N+7<-3 help lollllll

12 POINTS!!! Find x. (Hint. Draw an auxiliary line

padilas [110]

Answer:

30

Step-by-step explanation:

x+50=180(संगत कोन भएकाले )

x=180-50

X=130

अब,

130-100=30

4 0

3 years ago

A CHICKEN COOP HOLDS A TOTAL OF 10 HENS OR 20 CHICKS. IF 2/5THS OF THE HENS IN A FULL COOP ARE REMOVED, HOW MANY NEW CHICKS COUL

LiRa [457]

Answer:

8 new chicks can be fitted in the coop.

Step-by-step explanation:

A chicken coop holds 10 hens or 20 chicks.

That means space in the coop for 10 hens = space for 20 chicks

Or space for 1 hen = space for 2 chicks

Now $\frac{2}{5}$ th of the hens were removed from the coop.

So, number of hens removed from the coop = $\frac{2}{5}\times 10$

= 4 hens

And space for 4 hens in the coop = space for the 8 chicks

Therefore, 8 new chicks can be fitted in the coop.

5 0

3 years ago

A consumer organization inspecting new cars found that many had appearance defects (dents, scratches, paint chips, etc.). While

Murljashka [212]

Answer:

The expected number of appearance defects in a new car = 0.64

standard deviation = 0.93(approximately)

Step-by-step explanation:

Given -

7% had exactly three, 11% exactly two, and 21% only one defect in cars.

Let X be no of defects in a car

P(X=3) = .07 , P(X=2) = .11 , P(X=3) = .21

P(X=0 ) = 1 - .07 - .11 - .21 = 0.61

The expected number of appearance defects in a new car =

E(X) = $\nu$ = $= \sum X P(X) =3\times 0.07 + 2\times0.11 + 3\times0.21 + 0\times0.61$ = 0.64

$\sigma^{2}$ = variation of E(X) = $\sum (X - \nu) ^{2}P(X)$ = $(3 - 0.64) ^{2}\times0.07 + (2 - 0.64) ^{2}\times0.11 + (1 - 0.64) ^{2}\times0.21 + (0 - 0.64) ^{2}\times0.61$

= $(2.36) ^{2}\times0.07 + (1.36) ^{2}\times0.11 + (.36) ^{2}\times0.21 + ( 0.64) ^{2}\times0.61$ = 0.8704

standard deviation = $\sqrt{\sigma^{2}}$ = $\sqrt{0.8704^{2}}$ = 0.93(approximately)

4 0

3 years ago

a statistics professor finds that when he schedules an office hour at the 10:30a, time slot, an average of three students arrive

Veseljchak [2.6K]

Answer:

The probability that in a randomly selected office hour in the 10:30 am time slot exactly two students will arrive is 0.2241.

Step-by-step explanation:

Let X = number of students arriving at the 10:30 AM time slot.

The average number of students arriving at the 10:30 AM time slot is, λ = 3.

A random variable representing the occurrence of events in a fixed interval of time is known as Poisson random variables. For example, the number of customers visiting the bank in an hour or the number of typographical error is a book every 10 pages.

The random variable X is also a Poisson random variable because it represents the fixed number of students arriving at the 10:30 AM time slot.

The random variable X follows a Poisson distribution with parameter λ = 3.

The probability mass function of X is given by:

$P(X=x)=\frac{e^{-\lambda}\lambda^{x}}{x!};\ x=0,1,2,3...,\ \lambda>0$

Compute the probability of X = 2 as follows:

$P(X=2)=\frac{e^{-3}3^{2}}{2!}=\frac{0.0498\times 9}{2}=0.2241$

Thus, the probability that in a randomly selected office hour in the 10:30 am time slot exactly two students will arrive is 0.2241.

6 0

3 years ago

Read 2 more answers

During a winter storm, the temperature on the top of a

Iteru [2.4K]

According to the rate of change, it is found that the hourly rate of temperature change at the bottom of the mountain was of -2.5 ºC per hour.

<h3>What is the average rate of change?</h3>

The average rate of change of a function is given by the change in the output divided by the change in the input.

In this problem, it is stated that at the top of the mountain, the average rate of change is of -2.5 ºC per hour, and at the bottom of the mountain the rate is the same, hence the hourly rate of temperature change at the bottom of the mountain was of -2.5 ºC per hour.

You can learn more about average rate of change at brainly.com/question/26420851

5 0

2 years ago