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sashaice [31]
3 years ago
10

Tim ate 1/6 of the brownies and Abigail ate 1/8. How many all together?

Mathematics
2 answers:
BARSIC [14]3 years ago
4 0
B.7/24
hope this helped
sasho [114]3 years ago
4 0
7/24
Hope this helps(:
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I need answers for the question below to end a debate. Explain why that is the answer too.
IRINA_888 [86]
PEMDAS. Parenthesis comes first. 3+2 = 5 which is then multiplied to 6/2 which is 3 so 3 x 5 is 15. 
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What is the perimeter of the rectangle shown on the coordinate plane, to the nearest tenth of a unit?
Inga [223]

26.8 units

What is the perimeter of the rectangle shown on the coordinate plane, to the nearest tenth of a unit?


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5 0
3 years ago
Find the product of (x − 5)2
Igoryamba

Answer: The product is x^2-10x+25

Step-by-step explanation:

Since we have given that

(x-5)^2

We need to find the product:

Since we know that

(a-b)^2=a^2-2ab+b^2

Here, a = x

and

b = 5

So, it becomes,

(x-5)^2=x^2-2\times x\times 5+5^2=x^2-10x+25

Hence, the product is x^2-10x+25

4 0
3 years ago
is recommended that a ramp have least feet of horizontal distance for every 1 foot of horizontal rise along an incline. The ramp
lesya [120]

Answer: no

Step-by-step explanation:

8 0
3 years ago
Let Y denote a geometric random variable with probability of success p. a Show that for a positive integer a, P(Y > a) = qa .
lakkis [162]

Answer:

a) For this case we can find the cumulative distribution function first:

F(k) = P(Y \leq k) = \sum_{k'=1}^k P(Y =k')= \sum_{k'=1}^k p(1-p)^{k'-1}= 1-(1-p)^k

So then by the complement rule we have this:

P(Y>a) = 1-F(a)= 1- [1-(1-p)^a]= 1-1 +(1-p)^a = (1-p)^a = q^a

b) P(Y>a)= q^a

P(Y>b) = q^b

So then we have this using independence:

P(Y> a+b) = q^{a+b}

We want to find the following probability:

P(Y> a+b |Y>a)

Using the definition of conditional probability we got:

P(Y> a+b |Y>a)= \frac{P(Y> a+b \cap Y>a)}{P(Y>a)} = \frac{P(Y>a+b)}{P(Y>a)} = \frac{q^{a+b}}{q^a} = q^b = P(Y>b)

And we see that if a = 2 and b=5 we have:

P(Y> 2+5 | Y>2) = P(Y>5)

c) For this case we use independent identical and with the same distribution experiments.

And the result for part b makes sense since we are interest in find the probability that the random variable of interest would be higher than an specified value given another condition with a value lower or equal.

Step-by-step explanation:

Previous concepts

The geometric distribution represents "the number of failures before you get a success in a series of Bernoulli trials. This discrete probability distribution is represented by the probability density function:"

P(X=x)=(1-p)^{x-1} p

If we define the random of variable Y we know that:

Y\sim Geo (1-p)

Part a

For this case we can find the cumulative distribution function first:

F(k) = P(Y \leq k) = \sum_{k'=1}^k P(Y =k')= \sum_{k'=1}^k p(1-p)^{k'-1}= 1-(1-p)^k

So then by the complement rule we have this:

P(Y>a) = 1-F(a)= 1- [1-(1-p)^a]= 1-1 +(1-p)^a = (1-p)^a = q^a

Part b

For this case we can use the result from part a to conclude that:

P(Y>a)= q^a

P(Y>b) = q^b

So then we have this assuming independence:

P(Y> a+b) = q^{a+b}

We want to find the following probability:

P(Y> a+b |Y>a)

Using the definition of conditional probability we got:

P(Y> a+b |Y>a)= \frac{P(Y> a+b \cap Y>a)}{P(Y>a)} = \frac{P(Y>a+b)}{P(Y>a)} = \frac{q^{a+b}}{q^a} = q^b = P(Y>b)

And we see that if a = 2 and b=5 we have:

P(Y> 2+5 | Y>2) = P(Y>5)

Part c

For this case we use independent identical and with the same distribution experiments.

And the result for part b makes sense since we are interest in find the probability that the random variable of interest would be higher than an specified value given another condition with a value lower or equal.

8 0
4 years ago
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