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2 years ago

A block of silver is weighing 8 ounces is worth $104. What is the unit rate for the Value of silver in one ounce

Mathematics

1 answer:

xxMikexx [17]2 years ago

4 0

Answer:

Blockof silver's wieghing = 8 ounces

total rate = $104

therefore, one ounce=$104/8= $13

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I never really understood what this was could u help

ycow [4]

Answer:

1. f(4) or f(n)= 28

2. f(-2) or f(n)= -2

(please consider marking brainliest <3 )

Step-by-step explanation:

n=4 n=-2

f(n)= 5(n)+8 f(n)= 5(n)+8

f(n)= 5(4)+8 f(n)= 5(-2)+8

(n)= 20+8 f(n)= -10+8

f(n)=28 f(n)=-2

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2 years ago

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A pharmaticeutical company is making a new medicine capsule shaped like a cylinder with a half sphere at each end. The

DerKrebs [107]

Answer:

339.12 cubic millimeters

Step-by-step explanation:

The picture of the question in the attached figure

we know that

The volume of the figure is equal to the volume of the two hemispheres (one sphere) plus the volume of the cylinder

step 1

Find the volume of the cylinder

The volume is given by

$V=Bh$

where

B is the area of the base of cylinder

h is the height of cylinder

we have

$B=\pi r^{2}$

we have

$h=8\ mm$

$r=6/2=3\ mm$ ----> the radius is half the diameter

$B=(3.14)(3)^{2}$

$B=28.26\ mm^2$

substitute

$V=(28.26)(8)=226.08\ mm^3$

step 2

Find the volume of the sphere

The volume is given by

$V=\frac{4}{3}\pi r^{3}$

we have

$r=6/2=3\ mm$ ----> the radius is half the diameter

substitute

$V=\frac{4}{3}(3.14)(3)^{3}=113.04\ mm^3$

step 3

Adds the volumes

$226.08+113.04=339.12\ mm^3$

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2 years ago

Suppose we have an unfair coin that its head is twice as likely to occur as its tail. a)If the coin is flipped 3 times, what is

Alenkinab [10]

Answer: a) 0.2222, b) 0.3292, c) 0.1111

Step-by-step explanation:

Since we have given that

Let the probability of getting head be p.

Since, its head is twice as likely to occur as its tail.

$p+\dfrac{p}{2}=1\\\\\dfrac{3p}{2}=1\\\\p=\dfrac{2}{3}$

a)If the coin is flipped 3 times, what is the probability of getting exactly 1 head?

So, here, n = 3

$p=\dfrac{2}{3}$

$q=\dfrac{1}{3}$

Now,

$P(X=1)=^3C_1(\dfrac{2}{3})^1(\dfrac{1}{3})^2=0.2222$

b)If the coin is flipped 5 times, what is the probability of getting exactly 2 tails?

2 tails means 3 heads.

So, it becomes,

$P(X=3)=^5C_3(\dfrac{2}{3})^3(\dfrac{1}{3})^2=0.3292$

c)If the coin is flipped 4 times, what is the probability of getting at least 3 tails?

$P(X\leq 1)=\sum _{x=0}^1^4C_x(\dfrac{2}{3}^x(\dfrac{1}{3})^{4-x}=0.1111$

Hence, a) 0.2222, b) 0.3292, c) 0.1111

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2 years ago

How does x/2-12=-14 I don’t understand where the negative comes from

goblinko [34]

Solve x by simplifying both sides of the equation & then isolating the variable x=-4 & the negative comes in from the -12

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2 years ago

-54 + 108 =<br> Thank you!

julsineya [31]

Answer: 54 and welcome

Step-by-step explanation:

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2 years ago

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