes and quarters. 40% of the coins are quarters and 35% of the quarters are on heads. How ma

ny of the quarters are on tails? ​
Mathematics
1 answer:
erik [133]2 years ago
7 0

Answer:

13 quarters must be tails.

Step-by-step explanation:

Number of quarters = 40% of 50 coins

Number of quarters = 40/100 * 50 = 20 coins which are quarters.

35% of these 20 quarters on heads. We can proceed in 2 ways from here.

We can actually find the number of quarters that are on heads.

35/100 * 20 = 7 quarters are showing heads.

Since the other quarters must be tails 20 - 7 = 13 quarters must be tails.

The other way to do it is to realize that if 35% of the coins are heads, 65% of the coins must be tails 20 * 65/100 = 13 which is a bit more direct.

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Write an equation in slope intercept form that describes a line with a slope of 3 and containing the point (5,-1)
soldier1979 [14.2K]

Answer:

<h2>         y = 3x - 16</h2>

Step-by-step explanation:

Equation of line with a given slope of a and containing a given point (x₀, y₀):

y - y₀ = a(x - x₀)

a = 3,  x₀ = 5,  y₀ = -1

So:

y - (-1) = 3(x - 5)  

y = 3x - 15 - 1

<u>y = 3x - 16</u>

3 0
2 years ago
He daily milk production of guernsey cows is approximately normally distributed with a mean of 35 kg/day and a standard deviatio
soldier1979 [14.2K]

Answer:

25.1 kg

Step-by-step explanation:

A suitable probability calculator can tell you the 5th percentile of this distribution.

7 0
3 years ago
What is the midpoint of (-23,-14) and (-18,2)
zloy xaker [14]

Given two coordinates point 1 as (-23,-14) and point 2 as(-18,2).

To find the midpoint we would use the formula below;

\text{midpoint(x,y)}=(\frac{x_1+x_2_{}_{}}{2},\frac{y_1+y_2}{2})

Where x1=-23, x2=-18, y1=-14, y2=2

We would substitute the values into the midpoint formula.

\begin{gathered} \text{midpoint =(}\frac{-23-18}{2},\frac{-14+2}{2}) \\ =(-\frac{41}{2},-6) \end{gathered}

Therefore, the answer is

\text{midpoint}=(-\frac{41}{2},-6)

7 0
1 year ago
Geometry help please!!!
julia-pushkina [17]
Before we begin, let's identify what kind of angles these are and are they related in any way?

These angles are both acute and they are both corresponding angles.
Corresponding angles are equal to each other, and we can use this fact to our advantage.

Since they are equal to each other, we can set the equations of 1 and 2 equal to each other. Like so,

1 = 2
83 - 2x = 92 - 3x

Now, we can solve for X by isolating it on one side.

83 - 2x = 92 - 3x

Add 3x to each side: (This basically moves the X on the right side to the left.)
83 - 2x + 3x = 92 - 3x + 3x
83 + x = 92

Subtract 83 on each side to isolate the X.
83 + x - 83 = 92 - 83 
x = 92 - 83
x = 9

Therefore, X equals 9. To check our work, we can substitute X for 9.
83 - 2(9) = 92 - 3(9)
83 - 18 = 92 - 27
65 = 65  - 
TRUE

So to conclude, Angle 1 is 65 degrees, Angle 2 is 65 degrees, and X equals 9.

Hope I could help you out!
If my answer is incorrect, or I provided an answer you were not looking for, please let me know. However, if my answer is explained well and correct, please consider marking my answer as Brainliest!  :)

Have a good one.
God bless!
7 0
3 years ago
There is 1000cm3 of aluminum available to cast a trophy that will be in the shape of a right square pyramid. Is this enough alum
stealth61 [152]

The 1000 cubic centimeters of aluminium is enough for aluminium a trophy  that will be in the shape of a right square pyramid and has a base edge of 10 cm and a slant height of 13 cm.

Step-by-step explanation:

The given is,

                    Volume of aluminium available is 1000 cubic centimeters

                    Shape of trophy is right square pyramid

                    Trophy has a base edge of 10 cm and slant height of 13 cm

Step:1

                     Formula for volume of right square pyramid,

                                               Volume, V = a^{2}\frac{h}{3}.....................................(1)

                     Where, a - Base edge value

                                  h - Height of pyramid

                      From given,

                                        a = 10 cm

                                        h = 13 cm

                      Equation (1) becomes,

                                           = 10^{2}(\frac{13}{3}  )

                                           = (100)(4.333)

                                           = 433.33 cm^{3}

             Volume of trophy = 433.33 cubic centimeters

             Compare with the volume of available aluminium and volume of right square pyramid,                          

            Volume of available aluminium > Volume of right square pyramid

                                               1000 cm^{3} > 433.33 cm^{3}

            So, the given volume of aluminium is enough to make right square pyramid shaped trophy.

Result:

          The 1000 cubic centimeters of aluminium is enough for aluminium a trophy  that will be in the shape of a right square pyramid and has a base edge of 10 cm and a slant height of 13 cm.

6 0
3 years ago
Read 2 more answers
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