Ask question

Ask question

All categories

3 years ago

12

A farmer saw some chickens and pigs in a field. he counted 30 heads and 84 legs. determine exactly how many chickens and pigs he

saw.

1 answer:

loris [4]3 years ago

8 0

12 pigs×4 legs=48 legs
18 chickens×2 legs=36 legs
48 legs+36 legs=84 legs

12 are pigs and 18 are chickens

You might be interested in

3x + 8y= -5<br> -2x + 2y= 18

WINSTONCH [101]

Answer:

3x+8y=-5-equatin.1

-2x+2y=18-equation.2

now,

subtracting both equation.1 by equation.2,then

3x+8y=-5

-8x+8y=72

+ - -

<h3 />

6 0

3 years ago

An initial investment of $60.00 increases in value by 15% each year. Which of the following statements are true? Select all that

VikaD [51]

15% each yer
represented by
f(x)=60(1.15^x)

xintercept is when the function equals zero
since it increases only, it cannot equal zero and does not have an x intercept

after 4 years the investment value is 104
after 6 yaers value is 138.7
after 7.86 years the value is179.98 which is roughly 3 times of 60
8 years later183.54 close enough

answers
f(x)=60*1.15^x
after 7.86 years value is 3 times initioal value
after 8 years value is 184

4 0

4 years ago

Read 2 more answers

Of all the registered automobiles in a city, 12% fail the emissions test. Fourteen automobiles are selected at random to undergo

postnew [5]

Answer:

<u>a) 0.1542</u>

<u>b) 0.7685</u>

<u>c) 0.2315</u>

<u>d) No, it is not unusual</u>

Explanation:

The procedure to make the test meets the requirements of binomial experiments because:

there are two possible mutually exclusive outputs: fail the test, or pass the test.
the probability of each event remains constant during all the test (p=12% = 0.12, for failing the test, and 1-p = 88% = 0.88, for passing the test)
each trial (test) is independent of other trial.

Solution

(a) Find the probability that exactly three of them fail the test.

You want P(X=3)

Using the equation for discrete binomial experiments, the probability of exactly x successes is:

$P(X=x)=C(n,x)\cdot p^x\cdot (1-p)^{(n-x)}$

Substituting C(n,x) with its developed form, that is:

$P(X=x)=\dfrac{n!}{x!\cdot (n-x)!}\cdot p^x\cdot (1-p)^{(n-x)}$

Thus, you must use:

x = 3 (number of automobiles that fail the emissions test)
n = 14 (the number of automobiles selected to undergo the emissions test),
p = 0.12 (probability of failing the test; this is the success of the variable on our binomial experiment)
1 - p = 0.88 (probability of passing the test; this is the fail of the variable on our binomial experiment)

$P(X=3)=\dfrac{14!}{3!\cdot (14-3)!}\cdot 0.12^3\cdot 0.88^{11}=0.1542$

(b) Find the probability that fewer than three of them fail the test.

The probability that fewer than three of them fail the test is the probability that exactly 0, or exactly 1, or exactly 2 fail the test.

That is: P(X=0) + P(X=1) + P(X=2)

Using the same formula:

$P(X=0)=\dfrac{14!}{0!\cdot 14!}\times 0.12^0\cdot 0.88^{14}$

$P(X=0)=0.1670$

$P(X=1)=\dfrac{14!}{1!\cdot 13!}\cdot 0.12^1\cdot 0.88^{13}$

$P(X=1)=0.3188$

$P(X=2)=\dfrac{14!}{2!\cdot 12!}\codt0.12^2\cdot 0.88^{12}$

$P(X=2)=0.2826$

P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0.7685

(c) Find the probability that more than two of them fail the test.

The probability that more than two of them fail the test is equal to 1 less the probability that exactly 0, or exactly 1, or exactly 2 fail the test:

P( X > 2) = 1 - P( X = 0) - P(X = 1) - P(X = 2)

P X > 2) = 1 - [P(X=0) + P(X = 1) + P(X = 2)]

P (X > 2) = 1 - [0.7685]

P (X > 2) = 0.2315

(d) Would it be unusual for none of them to fail the test?

Remember that not failing the test is the fail of the binomial distribution. Thus, none of them failing the test is the same as all of them passing the test.

You can find the probability that all the automibles pass the emission tests by multiplying the probability of passing the test (0.88) 14 times.

Then, the probability that none of them to fail the test is equal to:

$(1-p)^{14}\\\\(0.88)^{14}=0.1671$

That means that the probability than none of the automobiles of the sample fail the test is 16.71%.

Unusual events are usually taken as events with a probability less than 5%. Thus, this event should not be considered as unusual.

5 0

3 years ago

Please help which of the following scenarios must represent two independent events?

sergejj [24]

Given probabilities P(A) AND P(B) are independent, the probability that both events to occur is given by:
P(A)∩P(B)
thus from the example, independent events will be:
P(blue)=70%, P(green)=30%, P(blue and green)=21%
Because:
P(blue)∩P(green)=0.7×0.3=0.21=21%

5 0

3 years ago

Can someone please help me with this problem, it is really confusing

levacccp [35]

Answer =-0.7x +0

Find the gradient by y2-y1/x2-x1
-14 -7/ 20- -10
-21/30
-0.7
And plug in the coordinates to find the y intercept
7= -0.7 (-10) + c
C=0

4 0

3 years ago