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3 years ago

6

The diameter of a circle is 7 in. Find the circumference to the to the nearest tenth

1 answer:

coldgirl [10]3 years ago

3 0

Answer: around 22 (21.99)

Step-by-step explanation:

SInce the diameter of a circle is 2piR, you just need to multiply the radius (half of the diameter ,3.5) by pi and then 2. The answer is around 22

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Can anyone answer this?

astra-53 [7]

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#14 y=1/2x+1
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2 years ago

if an ambulance is able to respond to a car accident located 5 miles away in 7 minutes, what is the speed of the ambulance in mi

loris [4]

Answer:

Step-by-step explanation:

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3 years ago

A) Is {a, e,i,o,u} a subset of {a,b,c,d,e,f,g,h,i) <br> Why?<br>

kozerog [31]

Answer:

No

Step-by-step explanation:

u and o arent elements of {a, b, c, d, e, f, g, h, i}

8 0

3 years ago

Read 2 more answers

HELPSolve for x<br> X =<br> A: 1<br> B: 3<br> C: 7

OverLord2011 [107]

X equals 7

Hope that helped

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3 years ago

Delicious Candy markets a two-pound box of assorted chocolates. Because of imperfec- tions in the candy making equipment, the ac

saveliy_v [14]

Answer:

The required probability is 0.533.

Step-by-step explanation:

Consider the provided information.

The actual weight of the chocolate has a uniform distribution ranging from 31 to 32.5 ounces.

Let x is the random variable for the actual weight of chocolate.

According to PDF function.

$P(a\leq x\leq b)=\int\limits^b_a {f(x)} \, dx$

Where $f(x)=\left\{\begin{matrix}\frac{1}{b-a} & a$

It is given that ranging from 31 to 32.5 ounces.

Substitute a=31 and b=32.5 in above function.

$f(x)=\left\{\begin{matrix}\frac{1}{32.5-31} & 31$

$f(x)=\left\{\begin{matrix}\frac{1}{1.5} & 31$

We need to find the probability that a box weighs less than 31.8 ounces

Now according to PDF:

$P(x$

$P(x$

Hence, the required probability is 0.533.

6 0

3 years ago