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zalisa [80]
3 years ago
12

A3=-27 and a5=-162. Find the common ratio ​

Mathematics
1 answer:
SpyIntel [72]3 years ago
3 0

Answer:

c = \sqrt{6}

Step-by-step explanation:

a3 = -27

a5 = -162

a5 = a3 * c^2

6 = c^2

c = \sqrt{6}

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If we have 2 bags containing marbles. bag a has 7 green and 6 black marbles. bag b has 8 yellow and 4 red marbles. if we pick ra
Elena L [17]
A.  (7/13)·(1/3)=7/39.       P=7/39
B.  (6/13)·(2/3)=12/39.    P=12/39
C.  (7/13)·(2/3)=14/39.    P=14/39
D.  (6/13)·(1/3)=6/39.      P=6/39

All the numbers from Bag B were written as thirds cause their probabilities could be simplified to make multiplying easier
8 0
3 years ago
Jack spends 1 2/5 as long on his homework as Jill. Last week, Jill spent 6 3/4 hours doing homework. How long did Jack spend doi
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6 3/4 hours = (6*60)+45 = 405 minutes
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405 minutes * 1,4 = 567 minutes
= Jack spent 9 3/4 hours

8 0
3 years ago
I need help. i don’t understand this question .
user100 [1]

So the question is asking to solve for y

You know RS=7y-4

You know ST=y+5

You know RT=28

RS+ST=RT as shown in the diagram, This means that you can use that equation and replace RS,ST, and RT with what you know they are equal too. So you end up with the equation:

(7y-4)+(y+5)=28, Now you have to solve for y

7y-4+y+5=28

8y+1=28

8y=27

y=3.375

6 0
3 years ago
Suppose that X has a discrete uniform distribution on the integers 0 through 9. Determine the mean, variance, and standard devia
Anna35 [415]
With 10 integers available, X has PMF

\mathbb P(X=x)=\begin{cases}\dfrac1{10}&\text{for }0\le x\le9,x\in\mathbb Z\\\\0&\text{otherwise}\end{cases}

We're interested in the statistics of the new random variable Y=5X. To do this, we need to know the PMF for Y. This isn't too hard to find.

\mathbb P(Y=y)=\mathbb P(5X=y)=\mathbb P\left(X=\dfrac y5\right)

Since the PMF for X gives a value of \dfrac1{10} whenever x is an integer between 0 and 9, it follows that \dfrac y5 must also be an integer for the PMF to give the identical value of \dfrac1{10}. This means

\mathbb P(Y=y)=\begin{cases}\dfrac1{10}&\text{for }y\in\{0,5,\ldots,45\}\\\\0&\text{otherwise}\end{cases}

Now the mean (expectation) is

\mathbb E(Y)=\displaystyle\sum_yy\mathbb P(Y=y)=\frac1{10}\sum_{y\in\{0,\ldots,45\}}y
\mathbb E(Y)=\dfrac{225}{10}=22.5

The variance would be

\mathbb V(X)=\mathbb E(X^2)-\mathbb E(X)^2
\mathbb V(X)=\displaystyle\sum_yy^2\mathbb P(Y=y)-\mathbb E(Y)^2
\mathbb V(X)=\displaystyle\frac1{10}\sum_{y\in\{0,\ldots,45\}}y^2-\left(\frac{225}{10}\right)^2
\mathbb V(X)=\dfrac{7125}{10}-\dfrac{50625}{100}=206.25

The standard deviation is the square root of the variance, so you have

\sqrt{\mathbb V(X)}=\sqrt{206.25}\approx14.3614
8 0
3 years ago
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