Find the circumference of the circle, round to the nearest tenth 21 m Answer:

Mathematics

2 answers:

Lorico [155]2 years ago

6 0

Answer:

Circumference is 131.94

Step-by-step explanation:

Circumference Formula is C=2πr

Enter 21 for r so it will look like C=2(π)(21)

That makes it equal 131.9468915 so since you are rounding to the nearest tenth, you go next to the 4 and since it is a 6, it will round the 4 up to a 5

so that will make the answer 131.94

Hope this helps!

Liono4ka [1.6K]2 years ago

4 0

Answer:

yeetyeetyeetyeetyeetyeet

Step-by-step explanation:

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the ratio of men to women working for a company is 4 to 7 if there are 231 employees total how many women work for the company.

Dvinal [7]

Answer:

297

Step-by-step explanation:

8 0

2 years ago

Find the indicated probability. round to three decimal places. a test consists of 10

Charra [1.4K]

To answer this problem, we use the binomial distribution formula for probability:

P (x) = [n! / (n-x)! x!] p^x q^(n-x)

Where,

n = the total number of test questions = 10

x = the total number of test questions to pass = >6

p = probability of success = 0.5

q = probability of failure = 0.5

Given the formula, let us calculate for the probabilities that the student will get at least 6 correct questions by guessing.

P (6) = [10! / (4)! 6!] (0.5)^6 0.5^(4) = 0.205078

P (7) = [10! / (3)! 7!] (0.5)^7 0.5^(3) = 0.117188

P (8) = [10! / (2)! 8!] (0.5)^8 0.5^(2) = 0.043945

P (9) = [10! / (1)! 9!] (0.5)^9 0.5^(1) = 0.009766

P (10) = [10! / (0)! 10!] (0.5)^10 0.5^(0) = 0.000977

Total Probability = 0.376953 = 0.38 = 38%

There is a 38% chance the student will pass.

7 0

2 years ago

Can I get help solving this graph please?

Daniel [21]

see the attached figure with the letters

1) find m(x) in the interval A,B
A (0,100) B(50,40) -------------- > p=(y2-y1(/(x2-x1)=(40-100)/(50-0)=-6/5
m=px+b---------- > 100=(-6/5)*0 +b------------- > b=100
mAB=(-6/5)x+100

2) find m(x) in the interval B,C
B(50,40) C(100,100) -------------- > p=(y2-y1(/(x2-x1)=(100-40)/(100-50)=6/5
m=px+b---------- > 40=(6/5)*50 +b------------- > b=-20
mBC=(6/5)x-20

3) find n(x) in the interval A,B
A (0,0) B(50,60) -------------- > p=(y2-y1(/(x2-x1)=(60)/(50)=6/5
n=px+b---------- > 0=(6/5)*0 +b------------- > b=0
nAB=(6/5)x

4) find n(x) in the interval B,C
B(50,60) C(100,90) -------------- > p=(y2-y1(/(x2-x1)=(90-60)/(100-50)=3/5
n=px+b---------- > 60=(3/5)*50 +b------------- > b=30
nBC=(3/5)x+30

5) find h(x) = n(m(x)) in the interval A,B
mAB=(-6/5)x+100
nAB=(6/5)x
then
n(m(x))=(6/5)*[(-6/5)x+100]=(-36/25)x+120
h(x)=(-36/25)x+120
find h'(x)
h'(x)=-36/25=-1.44

6) find h(x) = n(m(x)) in the interval B,C
mBC=(6/5)x-20
nBC=(3/5)x+30
then
n(m(x))=(3/5)*[(6/5)x-20]+30 =(18/25)x-12+30=(18/25)x+18
h(x)=(18/25)x+18
find h'(x)
h'(x)=18/25=0.72

for the interval (A,B) h'(x)=-1.44
for the interval (B,C) h'(x)= 0.72

 h'(x) = 1.44 ------------ > not exist