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3 years ago

Estimate 1 7/8 - 1 1/2

Mathematics

1 answer:

OLEGan [10]3 years ago

3 0

Answer:

1⅞ - 1½ ≈ ½

Step-by-step explanation:

Estimation is mental arithmetic. Practice until you can do this in your head.

Here's one way to estimate with mxed numbers.

Step 1. Separate into whole number and fractional parts

1⅞ - 1½ = 1 + ⅞ - 1 - ½

Step 2. Solve the whole number parts

1 - 1 = 0

Step 3. Solve the fraction parts by estimating.

We will round to the nearest ½, so we should round up a number greater than ¾.

(a) ⅞ is greater than ⁶/₈ (¾), so we round it up to ⁸/₈:

⅞ - ½ ≈ ⁸/₈ - ½

(b) Convert each fraction to a common denominator.

We can write ⁸/₈ as ²/₂

⁸/₈ - ½ = ²/₂ - ½

(c) Solve the fraction part

²/₂ - ½ = ½

Step 4. Combine the whole and fraction parts

0 + ½ = ½

So, 1⅞ - 1½ ≈ ½

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Question 2-4

Gnoma [55]

Step-by-step explanation:

-x/3 < -4

-x < -12

x > 12

5x +11 ≤ -4

5x ≤ -4-11

5x ≤ -15

x ≤ -3

therefore, the solution is

x ≤ -3 or x > 12

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3 years ago

Suppose an investment of $2000 doubles in value every 8 years. How much is the investment worth after 24 years? After 32 years?

guapka [62]

24 years = $16,000
33 years = $32,000

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3 years ago

Which of the following approaches can be used to construct a parallelogram? Select one: a. Measure equal segments on parallel li

Fittoniya [83]

Please remember that a parallelogram is 4 parallel lines and the opposite sides are congruent. With that information you should be able to pick the correct answer.

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3 years ago

Explain how you could mentally find 8 x 45 by using the distributive property

dangina [55]

First, break up 45 into 40 and 5 then multipy 8(40) which equals 320 and multipy 8(5) which equals 40. Thrn add 320+40 to get the product 360.
Hope this helped

8x45
8(40)+8(5)
320+40
360

6 0

4 years ago

A company uses three different assembly lines- A1, A2, and A3- to manufacture a particular component. Of thosemanufactured by li

hammer [34]

Answer:

The probability that a randomly selected component needs rework when it came from line A₁ is 0.3623.

Step-by-step explanation:

The three different assembly lines are: A₁, A₂ and A₃.

Denote <em>R</em> as the event that a component needs rework.

It is given that:

$P (R|A_{1})=0.05\\P (R|A_{2})=0.08\\P (R|A_{3})=0.10\\P (A_{1})=0.50\\P (A_{2})=0.30\\P (A_{3})=0.20$

Compute the probability that a randomly selected component needs rework as follows:

$P(R)=P(R|A_{1})P(A_{1})+P(R|A_{2})P(A_{2})+P(R|A_{3})P(A_{3})\\=(0.05\times0.50)+(0.08\times0.30)+(0.10\times0.20)\\=0.069$

Compute the probability that a randomly selected component needs rework when it came from line A₁ as follows:

$P (A_{1}|R)=\frac{P(R|A_{1})P(A_{1})}{P(R)}=\frac{0.05\times0.50}{0.069} =0.3623$

Thus, the probability that a randomly selected component needs rework when it came from line A₁ is 0.3623.

6 0

3 years ago