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ANTONII [103]
3 years ago
7

Tent sized portraits has four different backgrounds and five poses in which to photograph children how many different pictures a

re there to choose from
Mathematics
1 answer:
kenny6666 [7]3 years ago
3 0

Answer:

20

Step-by-step explanation:

Assuming each pose can be used with each background, there are 4×5 = 20 possible combinations.

That is, for each one of the 4 backgrounds, there can be 5 poses. Then the total number of pictures is ...

  5 + 5 + 5 + 5 = 4×5 = 20

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PLEASEE HELPPP
Leni [432]

Answer:

Mean x¯¯¯ 2.2

Median x˜ 1

Range = 9

Q1 --> 0

Q2 --> 1

Q3 --> 3

Interquartile range: 3

Mean Absolute Deviation (MAD): 2.08

Standard deviation : 2.71

Step-by-step explanation:

1. Number of absent students:

0,1,1,0,3,2,5,1,9,0

Find the:

Mean

0, 0, 0, 1, 1, 1, 2, 3, 5, 9/10

= 22/10

= 2.2

Median

0, 0, 0, 1), 1, 1, (2, 3, 5, 9

= 1 + 1/2

= 2/2

= 1

Lower quartile

Upper quartile

Range

= 9 - 0

= 9

Interquartile range = 3

Mean absolute deviation = 2.08

Standard deviation = 2.71

3 0
3 years ago
Hello can I have some help with this expand question 4(z+2)+4(z+3)
wel

Step-by-step explanation:

4(z+2)+4(z+3)

=4z+8+4z+12

=8z+20

by simplifying-

=2z+5

4 0
3 years ago
Read 2 more answers
F(x)= 2-1/x<br>g(x)= x/2<br><br>what is f(g(x))?
kompoz [17]
To find f(g(x)), plug g(x) into f(x) and simplify.

f(x/2) = 2 - 1/(x/2)

All you have to do is simplify the right side of the equation.

Take it from here.



8 0
3 years ago
Given the equation (x+a)^2(x-2)=x^3+bx^2+12x-72 find a and b
GaryK [48]
Isolate the variable by dividing each side by factors that don't contain the variable.

a=-  \frac{ x^{2}-  \sqrt{( x^{3} + x^{2} b+12x-72(x-2)-2x} }{x-2} ,-  \frac{ x^{2}+  \sqrt{( x^{3} + x^{2} b+12x-72(x-2)-2x} }{x-2}

Solve for b by simplifying both sides of the equation then isolating the variabel.

b= \frac{12}{x}+ \frac{72}{ x^{2} }-2+2a- \frac{4a}{x}+ \frac{ a^{2} }{x}- \frac{2a^{z} }{ x^{2} }

Hopefully i helped ^.^ Mark brainly if possible 
 

7 0
3 years ago
The half-life of a certain radioactive element is 35 hours. If there are 904 grams of this element, how much will be left after
Cloud [144]
<h3>Answer: 226 grams</h3>

===========================================================

Explanation:

The half-life is 35 hours. This means every 35 hours, the amount of the substance is cut in half. If we started with 100 grams, then 35 hours later, we will have 50 grams. Then another 35 hours later we will then have 25 grams, and so on.

In this case, we start with 904 grams and follow this sequence:

904, 452, 226, 113, ...

Each term is cut in half. Going from term to term it takes 35 hours each. The third term 226 is the result of waiting two half-lives, which is 2*35 = 70 hours that have elapsed total.

--------------

If you want to use a formula, then it would be

y = A*\left(\frac{1}{2}\right)^{x/H}

where,

H = half-life duration (in this case, it is 35)

A = starting amount (which is 904 in this case)

x = amount of time that has elapsed (we will plug in x = 70)

y = final amount after x units of time has elapsed

note that x and H must have the same time unit. Both are in terms of hours in this case.

------------

So we will plug A = 904, x = 70, H = 35 into the formula to get...

y = A*\left(\frac{1}{2}\right)^{x/H}\\\\y = 904*\left(\frac{1}{2}\right)^{70/35}\\\\y = 904*(0.5)^{2}\\\\y = 904*(0.25)\\\\y = 226\\\\

We see that x = 70 leads to y = 226. This means after 70 hours, there is 226 grams of the material left.

4 0
3 years ago
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