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san4es73 [151]
4 years ago
12

BRAINLY PLS HELP I BEG YOU I NEED YOYR HELP ASP !

Mathematics
1 answer:
trapecia [35]4 years ago
8 0

Answer:

3

Step-by-step explanation:

We can find the value of x using Euclidean theorem

√18^2 = 5 × x

18 = 5x

3 = x

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The student government snack shop sold 32 items this week
Firdavs [7]

Answer:

right the whole question

Step-by-step explanation:

if he sold 32 items this week last week he sold 20

5 0
3 years ago
Evaluate the expression -2​
DaniilM [7]

Answer:  2x for x=3: 2(3) = 6.

Step-by-step explanation:

6 0
3 years ago
Using the formula, solve for c, if a = 9 and b = 40
alisha [4.7K]

Answer:

<h2>The answer is option C</h2>

Step-by-step explanation:

c² = a² + b²

a = 9 , b = 40

Substitute the values of a and b into the equation and solve

That's

{c}^{2}   =  {9}^{2}  +  {40}^{2}  \\  {c}^{2}  = 81 + 1600 \\  {c}^{2}  = 1681 \\ c   = \sqrt{1681}

We have the final answer as

<h3>c = 41</h3>

Hope this helps you

5 0
3 years ago
Find the following product.<br> (4a +6) (3a? - 3a+6)<br> (4a + 6) (3a2- 3a +6) =
katen-ka-za [31]

Answer:

12a^3+6a^2+6a+36

Step-by-step explanation:

Sana makatulong!

3 0
3 years ago
In a recent​ year, an author wrote 169 checks. Use the Poisson distribution to find the probability​ that, on a randomly selecte
grigory [225]
We should first calculate the average number of checks he wrote per day.  To do that, divide 169 by 365 (the number of days in a year) and you get (rounded) 0.463.  This will be λ in our Poisson distribution.  Our formula is
P(X=k)= \frac{ \lambda ^{k}-e^{-\lambda} }{k!}.  We want to evaluate this formula for X≥1, so first we must evaluate our case at k=0.  
P(X=0)= \frac{0.463 ^{0}-e ^{-0.463} }{0!} \\ = \frac{1-e ^{-0.463} }{1} =0.3706
To find P(X≥1), we find 1-P(X<1).  Since the author cannot write a negative number of checks, this means we are finding 1-P(X=0).  Therefore we have 1-0.3706=0.6294.
There is a 63% chance that the author will write a check on any given day in the year.<em />
8 0
3 years ago
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