How do I do this? I’m not very good at geometry.

One pound of chicken cost $3. Jims buys 8 2/3 pounds. How much does he pay

seropon [69]

He will pay 26.1$ for 8 2/3 pounds of chicken.

Hope this helps!!

3 0

3 years ago

A "planet transit" is a rare celestial event in which a planet appears to cross in front of its star as seen from Earth. The pla

nikitadnepr [17]

Answer:

There is a 55.95% probability that, in the next 3,000 stars monitored by the Kepler mission, more than 4 planet transits will be seen.

Step-by-step explanation:

Assume that the number of planet transits discovered for every 3,000 stars follows a Poisson distribution with λ=5.

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

$P(X = x) = \frac{e^{-\lambda}*\lambda^{x}}{(x)!}$

In which

x is the number of sucesses

$e = 2.71828$ is the Euler number

$\lambda$ is the mean in the given time interval.

For this problem, we have that $\lambda = 5$

What is the probability that, in the next 3,000 stars monitored by the Kepler mission, more than 4 planet transits will be seen?

That is $P(X > 4)$ . We either see 4 or less planets, or we see more than 4. The sum of the probabilities is decimal 1. So

$P(X \leq 4) + P(X > 4) = 1$

$P(X > 4) = 1 - P(X \leq 4)$

In which

$P(X \leq 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)$

So

$P(X = x) = \frac{e^{-\lambda}*\lambda^{x}}{(x)!}$

$P(X = 0) = \frac{e^{-5}*5^{0}}{(0)!} = 0.0067$

$P(X = 1) = \frac{e^{-5}*5^{1}}{(1)!} = 0.0337$

$P(X = 2) = \frac{e^{-5}*5^{2}}{(2)!} = 0.0842$

$P(X = 3) = \frac{e^{-5}*5^{3}}{(3)!} = 0.1404$

$P(X = 4) = \frac{e^{-5}*5^{4}}{(4)!} = 0.1755$

$P(X \leq 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.0067 + 0.0337 + 0.0842 + 0.1404 + 0.1755 = 0.4405$

Finally

$P(X > 4) = 1 - P(X \leq 4) = 1 - 0.4405 = 0.5595$

There is a 55.95% probability that, in the next 3,000 stars monitored by the Kepler mission, more than 4 planet transits will be seen.

6 0

3 years ago

If you answer all of the and tell me the formula you will get brainiest (first answer)

arsen [322]

1a. 9qt= 2 gal 1 qt2a. 20 oz= 2 C 4 oz3a. 10 C4a. 9360 ft5a. 3 gal 2 qt6a. 137 cm7a. 16 oz8a. 101 oz9a. 12 c10a. 22 cm 6 mm11a. 44 oz12a. 0 gal 2 qt 21 oz
1b. 15 oz2b. 3 mi 160 yds3b. 4 qt4b. 3 mi 1560 ft5b. 8320 yd6b. 76 oz7b. 74 oz8b. 914 cm9b. 6240 yd10b. 1 pt 1 c11b. 2 yd 2 ft12b. 24 oz
1c. 2,250 ml2c. 8 km 280 m3c. 3 kg 950 g4c. 15720 ft5c. 0 gal 3 qt 3 oz6c. 15 qt7c. 3 km 70 m8c. 17 oz9c. 2 qt 30 oz10c. 16 qt11c. 2400 yd12c. 2 mi 3480 ft

6 0

4 years ago

Find the value of variable x. If your answer is not an integer, write it in simplest radical form with the denominator rationali

Otrada [13]

Answer:

the answer is 7

Step-by-step explanation:

angle 30°

x=14/2

x=7

8 0

4 years ago

Which congruence theorems or postulates could be given as reasons why ABC =XYZ HELP !!!!

lara31 [8.8K]

Answers:
A) SAS
B) ASA
E) LL

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Explanation:

Let's go through each possible answer choice

A. We can use SAS because we know that AC = XZ (horizontal sides) is one pair of congruent sides, the angle pairing is C = Z, and the other pair of sides is BC = YZ (vertical sides)

B. We can also use ASA. Note how A = X is one pair of angles, AC = XZ is the middle pair of sides, and C = Z is the second pair of angles. The side is between the angles.

C. We can't use SSS as we don't know AB = XY is true or not. There are no tick marks indicating congruence.

D. We can't use LA here because we don't know the altitudes (LA = leg altitude)

E. We can use LL here because we have two right triangles. The two pairs of legs (AC = XZ and BC = YZ) are congruent based on the tickmarks.

F. We can't use HL similar to the reasoning for choice C. We don't know the hypotenuses are congruent or not. If we knew that AB = XY, then we could use HL (hypotenuse leg).

4 0

4 years ago

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