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1 year ago

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Ten little monkeys were jumping on a bed. There is a 35% chance that one will fall off and bump his head. In the bedroom next do

or, five kangaroos were jumping on a bed. Being more adept at jumping, there is only a 20% that a kangaroo will fall off the bed.
Required:
a. What are the chances that a monkey and a kangaroo will fall off the bed?
b. What are the chances that a monkey will fall off the bed?
c. If the monkeys enjoy this activity every night for an entire week, what are the chances that a monkey falls off the bed every one of the seven nights?

1 answer:

valina [46]1 year ago

3 0

Answer:

Explanation is in a file

Step-by-step explanation:

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Kay buys 12 pounds of apples. Each pound costs $3.if she gives the cashier two $20 bills, how much change should she receive.

zlopas [31]

12 pounds x $3 = $36

$40 - $36 = $4

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1 year ago

An automotive manufacturer wants to know the proportion of new car buyers who prefer foreign cars over domestic. In an earlier s

anyanavicka [17]

Answer:

1,496 new car buyers

Step-by-step explanation:

The sample size n in Simple Random Sampling is given by

$\bf n=\frac{z^2p(1-p)}{e^2}$

where

z = 1.645 is the critical value for a 90% confidence level (*)

p= 0.33 is the population proportion.

e = 0.02 is the margin of error

so

$\bf n=\frac{(1.645)^20.33*0.67}{0.02^2}=1,495.76\approx 1,496$

<em>(*)</em><em>This is a point z such that the area under the Normal curve N(0,1) inside the interval [-z, z] equals 90% = 0.9</em>

It can be obtained in Excel or OpenOffice Calc with

<em>NORMSINV(0.95)</em>

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1 year ago

Ivan and Jeff buy a package of 8 pens for $4.00. Ivan wants 5 of the pens, and Jeff wants 3. How much should each student pay?

irga5000 [103]

$4 / 8 = 50 cents per pen

.5 * 5 = 2.50
.5 * 3 = 1.50

Ivan pays $2.50 and Jeff pays $1.50

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1 year ago

In a recent year, a sample of grade 8 Washington State public school students taking a mathematics assessment test had a mean sc

Daniel [21]

Answer:

The number is $N =1147$ students

Step-by-step explanation:

From the question we are told that

The population mean is $\mu = 281$

The standard deviation is $\sigma = 34.4$

The sample size is n = 2000

percentage of the would you expect to have a score between 250 and 305 is mathematically represented as

$P(250 < X < 305 ) = P(\frac{ 250 - 281}{34.4 } < \frac{X - \mu }{\sigma } < \frac{ 305 - 281}{34.4 } )$

Generally

$\frac{X - \mu }{\sigma } = Z (Standardized \ value \ of \ X )$

So

$P(250 < X < 305 ) = P(-0.9012< Z$

$P(250 < X < 305 ) = P(z_2 < 0.698 ) - P(z_1 < -0.9012)$

From the z table the value of $P( z_2 < 0.698) = 0.75741$

and $P(z_1 < -0.9012) = 0.18374$

$P(250 < X < 305 ) = 0.75741 - 0.18374$

$P(250 < X < 305 ) = 0.57$

The percentage is $P(250 < X < 305 ) = 57\%$

The number of students that will get this score is

$N = 2000 * 0.57$

$N =1147$

6 0

2 years ago

From past experience, a company has found that in carton of transistors: 92% contain no defective transistors 3% contain one def

jasenka [17]

Answer:

$E(X) = 0*0.92 + 1*0.03 +2*0.03 +3*0.02 = 0.1500$

In order to find the variance we need to find first the second moment given by:

$E(X^2) = \sum_{i=1}^n X^2_i P(X_i)$

And replacing we got:

$E(X^2) = 0^2*0.92 + 1^2*0.03 +2^2*0.03 +3^2*0.02 = 0.3300$

The variance is calculated with this formula:

$Var(X) = E(X^2) -[E(X)]^2 = 0.33 -(0.15)^2 = 0.3075$

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

$Sd(X) = \sqrt{0.3075}= 0.5545$

Step-by-step explanation:

Previous concepts

The expected value of a random variable X is the n-th moment about zero of a probability density function f(x) if X is continuous, or the weighted average for a discrete probability distribution, if X is discrete.

The variance of a random variable X represent the spread of the possible values of the variable. The variance of X is written as Var(X).

Solution to the problem

LEt X the random variable who represent the number of defective transistors. For this case we have the following probability distribution for X

X 0 1 2 3

P(X) 0.92 0.03 0.03 0.02

We can calculate the expected value with the following formula:

$E(X) = \sum_{i=1}^n X_i P(X_i)$

And replacing we got:

$E(X) = 0*0.92 + 1*0.03 +2*0.03 +3*0.02 = 0.1500$

In order to find the variance we need to find first the second moment given by:

$E(X^2) = \sum_{i=1}^n X^2_i P(X_i)$

And replacing we got:

$E(X^2) = 0^2*0.92 + 1^2*0.03 +2^2*0.03 +3^2*0.02 = 0.3300$

The variance is calculated with this formula:

$Var(X) = E(X^2) -[E(X)]^2 = 0.33 -(0.15)^2 = 0.3075$

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

$Sd(X) = \sqrt{0.3075}= 0.5545$

8 0

2 years ago