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kow [346]
3 years ago
14

Calculate (Round two decimal places for final answer): 650 mL = _______________ oz.

Mathematics
2 answers:
Serggg [28]3 years ago
6 0

Answer:

6.5

Step-by-step explanation:

Tcecarenko [31]3 years ago
4 0

For this case we must perform a conversion of milliliters to ounces.

By definition we have to:

1 US liquid ounce equals 29.5735 milliliters. Then, making a rule of three:

1 oz. ---------> 29.5735 milliliters

x -------------------> 650 milliliters

Where "x" represents the number of oz equivalent to 650 mL.

x = \frac {650 * 1} {29.5735}\\x = 21.9791

Thus, 650 mL equals 21.9791 oz.

Answer:

650 mL equals 21.9791 oz.

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8. A cent is 0.01 of a dollar
Oksanka [162]

Answer:

Yes

Step-by-step explanation:

A dollar = 100 cents

1 cent : 100 cents

= 1/100

= 0.01

8 0
2 years ago
Simplify the expression 12−17/−5−(−2).
OleMash [197]

12-17/-5-(-2) so divison first,it will become 12+3.4-(-2) =two minuses will become a "+"

So,12+3.4+2=17,4

ANSWER: 17.4

4 0
2 years ago
A class is made up of 10 boys and 4 girls. Half of the girls wear glasses. A student is selected at random from the class. What
Andreas93 [3]

Hi!

Add the student together to get the number of students

10+4 = 14

Divide the girls by half

4/2 = 2

2/14 is the answer for this question.

<em>Hope this helps! Have an amazing day <3</em>

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4 0
3 years ago
What is 16x9+67x9 can someone help me?
vazorg [7]

Answer: 747


Step-by-step explanation:

16x9+67x9

First multiply 16 and 9 which is equal to 144

144+67x9

now multiply 67 and 9 which is equal to 603

144+603

Now add 144 and 603 which is equal to 747

So the answer to 16x9+67x9 is 747




5 0
3 years ago
Read 2 more answers
The time intervals between successive barges passing a certain point on a busy waterway have an exponential distribution with me
lisov135 [29]

Answer:

a) <u>0.4647</u>

b) <u>24.6 secs</u>

Step-by-step explanation:

Let T be interval between two successive barges

t(t) = λe^λt where t > 0

The mean of the exponential

E(T) = 1/λ

E(T) = 8

1/λ = 8

λ = 1/8

∴ t(t) = 1/8×e^-t/8   [ t > 0]

Now the probability we need

p[T<5] = ₀∫⁵ t(t) dt

=₀∫⁵ 1/8×e^-t/8 dt

= 1/8 ₀∫⁵ e^-t/8 dt

= 1/8 [ (e^-t/8) / -1/8 ]₀⁵

= - [ e^-t/8]₀⁵

= - [ e^-5/8 - 1 ]

= 1 - e^-5/8 = <u>0.4647</u>

Therefore the probability that the time interval between two successive barges is less than 5 minutes is <u>0.4647</u>

<u></u>

b)

Now we find t such that;

p[T>t] = 0.95

so

t_∫¹⁰ t(x) dx = 0.95

t_∫¹⁰ 1/8×e^-x/8 = 0.95

1/8 t_∫¹⁰ e^-x/8 dx = 0.95

1/8 [( e^-x/8 ) / - 1/8 ]¹⁰_t  = 0.95

- [ e^-x/8]¹⁰_t = 0.96

- [ 0 - e^-t/8 ] = 0.95

e^-t/8 = 0.95

take log of both sides

log (e^-t/8) = log (0.95)

-t/8 = In(0.95)

-t/8 = -0.0513

t = 8 × 0.0513

t = 0.4104 (min)

so we convert to seconds

t = 0.4104 × 60

t = <u>24.6 secs</u>

Therefore the time interval t such that we can be 95% sure that the time interval between two successive barges will be greater than t is <u>24.6 secs</u>

6 0
3 years ago
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