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aleksandr82 [10.1K]

2 years ago

10

6) Andy is going kyacking. He has $341 to spend on parking and paddles. Parking costs $30 and paddles cost $45.20 per person. De

termine how many people can go kyacking

1 answer:

Lelechka [254]2 years ago

8 0

Answer:

6 people can go

Step-by-step explanation:

341-30=311

311-45.2x=0

45.2x=311

x=6.88

6.88 people can go but .88 of a person doesnt exist

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One letter is selected from the words "conditional probability." What is the probability that an "t" or "a" is chosen?

alexgriva [62]

Answer:

4 out of 22 = 4/22

Step-by-step explanation:

5 0

3 years ago

According to a 2014 Gallup poll, 56% of uninsured Americans who plan to get health insurance say they will do so through a gover

Airida [17]

Answer:

a) 24.27% probability that in a random sample of 10 people exactly 6 plan to get health insurance through a government health insurance exchange

b) 0.1% probability that in a random sample of 1000 people exactly 600 plan to get health insurance through a government health insurance exchange

c) Expected value is 560, variance is 246.4

d) 99.34% probability that less than 600 people plan to get health insurance through a government health insurance exchange

Step-by-step explanation:

To solve this question, we need to understand the binomial probability distribution and the binomial approximation to the normal.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

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

And p is the probability of X happening.

The expected value of the binomial distribution is:

$E(X) = np$

The variance of the binomial distribution is:

$V(X) = np(1-p)$

The standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that $\mu = E(X)$ , $\sigma = \sqrt{V(X)}$ .

56% of uninsured Americans who plan to get health insurance say they will do so through a government health insurance exchange.

This means that $p = 0.56$

a. What is the probability that in a random sample of 10 people exactly 6 plan to get health insurance through a government health insurance exchange?

This is P(X = 6) when n = 10. So

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 6) = C_{10,6}.(0.56)^{6}.(0.44)^{4} = 0.2427$

24.27% probability that in a random sample of 10 people exactly 6 plan to get health insurance through a government health insurance exchange

b. What is the probability that in a random sample of 1000 people exactly 600 plan to get health insurance through a government health insurance exchange?

This is P(X = 600) when n = 1000. So

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 600) = C_{1000,600}.(0.56)^{600}.(0.44)^{400} = 0.001$

0.1% probability that in a random sample of 1000 people exactly 600 plan to get health insurance through a government health insurance exchange

c. What are the expected value and the variance of X?

$E(X) = np = 1000*0.56 = 560$

$V(X) = np(1-p) = 1000*0.56*0.44 = 246.4$

d. What is the probability that less than 600 people plan to get health insurance through a government health insurance exchange?

Using the approximation to the normal

$\mu = 560, \sigma = \sqrt{246.4} = 15.70$

This is the pvalue of Z when X = 600-1 = 599. Subtract by 1 because it is less, and not less or equal.

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{599 - 560}{15.70}$

$Z = 2.48$

$Z = 2.48$ has a pvalue of 0.9934

99.34% probability that less than 600 people plan to get health insurance through a government health insurance exchange

4 0

3 years ago

A frog sits at one end of a table which is 2m long. In its first jump the frog goes a distance of 1m along the table, with its s

Phantasy [73]

Ah, this is an infinite sum question, or a sum of geometric sequence
when will the sum reach 1cm of the edge or about 1.99m
so

2m=200cm
within 1cm means at least 1.99m
so
we will use m and not cm for consitancy

sum of geometric sequence is
$S_n= \frac{a_1(1-r^n)}{1-r}$
a1=first term=initial jjump=1
r=common ratio=1/2
n=?, we ar solving for that

so
we want it to equal 1.99 so

$1.99= \frac{1(1- (\frac{1}{2})^n)}{1-\frac{1}{2}}$
$1.99= \frac{(1- (\frac{1}{2})^n)}{\frac{1}{2}}$
$1.99= 2(1- (\frac{1}{2})^n)$
divide both sides by 2
$\frac{1.99}{2} = 1- (\frac{1}{2})^n$
times -1
$\frac{-1.99}{2} = (\frac{1}{2})^n-1$
add 1 or 2/2 to both sides
$\frac{0.01}{2} = (\frac{1}{2})^n$
take the ln of both sides
$ln(\frac{0.01}{2}) = ln((\frac{1}{2})^n)$
$ln(\frac{0.01}{2}) = n ln(\frac{1}{2})$
divide both sides by ln(1/2)
$\frac{ln(\frac{0.01}{2})}{ln(\frac{1}{2})} =n$
use your calculatro to find that n≈7.64386
so on 7th jump, it is not yet at 1cm to the edge but at 8th jump, it is past
so 8th jump

7 0

3 years ago

N+7< -3 <br> Solve this inequality

Mice21 [21]

$n+7\ \textless \ -3 \\ \\ n \ \textless \ -3 - 7 \\ \\ n \ \textless \ -10 \\ \\$

The final result is: n < -10.

7 0

3 years ago

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Gabrielle is 12 years younger than Mikhail. The sum of their ages is 36 . What is Mikhail's age?

ololo11 [35]

X = age of Mikhail

x+x-12 = 36
2x= 36 + 12
2x = 48
x = 48/2
x= 24

3 0

3 years ago

Read 2 more answers