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timurjin [86]
3 years ago
7

A manhole measures 3' 4" in diameter. What is the area of a metal cover for the manhole, assuming a one-inch overhang around the

circumference of the cover?

Mathematics
1 answer:
Natasha_Volkova [10]3 years ago
8 0
I drew the entire problem. The final answer would be 441pi in²

I highly advise you to look at how it could work by looking at the drawing. Hope this helps :)

The blue line is the 1 inch overhang, and I add two the the diameter to get the full diameter including the overhang

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16% of 242 = ? <br> Please help me solve this
MrRa [10]

Answer:

16% of 242 = 38.72

Step-by-step explanation:

16% = 16/100 = 0.16

242 * 0.16 = 38.72

5 0
3 years ago
Read 2 more answers
Refer to the previous exercise. The researchers wanted a sufficiently large sample to be able to estimate the probability of pre
Maru [420]

Answer:

<em>The large  sample n = 117.07</em>

Step-by-step explanation:

<u><em>Step(i):-</em></u>

Given that the estimate error (M.E) = 0.08

The proportion (p) = 0.75

q =1-p = 1- 0.75 =0.25

Level of significance = 0.05

Z₀.₀₅ = 1.96≅ 2

<u><em>Step(ii):-</em></u>

The Marginal error is determined by

M.E = \frac{Z_{0.05} \sqrt{p(1-p)}  }{\sqrt{n} }

0.08 = \frac{2 X \sqrt{0.75(1-0.75)}  }{\sqrt{n} }

Cross multiplication , we get

\sqrt{n}  = \frac{2 X \sqrt{0.75(1-0.75)}  }{0.08 }

√n = \frac{2 X0.4330}{0.08} = 10.825

squaring on both sides , we get

n = 117.07

<u><em>Final answer:-</em></u>

<em>The large  sample n = 117.07</em>

6 0
2 years ago
Aneesha travels at a rate of 50 miles per hour. Morris is traveling 3 feet per second less than Aneesha. Which is the best estim
astra-53 [7]
It would be nice to know some options, but here's Morris's speed.
Since Morris is traveling at three feet per second, we want to change it to per hour.
3 * 3600 = 10800 feet per hour
And we would want to convert that to miles per hour.
10800/5280 is about 2, more specifically 2.05 ish.
So Morris is traveling 2 mph less then 50, or about 48 mph.
3 0
3 years ago
Read 2 more answers
Find the probability that a randomly generated bit string of length 10 does not contain a 0 if bits are independent and if a 0 b
jeka94

Answer:

Step-by-step explanation:

From the given information; Let's assume that  R should represent the set of all possible outcomes generated from  a bit string of length 10 .

So; as each place is fitted with either 0 or 1

\mathbf{|R|= 2^{10}}

Similarly; the event E signifies the randomly generated bit string of length 10 does not contain a 0

Now;

if  a 0 bit and a 1 bit are equally likely

The probability that a randomly generated bit string of length 10 does not contain a 0 if bits are independent and if a 0 bit and a 1 bit are equally likely is;

\mathbf{P(E) = \dfrac{|E|}{|R|}}

so ; if bits string should not contain a 0 and all other places should be occupied by 1; Then:

\mathbf{{|E|}=1 }   ; \mathbf{|R|= 2^{10}}

\mathbf{P(E) = \dfrac{1}{2^{10}}}

4 0
3 years ago
Can someone help me with this math problem please and thank you. :-)
jok3333 [9.3K]

Triangle ABE is isosceles  / Given

AB congruent to AE     / Def isosceles

angle ABE congruent to angle AEB   / Property of isosceles triangles

angle ABD congruent to angle AEC   / Subst different name for same angles

BD congruent to EC     / Given

triange ABD congruent to triange AEC    / Side Angle Side

5 0
3 years ago
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