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velikii [3]
3 years ago
9

38. Emily is buying some graduation pictures. She pays $25 for the sitting and $15 for each

Mathematics
1 answer:
goldfiish [28.3K]3 years ago
5 0

a. She pays $100 for 5 sheets

b. 25+15x dollars for x sheets

c. 8 sheets

Step-by-step explanation:

Given

Sitting cost = $25

Per sheet picture cost = $15

Let p be the number of sheets of pictures

Then the cost can be written as a function of p

c(p) = 25+15p

Now,

<u>a. How much does she pay for 5 sheets of pictures?! </u>

Putting p = 5 in the function

c(5) = 25 + 15(5)\\= 25+75\\=100

She will pay $100 for 5 sheets of pictures

<u>b. How much does she pay for "x" sheets? </u>

Putting x in place of p

c(x) = 25+15x

<u>c. How many sheets can she buy for $145? </u>

We know the cost now, we have to find p so,

145 = 25+15p\\145-25 = 25+15p-25\\120 = 15p

Dividing both sides by 15

\frac{15p}{15} = \frac{120}{15}\\p = 8

Hence,

She can buy 8 sheets for $145

Keywords: Linear equation, Algebraic functions

Learn more about functions at:

  • brainly.com/question/629998
  • brainly.com/question/6208262

#LearnwithBrainly

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FinnZ [79.3K]
The greatest common factor( GCF) of 36 and 84 is 12.
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What is the area of a semicircle whose perimeter is 72cm
Nuetrik [128]

Answer:

A = 307.9 sq/cm

Step-by-step explanation:

Find the radius of the semi-circle

diameter + half the circumference = 72 cm

2r + pi*r = 72

r(2+pi) = 72

 

r = 72%2F%28%282%2Bpi%29%29

r = 14

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what will be the area of that semicircle?

A =1%2F2*pi%2A14%5E2

A = 307.9 sq/cm

7 0
3 years ago
What is the ratio of the intensities of an earthquake PP wave passing through the Earth and detected at two points 19 kmkm and 4
Tamiku [17]

Answer:

The ratio of the intensities is roughly 6:1.  

Step-by-step explanation:

The intensity I() of an earthquake wave is given by:

I = \frac{P}{4\pi d^{2}}

<em>where P: is the power ans d: is the distance. </em>

Hence, the ratio of the intensities of an earthquake wave passing through the Earth and detected at two points 19 km and 46 km from the source is:

\frac{I_{1}}{I_{2}} = \frac{P/4\pi d_{1}^{2}}{P/4\pi d_{2}^{2}}

<em>where I₁ = P/4πd₁², d₁=19 km, I₁ = P/4πd₂² and d₂=46 km     </em>

\frac{I_{1}}{I_{2}} = \frac{d_{2}^{2}}{d_{1}^{2}}

\frac{I_{1}}{I_{2}} = \frac{46 km^{2}}{19 km^{2}}

\frac{I_{1}}{I_{2}} = 5.9:1

Therefore, the ratio of the intensities is roughly 6:1.  

 

I hope it helps you!    

3 0
3 years ago
I’ll give points and brainalist for a correct answer / explanation
wlad13 [49]

Answer:

Answer is C

r= 3

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Step-by-step explanation:

3 0
3 years ago
A bus travels on an east-west highway connecting two cities A and B that are 100 miles apart. There are 2 services stations alon
melamori03 [73]

Answer:

51/4

Step-by-step explanation:

To begin with you have to understand what is the distribution of the random variable. If X represents the point where the bus breaks down. That is correct.  

X~ Uniform(0,100)

Then the probability mass function is given as follows.

f(x) = P(X=x) = 1/100  \,\,\,\, \text{if} \,\,\,\, 0 \leq x \leq 100\\f(x) = P(X=x) = 0  \,\,\,\, \text{otherwise}

Now, imagine that the D represents the distance from the break down point to the nearest station. Think about this, the first service station is 20 meters away from city A, and the second station is located  70 meters away from city A then the mid point between 20 and 70  is (70+20)/2 = 45 then we can represent D as follows

D(x) =\left\{ \begin{array}{ll}  x  & \mbox{if } 0\leq x \leq 20 \\  x-20 & \mbox{if } 20\leq x < 45\\                70-x & \mbox{if } 45 \leq x \leq 70\\                x-70 & \mbox{if } 70 \leq x \leq 100\\ \end{array}\right.

Now, as we said before X represents the random variable where the bus breaks down, then we form a new random variable Y = D(X), Y is a random variable as well, remember that there is a theorem that says that

E[Y] = E[D(X)] = \int\limits_{-\infty}^{\infty} D(x) f(x) \,\, dx

Where f(x) is the probability mass function of X. Using the information of our problem

E[Y] = \int\limits_{-\infty}^{\infty}  D(x)f(x) dx \\= \frac{1}{100} \bigg[ \int\limits_{0}^{20} x dx +\int\limits_{20}^{45} (x-20) dx +\int\limits_{45}^{70} (70-x) dx +\int\limits_{70}^{100} (x-70) dx  \bigg]\\= \frac{51}{4} = 12.75

3 0
3 years ago
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