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3 years ago

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Lisa gets $ 50 for free. She buys colored pencils for $ 5.90, a book for $ 8, a CD for $ 12 50ct and 3 scoops of ice cream at 90

ct each. She takes the rest to the bank. Before that, calculate how much Lisa's new savings account balance

1 answer:

Nikolay [14]3 years ago

8 0

20.9 cause you just subtract everything from the original number

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For a certain river, suppose the drought length Y is the number of consecutive time intervals in which the water supply remains

AnnZ [28]

Answer:

a) There is a 9% probability that a drought lasts exactly 3 intervals.

There is an 85.5% probability that a drought lasts at most 3 intervals.

b)There is a 14.5% probability that the length of a drought exceeds its mean value by at least one standard deviation

Step-by-step explanation:

The geometric distribution is the number of failures expected before you get a success in a series of Bernoulli trials.

It has the following probability density formula:

$f(x) = (1-p)^{x}p$

In which p is the probability of a success.

The mean of the geometric distribution is given by the following formula:

$\mu = \frac{1-p}{p}$

The standard deviation of the geometric distribution is given by the following formula:

$\sigma = \sqrt{\frac{1-p}{p^{2}}$

In this problem, we have that:

$p = 0.383$

So

$\mu = \frac{1-p}{p} = \frac{1-0.383}{0.383} = 1.61$

$\sigma = \sqrt{\frac{1-p}{p^{2}}} = \sqrt{\frac{1-0.383}{(0.383)^{2}}} = 2.05$

(a) What is the probability that a drought lasts exactly 3 intervals?

This is $f(3)$

$f(x) = (1-p)^{x}p$

$f(3) = (1-0.383)^{3}*(0.383)$

$f(3) = 0.09$

There is a 9% probability that a drought lasts exactly 3 intervals.

At most 3 intervals?

This is $P = f(0) + f(1) + f(2) + f(3)$

$f(x) = (1-p)^{x}p$

$f(0) = (1-0.383)^{0}*(0.383) = 0.383$

$f(1) = (1-0.383)^{1}*(0.383) = 0.236$

$f(2) = (1-0.383)^{2}*(0.383) = 0.146$

Previously in this exercise, we found that $f(3) = 0.09$

So

$P = f(0) + f(1) + f(2) + f(3) = 0.383 + 0.236 + 0.146 + 0.09 = 0.855$

There is an 85.5% probability that a drought lasts at most 3 intervals.

(b) What is the probability that the length of a drought exceeds its mean value by at least one standard deviation?

This is $P(X \geq \mu+\sigma) = P(X \geq 1.61 + 2.05) = P(X \geq 3.66) = P(X \geq 4)$ .

We are working with discrete data, so 3.66 is rounded up to 4.

Either a drought lasts at least four months, or it lasts at most thee. In a), we found that the probability that it lasts at most 3 months is 0.855. The sum of these probabilities is decimal 1. So:

$P(X \leq 3) + P(X \geq 4) = 1$

$0.855 + P(X \geq 4) = 1$

$P(X \geq 4) = 0.145$

There is a 14.5% probability that the length of a drought exceeds its mean value by at least one standard deviation

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3 years ago

PLS HELP. WILL MARK BRAINLIEST In a capture/recapture program, 26% of the animals recaptured have tags. Suppose 118 animals we

Mars2501 [29]

Total Population= 11800÷26

Step-by-step explanation:

Number of animals captured,tagged and released= 118

In capture/recapture program 26% have tags

i.e, 26% of total animals= 118

Let total number of animals = P

(26÷100)× P =118

26 P =118×100

P=11800÷26

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3 years ago

Please answer the ixl math problem for 15 points and please fill in the blanks with the correct answer please i beg will mark br

aivan3 [116]

Answer:

y = 5x + 4

Step-by-step explanation:

oooofoooofoooof

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Find the variance for the given sample data, Round your answer to one more decimal place than the original data 7 725 1 A. 11.8

VladimirAG [237]

Answer:

D. 7.7

Step-by-step explanation:

7.725, 2 does not round up so its left to 7.7

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3 years ago

Nadia is a stockbroker. She earns 13% commission each week. Last week, shes sold $6,4006 worth of stocks. How much did she m

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Last week= $8320.78 2011= 432680.56

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4 years ago