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1 year ago

The length of a swimming pool is 8m longer than its width and the area is 105m2

2 answers:

8 0

$width : w \\ length : \ l = w + 8 \\ A = 105 \ m^2 \\ \\A= w \cdot l \\ \\w(w+l)=105 \\ \\w^2+w = 105 \\ \\w^2+w-105 =0 \\ \\(w+15)(w-7)=0 \\ \\w+15 =0 \ \ or \ \ w-7 = 0 \\ \\w = -15 \ \ or \ \ w=7 \\ \\ width \ cant \ be \ negative, \ so \\ \\ w = 7 \ and \ l = w+8 = 7 +8 =15 \\ \\chek : \\ \\ A=w\cdot l \\ \\A=7\cdot 15=105 \ m^2$

Talja [164]1 year ago

5 0

$l\ \rightarrow\ the\ length\\w\ \rightarrow\ the\ width \\\\l=w+8\ \ \ and\ \ \ l\cdot w=105\ \ \ and\ \ \ l>0,\ w>0\\\\(w+8)\cdot w=105\ \ \ \Rightarrow\ \ \ w^2+8w=105\\\\ w^2+2\cdot4w+4^2=105+4^2\\\\ (w+4)^2=105+16\ \ \ \Rightarrow\ \ \ (w+4)^2=121\\\\w+4=11\ \ \ or\ \ \ w+4=-11\\\\w=7\ \ \ \ \ \ \ \ \ \ or\ \ \ w=-15$

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The Post Office has established a record in a major midwestern city for delivering 90 percent of its local mail the next working

IgorLugansk [536]

Answer:

a) $P(X=8)=(8C8)(0.9)^8 (1-0.9)^{8-8}=0.43$

b) $E(X) = np = 8*0.9 =7.2$

c) $Var(X) = np(1-p) =8*0.9*(1-0.9) = 0.72$

$Sd(X) = \sqrt{Var(X)}= \sqrt{0.72}=0.85$

d) For this case two deviations from the mean represent $2\sigma = 2*0.85= 1.7$

For this case we can use the binomial approximation to the normal distribution since the value of p is greater than 0.5, we have that np>5.

So we can approximate the distribution of X with

$X \sim N(7.2, 0.85)$

And we want this probability:

$P(7.2 -2*0.85 \leq X \leq 7.2 +2*0.85)= P(5.5 \leq X \leq 8.9)$

And we apply the continuity correction and we got:

$P(5.5 \leq X \leq 8.9)= P(X \leq 8.9+0.5) -P(X\leq 5.5+0.5) =P(X \leq 9.4) -P(X\leq 6)$

And using the z score formula given by:

$z = \frac{x -\mu}{\sigma}$

We find the two z scores for thw two values and we have:

$z= \frac{9.4-7.2}{0.85}= 2.59$

$z= \frac{6-7.2}{0.85}= -1.41$

$P(Z$

Step-by-step explanation:

Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

The probability mass function for the Binomial distribution is given as:

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

Where (nCx) means combinatory and it's given by this formula:

$nCx=\frac{n!}{(n-x)! x!}$

Let X the random variable of interest, on this case we now that:

$X \sim Binom(n=8, p=0.9)$

Solution to the problem

Part a

For this case we want to find the following probability:

$P(X=8)$

And we can use the pmf function and we got:

$P(X=8)=(8C8)(0.9)^8 (1-0.9)^{8-8}=0.43$

Part b

The expected valie for the binomial distribution is given by:

$E(X) = np = 8*0.9 =7.2$

Part c

For this case the variance is given by:

$Var(X) = np(1-p) =8*0.9*(1-0.9) = 0.72$

And the standard deviation would be just the square root of the variance and we got:

$Sd(X) = \sqrt{Var(X)}= \sqrt{0.72}=0.85$

Part d

For this case two deviations from the mean represent $2\sigma = 2*0.85= 1.7$

For this case we can use the binomial approximation to the normal distribution since the value of p is greater than 0.5, we have that np>5.

So we can approximate the distribution of X with

$X \sim N(7.2, 0.85)$

And we want this probability:

$P(7.2 -2*0.85 \leq X \leq 7.2 +2*0.85)= P(5.5 \leq X \leq 8.9)$

And we apply the continuity correction and we got:

$P(5.5 \leq X \leq 8.9)= P(X \leq 8.9+0.5) -P(X\leq 5.5+0.5) =P(X \leq 9.4) -P(X\leq 6)$

And using the z score formula given by:

$z = \frac{x -\mu}{\sigma}$

We find the two z scores for thw two values and we have:

$z= \frac{9.4-7.2}{0.85}= 2.59$

$z= \frac{6-7.2}{0.85}= -1.41$

$P(Z$

5 0

1 year ago

Describe the sample space for each of these experiments (a) A coin is tossed and a single die is rolled. (b) A paper cup is toss

kondor19780726 [428]

Answer:

a) S={(head,1),(head,2),(head,3),(head,4),(head,5),(head,6),(tails,1),(tails,2),(tails,3),(tails,4),(tails,5),(tails,6)}

b) S={top,bottom,side}

c) S={(Ace of diamonds),(Two of diamonds),(Three of diamonds),(Four of diamonds),(Five of diamonds),(Six of diamonds),(Seven of diamonds), (Eight of diamonds),(Nine of diamonds),(Ten of diamonds), (Jack of diamonds), (Queen of diamonds), (King of diamonds), (Ace of hearts),(Two of hearts),(Three of hearts),(Four of hearts),(Five of hearts),(Six of hearts),(Seven of hearts), (Eight of hearts), (Nine of hearts),(Ten of hearts), (Jack of hearts), (Queen of hearts),(King of hearts),(Ace of clubs),(Two of clubs),(Three of clubs),(Four of clubs),(Five of clubs),(Six of clubs),(Seven of clubs), (Eight of clubs), (Nine of clubs),(Ten of clubs), (Jack of clubs),(Queen of clubs),(King of clubs),(Ace of spades),(Two of spades),(Three of spades),(Four of spades),(Five of spades),(Six of spades),(Seven of spades),(Eight of spades), (Nine of spades),(Ten of spades),(Jack of spades),(Queen of spades),(King of spades)}

Step-by-step explanation:

The sample space is all the possible outcomes from each experiment:

a) The sample has the pairs of the result of the coin (head or tails) and the dice (a number from 1 to 6):

S={(head,1),(head,2),(head,3),(head,4),(head,5),(head,6),(tails,1),(tails,2),(tails,3),(tails,4),(tails,5),(tails,6)}

b) A paper cup can land in three different ways, on its top, bottom or side:

S={top,bottom,side}

c) The sample space is all the cards in the deck:

S={(Ace of diamonds),(Two of diamonds),(Three of diamonds),(Four of diamonds),(Five of diamonds),(Six of diamonds),(Seven of diamonds), (Eight of diamonds),(Nine of diamonds),(Ten of diamonds), (Jack of diamonds), (Queen of diamonds), (King of diamonds), (Ace of hearts),(Two of hearts),(Three of hearts),(Four of hearts),(Five of hearts),(Six of hearts),(Seven of hearts), (Eight of hearts), (Nine of hearts),(Ten of hearts), (Jack of hearts), (Queen of hearts),(King of hearts),(Ace of clubs),(Two of clubs),(Three of clubs),(Four of clubs),(Five of clubs),(Six of clubs),(Seven of clubs), (Eight of clubs), (Nine of clubs),(Ten of clubs), (Jack of clubs),(Queen of clubs),(King of clubs),(Ace of spades),(Two of spades),(Three of spades),(Four of spades),(Five of spades),(Six of spades),(Seven of spades),(Eight of spades), (Nine of spades),(Ten of spades),(Jack of spades),(Queen of spades),(King of spades)}

3 0

2 years ago

Suppose that 3\%3%3, percent of over 200{,}000200,000200, comma, 000 books borrowed from a library in a year are downloaded. The

sleet_krkn [62]

Answer:

The mean of the sampling distribution of the proportion of downloaded books is 0.03 and the standard deviation is 0.0197.

Step-by-step explanation:

Central Limit Theorem

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean $\mu = p$ and standard deviation $s = \sqrt{\frac{p(1-p)}{n}}$