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butalik [34]
3 years ago
10

Last summer Karl went to the beach every 5 days Antonia went to the beach every 7 days . How often did they see each other?

Mathematics
1 answer:
Nata [24]3 years ago
7 0
They saw each other every 35 days because the gcf that they both go into is 35
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ASAP HELP ME GIVING BRAINLIEST
Natasha2012 [34]

Answer:B

Step-by-step explanation:

The quadratic formula can be used to solve an equation only if the highest degree in the equation is 2

7 0
3 years ago
If 1 inch = 50 miles how many miles is 8.5 in
GenaCL600 [577]

✧・゚: *✧・゚:*    *:・゚✧*:・゚✧

                  Hello!

✧・゚: *✧・゚:*    *:・゚✧*:・゚✧

❖ 8.5 in = 425 miles

If 1 mi = 50 inches, multiply by 8.5:

50 x 8.5 = 425

~ ʜᴏᴘᴇ ᴛʜɪꜱ ʜᴇʟᴘꜱ! :) ♡

~ ᴄʟᴏᴜᴛᴀɴꜱᴡᴇʀꜱ

7 0
3 years ago
The operator of a pumping station has observed that demand for water during early afternoon hours has an approximately exponenti
melomori [17]

Answer:

a) 0.1496 = 14.96% probability that the demand will exceed 190 cfs during the early afternoon on a randomly selected day.

b) Capacity of 252.6 cubic feet per second

Step-by-step explanation:

Exponential distribution:

The exponential probability distribution, with mean m, is described by the following equation:

f(x) = \mu e^{-\mu x}

In which \mu = \frac{1}{m} is the decay parameter.

The probability that x is lower or equal to a is given by:

P(X \leq x) = \int\limits^a_0 {f(x)} \, dx

Which has the following solution:

P(X \leq x) = 1 - e^{-\mu x}

The probability of finding a value higher than x is:

P(X > x) = 1 - P(X \leq x) = 1 - (1 - e^{-\mu x}) = e^{-\mu x}

The operator of a pumping station has observed that demand for water during early afternoon hours has an approximately exponential distribution with mean 100 cfs (cubic feet per second).

This means that m = 100, \mu = \frac{1}{100} = 0.01

(a) Find the probability that the demand will exceed 190 cfs during the early afternoon on a randomly selected day. (Round your answer to four decimal places.)

We have that:

P(X > x) = e^{-\mu x}

This is P(X > 190). So

P(X > 190) = e^{-0.01*190} = 0.1496

0.1496 = 14.96% probability that the demand will exceed 190 cfs during the early afternoon on a randomly selected day.

(b) What water-pumping capacity, in cubic feet per second, should the station maintain during early afternoons so that the probability that demand will exceed capacity on a randomly selected day is only 0.08?

This is x for which:

P(X > x) = 0.08

So

e^{-0.01x} = 0.08

\ln{e^{-0.01x}} = \ln{0.08}

-0.01x = \ln{0.08}

x = -\frac{\ln{0.08}}{0.01}

x = 252.6

Capacity of 252.6 cubic feet per second

5 0
3 years ago
Jo's collection contains US, Indian and British stamps. If the ratio of US to Indian stamps is 5 to 2 and the ratio of Indian to
andreev551 [17]

Answer:(E). 25:2

Step-by-step explanation:

First ratio:

Us : India is 5 : 2

Second ratio

India : British is 5 : 1

multiply(5:2) by 5 and multiply (5:1) by 2 to balance the India stamps.

Therefore now

Us : India is 25 : 10

India : British is 10 : 2

The ratio of us : British is 25:2

5 0
3 years ago
A store had 235 MP3 players in the month of January. Every month ,30% of the MP3 players were sold and 50 new MP3 players were s
larisa86 [58]
F(n)=235-(235*.3^n)+50n is your equation where * is multiplication and ^ it to the power to.
8 0
3 years ago
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