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3 years ago

What is the answer when u covert the fraction 9/20 to decimal

Mathematics

1 answer:

ozzi3 years ago

7 0

$\dfrac{9}{20}=\dfrac{45}{100}=0.45$

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Basile [38]

Answer: y = 3

Step-by-step explanation:

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2 years ago

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I dont understand what the question is asking me

kipiarov [429]

Answer:

$\huge\boxed{\sf$

Step-by-step explanation:

∠E = 1/2 (mCD - mAB)

<u>Where mCD = 110° and mAB = 30°</u>

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2 years ago

Travis takes out a loan from his bank for $3,500. At the end of the year, Travis will need to pay the loan back to his bank alon

Reil [10]

Answer:

$3,787

Step-by-step explanation:

frist find the amount of intrest

3,500 X 0.082 X 1 = 287

next add the oragial amount to the amount of intrest

3,500 + 287 = 3,787

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2 years ago

A delivery truck made 8 stops in one neighborhood. By stop 5, there were 11 packages delivered. This graph shows this informatio

IceJOKER [234]

2 delivery stops

Find the point that hits 5 on the "packaging delivery".

Then look down to find the 'x'

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3 years ago

Resistors are labeled 100 Ω. In fact, the actual resistances are uniformly distributed on the interval (95, 103). Find the mean

Zinaida [17]

Answer:

$E[R]$ = 99 Ω

$\sigma_R$ = 2.3094 Ω

P(98<R<102) = 0.5696

Step-by-step explanation:

The mean resistance is the average of edge values of interval.

Hence,

The mean resistance, $E[R] = \frac{a+b}{2} = \frac{95+103}{2} = \frac{198}{2}$ = 99 Ω

To find the standard deviation of resistance, we need to find variance first.

$V(R) = \frac{(b-a)^2}{12} =\frac{(103-95)^2}{12} = 5.333$

Hence,

The standard deviation of resistance, $\sigma_R = \sqrt{V(R)} = \sqrt5.333$ = 2.3094 Ω

To calculate the probability that resistance is between 98 Ω and 102 Ω, we need to find Normal Distributions.

$z_1 = \frac{102-99}{2.3094} = 1.299$

$z_2 = \frac{98-99}{2.3094} = -0.433$

From the Z-table, P(98<R<102) = 0.9032 - 0.3336 = 0.5696

5 0

3 years ago