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rodikova [14]
3 years ago
13

Give 5 dividing decimal problems in the hundreds such as 261.68 divided by 189.20 etc​

Mathematics
2 answers:
Andrew [12]3 years ago
7 0

Answer:

742.563/365.247=2.03304

321.01/654.32=0.4906

987.65/567.89=0.017

669.32/124.33=5.383

505.05/101.01=5

faltersainse [42]3 years ago
6 0
275.25 divided by 25.17
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Step-by-step explanation:

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3 years ago
seven out of every 8 students surveyed owns a bike. The difference between the number of students who own a bike and those who d
guajiro [1.7K]
So if x people were surveyed, \frac{7}{8} people owned a bike and \frac{1}{8}  did not.

The difference between them is 72:
\frac{7}{8} x-\frac{1}{8} [/tex] =72
so this means that
\frac{6}{8} x=72

let's simplify:

\frac{3}{4} x=72

and divide by 3:

\frac{1}{4} x=24

and mupliply by 4:

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so 96 people were surveyed







5 0
3 years ago
What is the solution set to the equation (2x−4)(4x−5)=0?
STALIN [3.7K]
Answer: A. (5/4, 2)
Explanation:
2x-4=0
4x-5=0

2x=4
4x=5

x=2
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3 0
2 years ago
Which expression is equivalent to 144 3/2
natka813 [3]

Answer:

144=12², so 144^3/2=(12²)^(3/2)=12³=1728.

this answer is not from the internet

go check yourselves

mark me brainliest

3 0
1 year ago
Read 2 more answers
Suppose X has an exponential distribution with mean equal to 23. Determine the following:
e-lub [12.9K]

Answer:

a) P(X > 10) = 0.6473

b) P(X > 20) = 0.4190

c) P(X < 30) = 0.7288

d) x = 68.87

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}

Mean equal to 23.

This means that m = 23, \mu = \frac{1}{23} = 0.0435

(a) P(X >10)

P(X > 10) = e^{-0.0435*10} = 0.6473

So

P(X > 10) = 0.6473

(b) P(X >20)

P(X > 20) = e^{-0.0435*20} = 0.4190

So

P(X > 20) = 0.4190

(c) P(X <30)

P(X \leq 30) = 1 - e^{-0.0435*30} = 0.7288

So

P(X < 30) = 0.7288

(d) Find the value of x such that P(X > x) = 0.05

So

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

0.05 = e^{-0.0435x}

\ln{e^{-0.0435x}} = \ln{0.05}

-0.0435x = \ln{0.05}

x = -\frac{\ln{0.05}}{0.0435}

x = 68.87

5 0
3 years ago
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