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dsp73
3 years ago
10

Four bells ring at intervals of 3, 7 , 12 and 14 minutes respectively. All four rang together at 4:00a.m. At what time will they

ring together again?
Mathematics
1 answer:
Vlada [557]3 years ago
8 0
Basically ur just looking for the LCM (lowest common multiple) of all of these numbers.

3,7,12,14....lowest number that all 4 of these numbers go into evenly is 84.

so they will ring together next in 84 minutes (1 hr 24 minutes) making it 5:24 a.m.
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Maximizing revenue. Edwards University wants to determine what price to charge for tickets to football games. At $18 per ticket,
Papessa [141]

Answer:

New price is $12.75

New attendance is 57500

Step-by-step explanation:

For starters, we use the relation.

18 - 3x

Since we don't know the number of times we're expected to deduct $3 from the price.

Again, we have 40000 + 10000x on another hand. Question says to add 10000 people everytime $3 is deducted.

Since (40000 + 10000x) is the number of people, we multiply it by $4.50, the average amount each person spends is then gotten to be

(18 - 3x) (40000 + 10000x) + (40000 + 10000x) (4.5) = 0

Expanding the bracket, we have

[720000 + 60000x - 30000x²] + [180000 + 45000x] = 0, solving further, we have

900000 + 105000x - 30000x² = 0 or

-30000x² + 105000x + 900000 = 0

Then, we find the maximum of a quadratic, by finding the axis of symmetry. Use x = -b/2a

x = -105000 / 2 * -30000

x = -105000 / -60000

x = 1.75

From our earlier equations, the new price then is

18 - 3(1.75) = 18 - 5.25 = $12.75

The new attendance also is

40000 + 10000(1.75) =

40000 + 17500 = 57500

7 0
3 years ago
Hello:) anyone able to explain how to solve the equestrian for part (d) ? Thank youu
Anna007 [38]

Answer:

see explanation

Step-by-step explanation:

Given

4a^{4} - 5a² + 1 = 0

Use the substitution u = a², then equation is

4u²  - 5u + 1 = 0

Consider the product of the coefficient of the u² term and the constant term

product = 4 × 1 = 4 and sum = - 5

The factors are - 4 and - 1

Use these factors to split the u- term

4u² - 4u - u + 1 = 0 ( factor the first/second and third/fourth terms )

4u(u - 1) - 1(u - 1) = 0 ← factor out (u - 1) from each term

(u - 1)(4u - 1) = 0

Equate each factor to zero and solve for u

u - 1 = 0 ⇒ u = 1

4u - 1 = 0 ⇒ 4u = 1 ⇒ u = \frac{1}{4}

Convert u back into terms of a, that is

a² = 1 ⇒ a = ± 1

a² = \frac{1}{4} ⇒ a = ± \frac{1}{2}

Solutions are a = ± 1 , a = ± \frac{1}{2}

7 0
3 years ago
The number of hours a lightbulb burns before failing varies from bulb to bulb. The population distribution of burnout times is s
liubo4ka [24]

ANSWER:

The average burnout time of a large number of bulbs has a sampling distribution that is close to Normal.

STEP-BY-STEP EXPLANATION:

The cental limit theorem states, that id the sample size is large (30 or more), then the sampling distribution of the sample means is approximately normal with mean ц and standar deviation б/\sqrt{n}

Thus the correct answer is the average burnout time of a large number of bulbs has a sampling distribution that is close to Normal.

6 0
4 years ago
What is the prime factorization of 68
umka21 [38]
68 is not a prime number. The prime factorization of 68 would be 2 x 2 x 17.
6 0
3 years ago
Read 2 more answers
X+40+4x-5=6x+20 can I get some help with this problem please and thanks.
laila [671]

Answer:

the answer is c=−1/20x+7/4

Step-by-step explanation:

6 0
3 years ago
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