All categories

3 years ago

A large hose can fill 7/8 of a water tank in one hour. A small hose

Mathematics

1 answer:

luda_lava [24]3 years ago

4 0

Answer: 1/8

Step-by-step explanation:

From the question, we are informed that a large hose can fill 7/8 of a water tank in one hour while a small hose

can fill 3/4 of a water tank in one hour.

To get the solution to the question above, we have to find the difference between the amount if water filled by the large and the smaller hose. This will be:

= 7/8 - 3/4

= 7/8 - 6/8

= 1/8

You might be interested in

Whats 2+2*89*546/5^2+768/678*100000

Natasha_Volkova [10]

330987894/2825 you can check your answers by using a calculator such as online cal.

7 0

3 years ago

The half-life of a radioactive isotope is 10 years. About how long would it take 5 grams of the isotope to decay until there wer

valkas [14]

Answer:

Option A. 10 years

Step-by-step explanation:

we know that

Half-life $(t^{\frac{1}{2}})$ is the amount of time required for a quantity to fall to half its value as measured at the beginning of the time period.

In this problem $(t^{\frac{1}{2}})$ of a radioactive isotope is 10 years, which means that after 10 years half of the sample would have decayed and half would be left as it is.

After 10 years ( first half life) 5/2 = 2.5 g decays and 2.5 g remains left

8 0

3 years ago

The average lifetime of a certain new cell phone is 3.4 years. The manufacturer will replace any cell phone failing within three

melisa1 [442]

Answer:

68% of these phones last 3.87 years.

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 average lifetime of a certain new cell phone is 3.4 years.

This means that $m = 3.4, \mu = \frac{1}{3.4} = 0.2941$

$P(X \leq x) = 1 - e^{-0.2941x}$

68% of these phones last how long (in years)?

This is x for which:

$P(X \leq x) = 0.68$

$P(X \leq x) = 1 - e^{-0.2941x}$

Then

$0.68 = 1 - e^{-0.2941x}$

$e{-0.2941x} = 0.32$

$\ln{e{-0.2941x}} = \ln{0.32}$

$-0.2941x = \ln{0.32}$

$x = -\frac{\ln{0.32}}{0.2941}$

$x = 3.87$

68% of these phones last 3.87 years.

4 0

3 years ago

Given: ⅓ (18x + 27) = 81 Prove: x = 12

sasho [114]

Answer:

Firstly, rewrite the equation:

⅓ (18 + 27) = 81

Substitute x for the given number of it's supposed equivalent.

In this case x = 12.

⅓ (18(12) + 27) = 81

Solve using PEMDAS and simplify what is in the parenthesis first. Then, multiply.

(18 x 12) + 27 = 243

Now, solving using PEMDAS, multiply the total of what you got that was originally in the parenthesis by ⅓ .

⅓ (243) = 81

When you multiple these number they are equivalent to 81.

81 = 81

Since the equation given, when substituted x for 12, is equivalent to 81, this proves that substituting x for 12 makes this equation true.

4 0

3 years ago

(URGENT)

White raven [17]

A^2 + b^2 = c^2...a and b are the legs and c is the hypotenuse
8^2 + 5^2 = c^2
64 + 25 = c^2
89 = c^2...take the sqrt of both sides...this eliminates the ^2
sqrt 89 = c
9.4 = c....shortest path is 9.4 km

7 0

3 years ago