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2 years ago

Can someone help me with this?

Mathematics

2 answers:

Liula [17]2 years ago

8 0

Answer:

Step-by-step explanation:

4.5/C

olga nikolaevna [1]2 years ago

3 0

Its A 4.5/C you just had to figure out what is half of 9

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20 POINT WILL GIVE BRAINLIEST <br> QUESTION #3

Serjik [45]

36x just got done taking this

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3 years ago

Approximate the sum of the convergent infinite series the least usingnumber of terms so that the error is less than .0001. What

____ [38]

Solution :

$\sum_{n=1 }^{\infty } \frac{(-1)^n}{n ! 2^n}$ $a_n= \frac{(-1)^n}{n ! 2^n} \ \ \ \ \ a_{n+1}= \frac{(-1)^{n+1}}{(n+1) ! 2^{n+1}}$

Error = $|a_{n+1}|$

Error ≤ 0.0001

$|\frac{(-1)^{n+1}}{(n+1)!2^{n+1}}| \leq 0.0001$

$|\frac{1}{(n+1)!2^{n+1}}| \leq 10^{-4}$

$(n+1)! 2^{n+1} \geq 10000$

Now try, n ≥ 5

$\sum_{n=1 }^{\infty } \frac{(-1)^n}{n ! 2^n}$ = $\sum_{n=1 }^{5 } \frac{(-1)^n}{n ! 2^n}$ (with error 0.0001)

$\sum_{n=1 }^{\infty } \frac{(-1)^n}{n ! 2^n}$ = $\frac{-1}{1!2}+\frac{1}{2!2^2}-\frac{1}{3! 2^3}+\frac{1}{4! 2^4}-\frac{1}{5! 2^5}$

$\sum_{n=1 }^{\infty } \frac{(-1)^n}{n ! 2^n}$ = 0.6065104

3 0

2 years ago

Rhonda has $4.85 in coins. If she has six more nickels than dimes and twice as many quarters as dimes, how many coins of each ty

Novay_Z [31]

Answer: nickels = 13, dimes = 7, quarters = 14

<u>Step-by-step explanation:</u>

First, set up the equations for each coin:

$\begin {array}{c|c|c|l}\underline{\quad Type\quad }&\underline{Value}&\underline{Quantity}&\underline{\qquad \qquad \qquad Equation\qquad }\\ Nickel& \$0.05&6+d&0.05(6+d) = 0.30 + 0.05d\\Dime& \$0.10&d&\qquad 0.10(d)=0.10d\\Quarter& \$0.25&2d&\qquad 0.25(2d)=0.50d\\\end{array}$

Next, the sum of the coins is $4.85 so substitute and solve for the variable:

Nickels + Dimes + Quarters = Sum

(0.30 + 0.05d) + 0.10d + 0.50d = 4.85

0.30 + 0.65d = 4.85

0.65d = 4.55

d = 7

Lastly, plug the d-value into the quantity equation for the nickels and quarters to find their quantity.

nickels: 6 + d = 6 + (7) = 13

quarters: 2d = 2(7) = 14

6 0

2 years ago

Does someone know this one please :)

GarryVolchara [31]

6/11(5/6)+2/11(5)
=5/11+10/11
=15/11

3 0

2 years ago

The X- and y-coordinates of point P are each to be chosen at random from the set of integers 1 through 10.

bagirrra123 [75]

Answer:

Ok, as i understand it:

for a point P = (x, y)

The values of x and y can be randomly chosen from the set {1, 2, ..., 10}

We want to find the probability that the point P lies on the second quadrant:

First, what type of points are located in the second quadrant?

We should have a value negative for x, and positive for y.

But in our set; {1, 2, ..., 10}, we have only positive values.

So x can not be negative, this means that the point can never be on the second quadrant.

So the probability is 0.

3 0

2 years ago