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3 years ago

10

There are 3 ruby throated hummingbirds and two of another type at the feeder. The birds have a mass of 23 grams in all. What oth

er type of hummingbird is at the feeder? Explain

1 answer:

frutty [35]3 years ago

7 0

You have to divide by 23/3 which goes to 7. Which is 21. 2 ruby are left to feed.

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Y=-4x-10 for all prime numbers less than 15. What is the domain and range for this?

Bad White [126]

$The\ prime\ numbers\ less\ than\ 15\ is\ domain:\\\\D=\{2;\ 3;\ 5;\ 7;\ 11;\ 13\}\\\\Range:\\\\f(2)=-4(2)-10=-18\\f(3)=-4(3)-10=-22\\f(5)=-4(5)-10=-30\\f(7)=-4(7)-10=-38\\f(11)=-4(11)-10=-54\\f(13)=-4(13)-10=-62\\\\Range:\{-18;-22;-30;-38;-54;-62\}$

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3 years ago

What is 500,000,115 written in standard form

neonofarm [45]

<u>Answer:</u>

<u>The answer to five hundred million, one hundred fifteen in standard form is actually 500,000,115 because five hundred million in standard form is 500,000,000 and one hundred fifteen in standard form is 115.</u>

3 0

3 years ago

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Can someone please tell me the answer

Vilka [71]

Answer:

36 since they are similer triangles

Longer process:

24/24 = 36/x

solve this and you get 36

8 0

3 years ago

2. The quality assurance department inspects its production line. The product either fails or passes the inspection. Past experi

Oksana_A [137]

Answer:

(a) $E(X) = 950$

(b) $COV = 0.007255$

(c) $P(X > 980) = 0.00001\\\\$

Step-by-step explanation:

The given problem can be solved using binomial distribution since the product either fails or passes, the probability of failure or success is fixed and there are n repeated trials.

probability of failure = q = 0.05

probability of success = p = 1 - 0.05 = 0.95

number of trials = n = 1000

(a) What is the expected number of non-defective units?

The expected number of non-defective units is given by

$E(X) = n \times p \\\\E(X) = 1000 \times 0.95 \\\\E(X) = 950$

(b) what is the COV of the number of non-defective units?

The coefficient of variance is given by

$COV = \frac{\sigma}{E(X)}$

Where the standard deviation is given by

$\sigma = \sqrt{n \times p\times q} \\\\\sigma = \sqrt{1000 \times 0.95\times 0.05} \\\\\sigma = 6.892$

So the coefficient of variance is

$COV = \frac{6.892}{950}$

$COV = 0.007255$

(c) What is the probability of having more than 980 non-defective units?

We can use the Normal distribution as an approximation to the Binomial distribution since n is quite large and so is p.

$P(X > 980) = 1 - P(X < 980)\\\\P(X > 980) = 1 - P(Z < \frac{x - \mu}{\sigma} )\\\\$

We need to consider the continuity correction factor whenever we use continuous probability distribution (Normal distribution) to approximate discrete probability distribution (Binomial distribution).

$P(X > 980) = 1 - P(Z < \frac{979.5 - 950}{6.892} )\\\\P(X > 980) = 1 - P(Z < \frac{29.5}{6.892} )\\\\P(X > 980) = 1 - P(Z < 4.28)\\\\$

The z-score corresponding to 4.28 is 0.99999

$P(X > 980) = 1 - 0.99999\\\\P(X > 980) = 0.00001\\\\$

So it means that it is very unlikely that there will be more than 980 non-defective units.

8 0

3 years ago

A birthday cake is being made using a cylindrical baking pan. The radius of the pan is 4 inches, and the height is 5 inches whic

Aleonysh [2.5K]

Answer:

This formula:

Cylinder Volume = π • r² • height

Volume = PI * 4 ^2 * 5

Volume = PI * 16 * 5

Volume = 251.327412 cubic inches

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers