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3 years ago

I need help on this please help :)

Mathematics

1 answer:

GenaCL600 [577]3 years ago

4 0

Answer:

Because they are different and have different MCQ scqare UNIT.

Step-by-step explanation:

Make me brainliest please :)

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I need help with this

Sergio [31]

Answer:

Step-by-step explanation:

∠5 = ∠1 {Alternate exterior angles are congruent}

8x - 19 = 7x - 3

Add 19 to both sides

8x = 7x - 3 + 19

8x = 7x + 16

Subtract 7x from both sides

8x - 7x = 16

x = 16

∠5 = 8x - 19

= 8*16 - 19

= 128 - 19

∠5 = 109

∠5 + ∠4 = 180 {linear pair]

109 + ∠4 = 180

∠4 = 180 - 109

∠4 = 71°

4 0

3 years ago

Jane is is charge of making a banner for the basketball game this Saturday. She measures how long the banner is before painting

RSB [31]

Answer:

The measurement which Jane finds to be 10 meters is the length of the banner.

Step-by-step explanation:

The measurement of 10 meters which Jane found after measuring how long the banner is before painting is the LENGTH of the banner.

This is clear from the unit of what she finds (meters). It only indicates the measurement of one part of the banner, even though a banner has two parts, the length and width.

It is possible to find the AREA, or PERIMETER, or LENGTH.

But what she finds is the LENGTH of the banner. If it was Area or Perimeter, the unit would have been square meters.

7 0

3 years ago

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

postnew [5]

Answer:

a) 0.1542

b) 0.7685

c) 0.2315

d) No, it is not unusual

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

11. Which of the following shows the first 5 terms in

iren2701 [21]

G is the answer to question

5 0

3 years ago

A major credit card company has determined that customers charge between $100 and $1100 per month. If the monthly amount charged

Alexxx [7]

Answer:

a) $E(X) = \frac{a+b}{2} = \frac{100+1100}{2}=600$

b) First we need to calculate the variance given by this formula:

$Var(X) = \frac{(b-a)^2}{12}= \frac{(1100-100)^2}{12} = 83333.33$

And the deviation would be:

$Sd(X) = \sqrt{83333.33}= 288.675$

c) $P(600 < X< 889) = P(X$

And using the cdf we got:

$P(600 < X< 889)= F(889) -F(600) = \frac{889-100}{1000} -\frac{600-100}{1000}= 0.789- 0.5= 0.289$

d) $P(311 < X< 889)= F(889) -F(311) = \frac{889-100}{1000} -\frac{311-100}{1000}= 0.789- 0.211= 0.578$

Step-by-step explanation:

For this case we define the random variable X who represent the customers charge, and the distribution for X on this case is:

$X \sim Unif (a= 100, b=1100)$

Part a

For this case the average is given by the expected value and we can use the following formula:

$E(X) = \frac{a+b}{2} = \frac{100+1100}{2}=600$

Part b

First we need to calculate the variance given by this formula:

$Var(X) = \frac{(b-a)^2}{12}= \frac{(1100-100)^2}{12} = 83333.33$

And the deviation would be:

$Sd(X) = \sqrt{83333.33}= 288.675$

Part c

For this case we want to find the percent between 600 and 889, so we can use the cumulative distribution function given by:

$F(x) = \frac{X -100}{1100-100}= \frac{x-100}{1000}, 100 \leq X \leq 1100$

And we can find this probability:

$P(600 < X< 889) = P(X$

And using the cdf we got:

$P(600 < X< 889)= F(889) -F(600) = \frac{889-100}{1000} -\frac{600-100}{1000}= 0.789- 0.5= 0.289$

Part d

$P(311 < X< 889) = P(X$

And using the cdf we got:

$P(311 < X< 889)= F(889) -F(311) = \frac{889-100}{1000} -\frac{311-100}{1000}= 0.789- 0.211= 0.578$

6 0

3 years ago