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Fudgin [204]
2 years ago
15

Is fast please help me is a exam

Mathematics
1 answer:
qaws [65]2 years ago
5 0

Answer:

100 minutes

Step-by-step explanation:

1 hour = 60 minutes.

(2 / 3) * (1 hour) =

40 minutes

So the movie was 1 hour 40 minutes

REMEMBER:

1hr = 60 mins

60 mins + 40 mins = 100 mins

Hence, the answer is 100 mins

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(0.3d-1.5)/3 1/3=0.84:7/15
Anika [276]

Answer:

  d = 25

Step-by-step explanation:

Perhaps you want to solve for d:

  (0.3d -1.5)/(3 1/3) = 0.84/(7/15)

Change the division by a fraction to multiplication by its inverse.

  (0.3d -1.5)·(3/10) = (0.84)·(15/7)

  0.09d -0.45 = 1.8 . . . . . . . . . . . . simplify

  d - 5 = 20 . . . . . . . . . . . . . . . . . . divide by 0.09

 d = 25 . . . . . . . . . . . . . . . . . . . . . add 5

_____

Grouping symbols help show what is being divided by what. Especially where mixed numbers are involved, parentheses help show that the number 3 1/3 is not supposed to be interpreted as the product of 3 and 1/3.

4 0
3 years ago
What happens to the image of a figure that is rotated 90° counterclockwise?
Paul [167]
I think the answer is A
6 0
3 years ago
X² – 8x+9<br> Name the polynomial based on its degree and number of terms.
Vaselesa [24]
Quadratic Trinomial
2nd degree and three terms
5 0
3 years ago
The normal distribution An automobile battery manufacturer offers a 31/54 warranty on its batteries. The first number in the war
seraphim [82]

Answer:

1) if the manufacturer's assumptions are correct, it would reed to replace 0.62% of its batteries free.

2) a standard deviation of 6.0843 results in a 1.07% replacement rate

3) using the revised standard deviation for battery life, 91.9% of the manufacturer's batteries don't get free replacement but qualifies for the prorated credit

Step-by-step explanation:

based on the given data;

x will represent the random variable such that the lifetime of its auto batteries, is normally distributed with a mean of 45 months and a standard deviation of 5.6 months

so

x → N( U = 45, ∝ = 5.6)

Under the warranty, if a battery fails within 31 months of purchase, the manufacturer replaces the battery at no charges to the consumer.

if the battery fails after 31 months but within 54 months, the manufacturer provides a prostrated credit towards the purchase of anew battery

1) If the manufacturer's assumptions are correct,

p(x < 3) = p( [x-u / ∝ ] < [ 31-45 / 5.6] )

= p( z < -2.5 )

using the standard normal table,

value of z = 0.0062 ≈ 0.62%

so if the manufacturer's assumptions are correct, it would reed to replace 0.62% of its batteries free.

2)

The company finds that it is replacing 1.07% of its batteries free of charge. It suspects that its assumption standard deviation of the life of its batteries is incorrect, so a standard deviation of ? results in a 1.07%

so lets say;

p ( x < 31 ) = ( 1.07%) = 0.0107

p ( [x-u / ∝ ] < [ 31-45 / ∝] ) = 0.0107

now from the standard table

-2.301 is 1.07%

so

( 31 - 45 / ∝ ) = -2.301

-14 / ∝ = -2.301

∝ = -14 / - 2.301

∝ = 6.0843

therefore a standard deviation of 6.0843 results in a 1.07% replacement rate

3)

Using the revised standard deviation for battery life, what percentage of the manufacturer's batteries don't free replacement but do qualify for the prorated credit?

p( 31 < x < 54 ) = p ( [31 - u / ∝ ] < [ x-u / ∝]  < [ 54 - 45 / ∝] )

= p ( [31 - 45 / 6.0843 ] < [ x-u / ∝]  < [ 54 - 45 / 6.0843] )

= p ( -2.301 < z < 1.4792 )

= p(Z < 1.5) - p(Z < -2.3)

= 0.9393 - 0.0108

= 0.919 ≈ 91.9%

therefore using the revised standard deviation for battery life, 91.9% of the manufacturer's batteries don't get free replacement but qualifies for the prorated credit

8 0
3 years ago
See attachment below. Please answer. Thank you.
KonstantinChe [14]

Hi

The probability of landing a green side up is 2 / 6 .

Hope this helps

3 0
3 years ago
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