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ipn [44]
3 years ago
11

Harold uses a cone shaped feeder, with a diameter of 7.5 centimeters and a height of 3.5 centimeters, to feed his tropical fish.

How many cubic centimeters of feed can this feeder hold? Use 3.14 for pi. Enter your answer in the box as a decimal rounded to the nearest tenth
Mathematics
2 answers:
Jlenok [28]3 years ago
8 0
This requires you to find the volume of the cone.
The volume of a cone is 1/3 the area of the base (a circle) times the height of the cone, or 1/3(3.14)r^2 * h.

The radius of the base is 7.5/2 or 3.75.

V = 1/3(3.14 * 3.75^2) * 3.5

V = 1/3(44.16) * 3.5

V = 1/3*154.56

V = 51.5 cm^3 of feed
PtichkaEL [24]3 years ago
8 0

Answer:

51.5\ cm^{3}

Step-by-step explanation:

step 1

Find the volume of a cone

we know that

The volume of a cone is equal to

V=\frac{1}{3}\pi r^{2}h

we have

r=7.5/2=3.75\ cm -----> the radius is half the diameter

h=3.5\ cm

substitute the values

V=\frac{1}{3}(3.14)(3.75^{2})(3.5)=51.5\ cm^{3}

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Birth weights of babies born to full-term pregnancies follow roughly a Normal distribution. At Meadowbrook Hospital, the mean we
Marina86 [1]

Answer:

Required Probability = 0.1283 .

Step-by-step explanation:

We are given that at Meadow brook Hospital, the mean weight of babies born to full-term pregnancies is 7 lbs with a standard deviation of 14 oz.

Firstly, standard deviation in lbs = 14 ÷ 16 = 0.875 lbs.

Also, Birth weights of babies born to full-term pregnancies follow roughly a Normal distribution.

Let X = mean weight of the babies, so X ~ N(\mu = 7 lbs , \sigma^{2}  = 0.875^{2}  lbs)

The standard normal z distribution is given by;

              Z = \frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } } ~ N(0,1)

where, X bar = sample mean weight

             n = sample size = 4

Now, probability that the average weight of the four babies will be more than 7.5 lbs = P(X bar > 7.5 lbs)

P(X bar > 7.5) = P( \frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } } > \frac{7.5-7}{\frac{0.875}{\sqrt{4} } }  ) = P(Z > 1.1428) = 0.1283 (using z% table)

Therefore, the probability that the average weight of the four babies will be more than 7.5 lbs is 0.1283 .

8 0
3 years ago
What's the correct answer for this?
11Alexandr11 [23.1K]
\bf \qquad \qquad \textit{inverse proportional variation}\\\\
\textit{\underline{y} varies inversely with \underline{x}}\qquad \qquad  y=\cfrac{k}{x}\impliedby 
\begin{array}{llll}
k=constant\ of\\
\qquad  variation
\end{array}\\\\
-------------------------------\\\\

\bf \textit{\underline{N} is inversely proportional to \underline{A}}\qquad N=\cfrac{k}{A}
\\\\\\
\textit{we also know that }
\begin{cases}
A=3\\
N=16
\end{cases}\implies 16=\cfrac{k}{3}\implies 16\cdot 3=k
\\\\\\
48=k\qquad thus\qquad \boxed{N=\cfrac{48}{A}}\\\\
-------------------------------\\\\
\textit{if \underline{A} is 4, what is \underline{N}?}\qquad N=\cfrac{48}{4}
4 0
3 years ago
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