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4 years ago

There are 31 days in December. After December 25, what fraction of the month remains

Mathematics

2 answers:

il63 [147K]4 years ago

6 0

Answer:

$\frac{6}{31}$

Step-by-step explanation:

Remember that in writing fractions, the denominator is the number of portions a certain object is cut equally and the numerator is the number of portions considered for the particular item.

6(31-25)

The numerator as the remaining number of days and and us the denominator as the number of days in December

7nadin3 [17]4 years ago

4 0

Answer:

After December 25,there are <em>6 days</em> to complete that year.

$\frac{6 \: days }{31\: days} \\ = > \frac{6}{31} fraction \: of \: the \: month \: remains$

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What is 6^4 x 6 1/4????

Natalka [10]

Answer:

8100

Step-by-step explanation:

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3 years ago

How do u work out how much 8 goes into 53>

Rufina [12.5K]

You would have to divide and either get a remainder or a decimal

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3 years ago

4. Find the standard from of the equation of a hyperbola whose foci are (-1,2), (5,2) and its vertices are end points of the dia

Ad libitum [116K]

The <em>standard</em> form of the equation of the hyperbola that satisfies all conditions is (x - 2)²/4 - (y - 2)²/5 = 1 .

<h3>How to find the standard equation of a hyperbola</h3>

In this problem we must determine the equation of the hyperbola in its <em>standard</em> form from the coordinates of the foci and a <em>general</em> equation of a circle. Based on the location of the foci, we see that the axis of symmetry of the hyperbola is parallel to the x-axis. Besides, the center of the hyperbola is the midpoint of the line segment with the foci as endpoints:

(h, k) = 0.5 · (- 1, 2) + 0.5 · (5, 2)

(h, k) = (2, 2)

To determine whether it is possible that the vertices are endpoints of the diameter of the circle, we proceed to modify the <em>general</em> equation of the circle into its <em>standard</em> form.

If the vertices of the hyperbola are endpoints of the diameter of the circle, then the center of the circle must be the midpoint of the line segment. By algebra we find that:

x² + y² - 4 · x - 4 · y + 4= 0

(x² - 4 · x + 4) + (y² - 4 · y + 4) = 4

(x - 2)² + (y - 2)² = 2²

The center of the circle is the midpoint of the line segment. Now we proceed to determine the vertices of the hyperbola:

V₁(x, y) = (0, 2), V₂(x, y) = (4, 2)

And the distance from the center to any of the vertices is 2 (<em>semi-major</em> distance, a) and the semi-minor distance is:

b = √(c² - a²)

b = √(3² - 2²)

b = √5

Therefore, the <em>standard</em> form of the equation of the hyperbola that satisfies all conditions is (x - 2)²/4 - (y - 2)²/5 = 1 .

To learn more on hyperbolae: brainly.com/question/27799190

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5 0

2 years ago

. A salesman sold 300 bags of maize to a retailer at Kshs .2000 each .He was given a commission of 3%.The salesman allowed a dis

nalin [4]

Answer:

The discount allowed per bag is kshs 4

while the discount allowed on all the 300 bags is kshs 1,200

Step-by-step explanation:

Here, we are interested in calculating the discount allowed on the sales of the bag of maize.

From the question, we are told that the sales man allowed a discount of 0.2% on the maize sold.

Now, to find the amount of this, we proceed as follows;

What we simply need to do is to find 0.2% of the each of the price of the maize bags, and then we can proceed to find the total discount given on all maize bags sold.

The discount on each bag of maize would be;

0.2% of kshs 2000

That would be;

0.2/100 * 2,000 = 400/100 = kshs 4

Since there are 300 bags, the total amount of discount allowed is 300 * kshs 4 = kshs 1,200

8 0

4 years ago

I. If 40% of all commuters ride to work in carpools, if eight workers are selected, find the probability that exactly five will

snow_tiger [21]

Complete question is;

I. If 40% of all commuters ride to work in carpools, if eight workers are selected, find the probability that exactly five will ride in carpools.

II. Traffic lights with a turning arrow are red 70% of the time. If you approach 6 lights in a row, find the probability of stopping: a. Exactly four times. b. At least three times.

Answer:

I)P(X = 5) = 0.1239

IIa)P(X = 4) = 0.3241

IIb)P(X ≥ 3) = 0.9294

Step-by-step explanation:

I) This is a binomial probability problem.

Formula is;

P(X = x) = nCx(p^(x) × (1 - p)^(n - x)

40% of all commuters ride to work in carpools. Thus, p = 0.4

eight workers are selected and we are to find the probability that exactly five will ride in carpools.

Thus, n = 8 and x = 5

P(X = 5) = 8C5 × 0.4^(5) × (1 - 0.4)^(8 - 5)

P(X = 5) = 56 × 0.01024 × 0.6³

P(X = 5) = 0.1239

II) Traffic lights with a turning arrow are red 70% of the time and red means stop.

Thus; p = 0.7

Since 6 lights in a row;

n = 6

a) Probability of stopping exactly 4 times;

P(X = 4) = 6C4 × 0.7⁴ × (1 - 0.7)^(6 - 4)

P(X = 4) = 15 × 0.2401 × 0.09

P(X = 4) = 0.3241

b) probability of at least 3 times is;

P(X ≥ 3) = P(3) + P(4) + P(5) + P(6)

P(3) = 6C3 × 0.7³ × (1 - 0.7)^(6 - 3)

P(3) = 0.1852

P(4) = 0.3241

P(5) = 6C5 × 0.7^(5) × (1 - 0.7)^(6 - 5)

P(5) = 0.3025

P(6) = 6C6 × 0.7^(6) × (1 - 0.7)^(6 - 6)

P(6) = 0.1176

P(X ≥ 3) = 0.1852 + 0.3241 + 0.3025 + 0.1176

P(X ≥ 3) = 0.9294

5 0

3 years ago