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Maru [420]
3 years ago
15

Henry and Bailey plan to make key chains. They need foot of string and 5 beads for each key chain. Henry has $8 to buy string, a

nd he can buy 120 feet of string for
$3.25. Bailey has $27 to purchase beads, and she can buy a pack of 35 beads for $2.59.

If Henry buys as much string as he can, and Bailey buys as many beads as she can, how many key chains can Henry and Bailey make?
Mathematics
1 answer:
Alona [7]3 years ago
5 0

Answer:

72 Keychains

Step-by-step explanation:

120 feet = $3.25

divide 120 by 3.25 to find how much string can be bought with $1. The result is 36.923. Then multiply that number by the number of dollars allotted to buying string. 8*36.923= 295.384. This is the total number of feet of string they have. They use 1 foot for each key chain so they can make 295 keychains based on how much string they have.

35 beads = $2.59

Divide 35 by 2.59 to find how many beads can be bought with $1. The result is 13.514. Then multiply that number by the number of dollars allotted to buying beads. This comes out to 364.878. This means She can buy 364 beads. Each Key chain needs 5 beads so divide 364/5 getting a total of 72 keychains.

Since there aren't enough beads to make any more than 72 keychains, they can make a total of 72 keychains (even though there is extra string)

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A really bad carton of eggs contains spoiled eggs. An unsuspecting chef picks eggs at random for his ""Mega-Omelet Surprise."" F
Dima020 [189]

Answer:

(a) The probability that of the 5 eggs selected exactly 5 are unspoiled is 0.0531.

(b) The probability that of the 5 eggs selected 2 or less are unspoiled is 0.3959.

(c) The probability that of the 5 eggs selected more than 1 are unspoiled is 0.8747.

Step-by-step explanation:

The complete question is:

A really bad carton of 18 eggs contains 8 spoiled eggs. An unsuspecting chef picks 5 eggs at random for his “Mega-Omelet Surprise.” Find the probability that the number of unspoiled eggs among the 5 selected is

(a) exactly 5

(b) 2 or fewer

(c) more than 1.

Let <em>X</em> = number of unspoiled eggs in the bad carton of eggs.

Of the 18 eggs in the bad carton of eggs, 8 were spoiled eggs.

The probability of selecting an unspoiled egg is:

P(X)=p=\frac{10}{18}=0.556

A randomly selected egg is unspoiled or not is independent of the others.

It is provided that a chef picks 5 eggs at random.

The random variable <em>X</em> follows a Binomial distribution with parameters <em>n</em> = 5 and <em>p</em> = 0.556.

The success is defined as the selection of an unspoiled egg.

The probability mass function of <em>X</em> is given by:

P(X=x)={5\choose x}(0.556)^{x}(1-0.556)^{5-x};\ x=0,1,2,3...

(a)

Compute the probability that of the 5 eggs selected exactly 5 are unspoiled as follows:

P(X=5)={5\choose 5}(0.556)^{5}(1-0.556)^{5-5}\\=1\times 0.05313\times 1\\=0.0531

Thus, the probability that of the 5 eggs selected exactly 5 are unspoiled is 0.0531.

(b)

Compute the probability that of the 5 eggs selected 2 or less are unspoiled as follows:

P (X ≤ 2) = P (X = 0) + P (X = 1) + P (X = 2)

              =\sum\imits^{2}_{x=0}{{5\choose 5}(0.556)^{5}(1-0.556)^{5-5}}\\=0.0173+0.1080+0.2706\\=0.3959

Thus, the probability that of the 5 eggs selected 2 or less are unspoiled is 0.3959.

(c)

Compute the probability that of the 5 eggs selected more than 1 are unspoiled as follows:

P (X > 1) = 1 - P (X ≤ 1)

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

              =1-\sum\limits^{1}_{x=0}{{5\choose 5}(0.556)^{5}(1-0.556)^{5-5}}\\=1-0.0173-0.1080\\=0.8747

Thus, the probability that of the 5 eggs selected more than 1 are unspoiled is 0.8747.

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3 years ago
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1.False
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3 years ago
Find the center of a circle with the equation: x2 y2−32x−60y 1122=0 x 2 y 2 − 32 x − 60 y 1122 = 0
mixas84 [53]

The equation of a circle exists:

$(x-h)^2 + (y-k)^2 = r^2, where (h, k) be the center.

The center of the circle exists at (16, 30).

<h3>What is the equation of a circle?</h3>

Let, the equation of a circle exists:

$(x-h)^2 + (y-k)^2 = r^2, where (h, k) be the center.

We rewrite the equation and set them equal :

$(x-h)^2 + (y-k)^2 - r^2 = x^2+y^2- 32x - 60y +1122=0

$x^2 - 2hx + h^2 + y^2 - 2ky + k^2 - r^2 = x^2 + y^2 - 32x - 60y +1122 = 0

We solve for each coefficient meaning if the term on the LHS contains an x then its coefficient exists exactly as the one on the RHS containing the x or y.

-2hx = -32x

h = -32/-2

⇒ h = 16.

-2ky = -60y

k = -60/-2

⇒ k = 30.

The center of the circle exists at (16, 30).

To learn more about center of the circle refer to:

brainly.com/question/10633821

#SPJ4

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2 years ago
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