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2 years ago

The sum of three consecutive integers is -369

Mathematics

2 answers:

dezoksy [38]2 years ago

3 0

Let the middle number be x.

Then the other numbers are x - 1 and x + 1.

x - 1 + x + x + 1 = -369

3x = -369

x = -123

x - 1 = -124

x + 1 = -122

Answer: The numbers are -124, -123, -122

Serga [27]2 years ago

3 0

17 is the answer

divide the sum of the three consecutive odd integers by 3: 51 / 3 = 17

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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.

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

Let X = 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 X follows a Binomial distribution with parameters n = 5 and p = 0.556.

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

The probability mass function of X 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.

6 0

1 year ago

For the wheat-yield distribution of Exercise 4.3.5, find

Setler79 [48]

Answer:

a) 90.695 lb

b) 85.305 lb

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

$\mu = 88, \sigma = 7$

(a) The 65th percentile

X when Z has a pvalue of 0.65. So X when Z = 0.385.

$Z = \frac{X - \mu}{\sigma}$

$0.385 = \frac{X - 88}{7}$

$X - 88 = 7*0.385$

$X = 90.695$

(b) The 35th percentile

X when Z has a pvalue of 0.35. So X when Z = -0.385.

$Z = \frac{X - \mu}{\sigma}$

$-0.385 = \frac{X - 88}{7}$

$X - 88 = 7*(-0.385)$

$X = 85.305$

6 0

1 year ago

Circle 1 has center (−4, −7) and a radius of 12 cm. Circle 2 has center (3, 4) and a radius of 15 cm.

CaHeK987 [17]

<h3>Answer with explanation:</h3>

It is given that:

Circle 1 has center (−4, −7) and a radius of 12 cm.

Circle 2 has center (3, 4) and a radius of 15 cm.

Two circles are said to be similar if by some translation and dilation it could be placed over the other to form the same circle.

The circles are similar because the transformation rule ( x,y ) → (x+7,y+11) can be applied to Circle 1 and then dilate it using a scale factor of 5/4

( Since, as the center of circle 1 is (-4,-7)

so,

(-4+7,-7+11) → (3,4)

( Since, the radius of circle 1 is 12 and that of circle 2 is 15 cm.

so, let the scale factor be k .

that means :

$12\times k=15\\\\k=\dfrac{15}{12}\\\\k=\dfrac{5}{4}$ )

7 0

1 year ago

Read 2 more answers

James is four years younger than Austin. If three times James' age is increased by the square of Austin's age, the result is 28.

lara31 [8.8K]

Let's use J for James's age and A for Austin's age. The equations are:

J = A - 4
3J + A² = 28

Just plug (A - 4) in the place of J in the second equation. This gives you:

3(A - 4) + A² = 28
-->
A² + 3A - 12 = 28
-->
A² + 3A - 40 = 0
-->
(A - 5)(A + 8) = 0
-->
A = 5 or -8

-8 is nonsense, so Austin is 5 years old. Therefore, James is 1 year old.

7 0

1 year ago

On a coordinate plane, square P Q R S is shown. Point P is at (4, 2), point Q is at (8, 5), point R is at (5, 9), and point S is

sergey [27]

Answer:

(D)The midpoint of both diagonals is (4 and one-half, 5 and one-half), the slope of RP is 7, and the slope of SQ is Negative one-sevenths.

Step-by-step explanation:

Point P is at (4, 2),
Point Q is at (8, 5),
Point R is at (5, 9), and
Point S is at (1, 6)

Midpoint of SQ $=\frac{1}{2}(1+8,5+6)=(4.5,5.5)$

Midpoint of PR $=\frac{1}{2}(4+5,2+9)=(4.5,5.5)$

Now, we have established that the midpoints (point of bisection) are at the same point.

Two lines are perpendicular if the slope of one is the negative reciprocal of the other.

In option D

Slope of RP =7

Slope of SQ $=-\dfrac17$

Therefore, lines RP and SQ are perpendicular.

Option D is the correct option.

6 0

2 years ago