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ICE Princess25 [194]
3 years ago
11

What is the area of this circle?

Mathematics
2 answers:
kykrilka [37]3 years ago
6 0

Answer:

It should be 1,256.6 in square inches

Step-by-step explanation:

Degger [83]3 years ago
5 0

Answer:

100 pi (314.16)

Step-by-step explanation:

Because the diameter is 20 in, that makes the radius = 20/2 inches = 10 inches.

10^2 * pi = area

100 pi =  area

314.16 is approx. area.

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Which mixed number is equal to 7.6
Natali [406]
A mixed number is a number that has a whole number, and a fraction, so the mixed number that is equal to 7.6 is 7 6/10, simplified it would be 7 3/5
8 0
3 years ago
Read 2 more answers
Need help with number 6 plssssssssssassssa
Advocard [28]
C I think sorry I will try I will get other answers later
8 0
3 years ago
I will show you the qustion
Nutka1998 [239]

Question:

Solution:

The solution to a system of linear equations is where the graphs of each linear equation intersect. In this case, we can see that the lines intersect at the point:

(-3,2)

thus, the correct answer is (-3,2).

3 0
1 year ago
Super fine 40-gauge copper wire has a diameter of only 0.080mm and weighs only 44.5/gkm . Suppose a spool of 40-gauge wire weigh
guajiro [1.7K]

Answer:

Math expression=      L[m]=   W[g]   /   R[g/m]

answer=4.606Km

Step-by-step explanation:

It is known that the cable weighs 44kg / km therefore if the weight is divided by this amount we can find how much cable length was spent

L=W/R

L= leght of wire used(m)

W=weight used(g)=205g

R= relation between weight and leght(g/m)=44.5g/km=0.0445g/m

L=205g/0.0445g/m=  4606.74m=  4.606Km

6 0
3 years ago
Consider 3 trials, each having the same probability of successes. Let X denote the total number of successes in these trials.
Mariulka [41]

Since each trial has the same probability of success, 

Let, <span><span><span>Xi</span>=1</span></span> if the <span><span>i<span>th</span></span></span> trial is a success (<span>0</span> otherwise). Then, <span><span>X=<span>∑3<span>i=1</span></span><span>Xi</span></span><span>X=<span>∑<span>i=1</span>3</span><span>Xi</span></span></span>, 

and <span><span>E[X]=E[<span>∑3<span>i=1</span></span><span>Xi</span>]=<span>∑3<span>i=1</span></span>E[<span>Xi</span>]=<span>∑3<span>i=1</span></span>p=3p=1.8</span><span>E[X]=E[<span>∑<span>i=1</span>3</span><span>Xi</span>]=<span>∑<span>i=1</span>3</span>E[<span>Xi</span>]=<span>∑<span>i=1</span>3</span>p=3p=1.8</span></span>

So, <span><span>p=0.6</span><span>p=0.6</span></span>, and <span><span>P{X=3}=<span>0.63</span></span><span>P{X=3}=<span>0.63</span></span></span>

I thought what I did was sound, but the textbook says the answer to (a) is <span>0.60.6</span> and (b) is <span>00</span>.

Their reasoning (for (a)) is as follows:

3 0
3 years ago
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