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madreJ [45]
3 years ago
9

Please help!!!! solve for x showing EACH STEP please!!

Mathematics
2 answers:
brilliants [131]3 years ago
7 0

cant see the pic

please type out the question

so that i can help u

vovikov84 [41]3 years ago
6 0

Answer:

x = zk + z +  \mu

Step-by-step explanation:

z =  \frac{x -  \mu}{k + 1}  \\  \\ z(k + 1) = x -  \mu \\  \\ z(k + 1) +  \mu =x  \\  \\ \red{ \bold{ x = zk + z +  \mu}}

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What is the area of this triangle? (Image Attached)
Snezhnost [94]
\bf \textit{Law of Cosines}\\ \quad \\
c^2 = {{ a}}^2+{{ b}}^2-(2{{ a}}{{ b}})cos(C)\implies 
c = \sqrt{{{ a}}^2+{{ b}}^2-(2{{ a}}{{ b}})cos(C)}\\\\
-----------------------------\\\\
\textit{so, let's find the missing side}
\\\\\\
c=\sqrt{2.7^2+3.4^2-2(2.7)(3.4)cos(40^o)}
\\\\\\
c=\sqrt{4.78542402433556327368}\implies c\approx 2.187561204706182220

or we can round it, to say c = 2.19, so  hmm that's the missing side
now, we use Heron's Formula, which uses all 3 sides only

\bf \textit{Heron's Area Formula}\\\\
A=\sqrt{s(s-a)(s-b)(s-c)}\qquad 
\begin{cases}
a=2.7\\
b=3.4\\
c\approx 2.19\\\\
s=\cfrac{a+b+c}{2}
\end{cases}

and that'd be the area of it
6 0
3 years ago
Read 2 more answers
Six cards numbered from 1 to 6 are placed in an empty bowl. First one card is drawn and then put back into the bowl; then a seco
kari74 [83]

Answer:

C. \frac{1}{18}

Step-by-step explanation:

Given: Six cards numbered from 1 to 6 are placed in an empty bowl. First one card is drawn and then put back into the bowl then a second card is drawn.

To Find: If the cards are drawn at random and if the sum of the numbers on the cards is 8, what is the probability that one of the two cards drawn is numbered 5.

Solution:

Sample space for sum of cards when two cards are drawn at random is \{(1,1),(1,2),(1,3)......(6,6)\}

total number of possible cases =36

Sample space when sum of cards is 8 is \{(3,5),(5,3),(6,2),(2,6),(4,4)\}

Total number of possible cases =5

Sample space when one of the cards is 5 is \{(5,3),(3,5)\}

Total number of possible cases =2

Let A be the event that sum of cards is 8

p(\text{A}) =\frac{\text{total cases when sum of cards is 8}}{\text{all possible cases}}

p(\text{A})=\frac{5}{36}

Let B be the event when one of the two cards is 5

probability than one of two cards is 5 when sum of cards is 8

p(\frac{\text{B}}{\text{A}})=\frac{\text{total case when one of the number is 5}}{\text{total case when sum is 8}}

p(\frac{\text{B}}{\text{A}})=\frac{2}{5}

Now,

probability that sum of cards 8 is and one of cards is 5

p(\text{A and B}=p(\text{A})\times p(\frac{\text{B}}{\text{A}})

p(\text{A and B})=\frac{5}{36}\times\frac{2}{5}

p(\text{A and B})=\frac{1}{18}

if sum of cards is 8 then probability that one of the cards is 8 is \frac{1}{18}, option C is correct.

3 0
3 years ago
Mika wants to buy a condominium. He has the choice of buying it now or renting it with the option to buy at the end of 3 years.
mixer [17]

Answer:

Step-by-step explanation:

 

$828

5 0
3 years ago
Read 2 more answers
Need help on #7 help if u know how to solve thanks!
mash [69]
7a) 302.5
7b) 302500
Hope this helps! :)
8 0
3 years ago
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Write a word phrase that can be represented by b-6
Bess [88]
Six less than "b"
i think this would be the answer
good luck!!
7 0
3 years ago
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